Galvanic (voltaic) cells

WEBVTT
Kind: captions
Language: en

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- [Instructor] Galvanic cells,

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which are also called voltaic cells,

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use a thermodynamically favorable reaction

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to generate an electric current.

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Before we look at a diagram

00:00:09.440 --> 00:00:11.570
of a galvanic or voltaic cell,

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let's first look at the half reactions

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that are going to be used in the cell.

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In the first half reaction,

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zinc metal loses two electrons

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to turn into the zinc 2+ cation.

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Loss of electrons is oxidation.

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Therefore, this is the
oxidation half reaction.

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As a quick review, the other way to tell

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that this is the oxidation half reaction

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is we go from an oxidation number

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for solid zinc of zero

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to an oxidation number
for zinc 2+ of plus two

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and increase in the oxidation
number is oxidation.

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Therefore, zinc metal is oxidized.

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And here is the other half reaction

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that we're going to see in
our galvanic cell diagram.

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In this half reaction,

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the copper 2+ cation gains two electrons

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to turn into solid copper.

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Gain of electrons is reduction.

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Therefore, this is the
reduction half reaction.

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The other way to tell
that this is the reduction

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half reaction is to look
at the oxidation numbers.

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So the copper 2+ cation has an oxidation

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number of plus two and solid copper

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has an oxidation number of zero.

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Since there's a decrease or a reduction

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in the oxidation number
from plus two to zero,

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this is the reduction half reaction.

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A good way to remember
which half reaction is which

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is to think about LEO the lion goes GER.

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So loss of electrons is oxidation

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and gain of electrons is reduction.

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When we add the two
half reactions together,

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we have two electrons on the reactant side

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and two electrons on the product side.

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So those cancel out and
that gives us a solid zinc

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plus copper 2+ ions
goes to zinc 2+ cations

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and solid copper.

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In this redox reaction,

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zinc metal is oxidized to zinc 2+ cations

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and copper 2+ cations are
reduced to solid copper.

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Delta G naught for this reaction

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at 25 degrees Celsius or room temperature

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is less than zero,

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which means this reaction

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is thermodynamically favorable.

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Therefore, if we were to
put a piece of solid zinc

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in an aqueous solution of copper 2+ ions,

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we would see copper metal

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form on the zinc metal and zinc 2+ ions

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would form in solution.

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So that's what would happen

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if we did this reaction

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in only a single compartment.

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However, in a galvanic or voltaic cell,

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each half reaction gets
its own compartment.

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And the two compartments
are connected with a wire.

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Therefore, a thermodynamically favorable

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redox reaction is used to generate

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an electric current in the wire.

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And by electric current,

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we're talking the flow of electrons.

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Now that we've gone
over the half reactions

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in detail for this zinc-copper
galvanic or voltaic cell,

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let's look at a diagram of this cell

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and see how things actually work.

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Let's start by looking at
the compartment on the left,

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which is a beaker that contains

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a one molar aqueous
solution of zinc sulfate.

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Therefore, there are zinc 2+ ions

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and sulfate anions in aqueous
solution in the beaker.

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We saw from the oxidation half reaction

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that solid zinc is oxidized

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and turns into zinc 2+ cations.

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When zinc turns into zinc 2+,

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two electrons are lost.

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So those electrons will move

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in this wire that connects
the two compartments.

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So imagine we have movement of electrons

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going from the compartment on the left

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towards the compartment on the right.

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This piece of solid zinc
is called an electrode.

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The electrode at which
oxidation takes place

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is called the anode.

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So let me go ahead and
write anode in here.

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Next, let's look at the
compartment on the right,

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which is a beaker that contains

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a one molar solution of copper sulfate.

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So in this beaker,
there are copper 2+ ions

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and sulfate anions in aqueous solution.

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We know that electrons
are moving in this wire

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that connects the two compartments.

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At the surface of this copper electrode,

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the copper 2+ ions come in contact

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with two electrons and are
reduced to form solid copper.

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The electrode where reduction takes place

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is called the cathode.

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So the copper electrode is
the cathode for this cell.

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I went ahead and rewrote
each half reaction

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to remind us of what's going
on in each compartment.

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So we can think about what's gonna happen

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to our cell over time.

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The anode is where oxidation takes place.

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So solid zinc turns into zinc 2+ cations.

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So as time goes on,
the amount of the solid

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zinc electrode will decrease.

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And the amount of zinc 2+ cations

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in solution will increase.

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And the cathode is where
reduction takes place.

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So as copper 2+ cations gained

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two electrons to turn into solid copper,

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over time, the amount of copper,

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solid copper electrode will increase.

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And the concentration of copper 2+ ions

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in solution will decrease.

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A good way to remember that oxidation

00:05:28.590 --> 00:05:32.070
occurs at the anode and
reduction occurs at the cathode

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is to think about two
animals, an ox and a cat.

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So thinking about an ox reminds us

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that oxidation occurs at the anode

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and thinking about a red cat

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reminds us that reduction
occurs at the cathode.

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The next part of the cell

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that we have to think about

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is the salt bridge that connects

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these two compartments.

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Inside the salt bridge,

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there's an electrolyte solution
such as sodium nitrate.

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The electrolyte is often in a gel

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or paste form to prevent early mixing

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with the other compartments.

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These solutions in each compartment

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also called a half cell must
remain electrically neutral.

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And the purpose of the salt bridge

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is to balance the charges.

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Let's start by thinking
about the compartment

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or the half cell on the left.

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When we started, we had
equal concentrations

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of zinc 2+ and sulfate anion,

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so the charges were balanced.

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However, over time, the concentration

00:06:31.910 --> 00:06:35.250
of zinc 2+ ions increases.

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To balance out the
increased positive charge,

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the negatively charged anion,

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and in this case, the nitrate anion,

00:06:42.360 --> 00:06:43.640
moves from the salt bridge

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into the half cell on the left.

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Next, let's think about the compartment

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or the half cell on the right.

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We started off with equal concentrations

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of copper 2+ cation and sulfate anion.

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However, over time, there's a
decrease in copper 2+ cation.

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Therefore, there's too
much negative charge

00:07:03.810 --> 00:07:05.320
in this half cell.

00:07:05.320 --> 00:07:08.400
To balance out the negatively
charged sulfate anions,

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the positively charged
ion and the salt bridge,

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in this case, the sodium cation,

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moves from the salt bridge

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into the half cell on the right.

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A good way to remember which direction

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the ions go in the salt bridge

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is to think about anions
going toward the anode

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and cations going toward the cathode.

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The salt bridge is necessary

00:07:28.460 --> 00:07:30.320
for the galvanic cell to work.

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If the salt bridge were removed,

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the electrons would
stop moving in the wire,

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And that brings us to the last component

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of our galvanic cell, which
is this volt meter here.

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For this particular galvanic cell

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with one molar concentration
of zinc sulfate,

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one more concentration of copper sulfate

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at 25 degrees Celsius,

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the initial voltage would be 1.10 volts.

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So that's what would show
up on the volt meter.

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The volt meter measures the difference

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in electric potential between two points.

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And as long as there's a difference

00:08:01.810 --> 00:08:03.180
in electric potential,

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electrons will flow in the wire.

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However, this thermodynamically favorable

00:08:07.570 --> 00:08:10.190
redox reaction doesn't last forever.

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Eventually, the reaction
reaches equilibrium.

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And at that point, the
voltage is equal to zero.

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And when the voltage goes to zero,

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the current also goes to zero,

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which means the electrons

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are no longer moving in the wire.

00:08:25.210 --> 00:08:29.220
You could also call a galvanic
or voltaic cell a battery.

00:08:29.220 --> 00:08:31.930
So at that point, the battery is dead.

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Let's do a quick summary of
galvanic or voltaic cells.

00:08:35.470 --> 00:08:37.650
These cells use a thermodynamically

00:08:37.650 --> 00:08:40.210
favored redox reaction.

00:08:40.210 --> 00:08:42.040
This redox reaction generates

00:08:42.040 --> 00:08:44.180
an electric current that flows

00:08:44.180 --> 00:08:47.260
in the wire between the two electrodes.

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The electrode where oxidation takes place

00:08:49.720 --> 00:08:50.990
is called the anode,

00:08:50.990 --> 00:08:53.340
and the electrode where
reduction takes place

00:08:53.340 --> 00:08:55.000
is called the cathode.

00:08:55.000 --> 00:08:57.750
A salt bridge is used
to balance the charges

00:08:57.750 --> 00:08:59.550
in the two compartments.

00:08:59.550 --> 00:09:02.120
Eventually, the reaction
reaches equilibrium

00:09:02.120 --> 00:09:05.310
and at equilibrium, the
voltage is equal to zero,

00:09:05.310 --> 00:09:07.430
which means there's no more current.

00:09:07.430 --> 00:09:10.853
And the galvanic or
voltaic cell stops working.


Coupled reactions

WEBVTT
Kind: captions
Language: en

00:00:00.420 --> 00:00:01.470
- [Instructor] Coupled reactions

00:00:01.470 --> 00:00:03.880
use a thermodynamically favorable reaction

00:00:03.880 --> 00:00:07.910
to drive a thermodynamically
unfavorable reaction.

00:00:07.910 --> 00:00:10.960
For example, let's look
at a hypothetical reaction

00:00:10.960 --> 00:00:15.960
where reactants A and B combine
to form products C and i.

00:00:16.550 --> 00:00:20.370
The standard change in free
energy for this reaction,

00:00:20.370 --> 00:00:23.970
delta G naught, let's
say, is greater than zero.

00:00:23.970 --> 00:00:26.360
And when delta G naught
is greater than zero,

00:00:26.360 --> 00:00:30.050
that's a thermodynamically
unfavorable reaction.

00:00:30.050 --> 00:00:31.770
And at equilibrium,

00:00:31.770 --> 00:00:35.150
the equilibrium constant,
K, is less than one.

00:00:35.150 --> 00:00:36.710
When K is less than one,

00:00:36.710 --> 00:00:39.140
that means there are more
reactants than products

00:00:39.140 --> 00:00:40.420
at equilibrium.

00:00:40.420 --> 00:00:44.560
And if our goal is to make
a lot of the product C,

00:00:44.560 --> 00:00:46.820
that won't happen as long as this reaction

00:00:46.820 --> 00:00:49.370
is thermodynamically unfavorable.

00:00:49.370 --> 00:00:52.120
Next, let's look at another
hypothetical reaction

00:00:52.120 --> 00:00:56.990
where i reacts with D to
form E and F as the products.

00:00:56.990 --> 00:00:58.590
For this hypothetical reaction,

00:00:58.590 --> 00:01:01.740
the standard change in free
energy, delta G naught,

00:01:01.740 --> 00:01:04.820
is much less than zero.

00:01:04.820 --> 00:01:07.850
Since delta G naught for the
second hypothetical reaction

00:01:07.850 --> 00:01:09.350
is less than zero,

00:01:09.350 --> 00:01:12.800
reaction two is
thermodynamically favorable.

00:01:12.800 --> 00:01:14.280
And at equilibrium,

00:01:14.280 --> 00:01:18.220
the equilibrium constant,
K, is much greater than one.

00:01:18.220 --> 00:01:20.200
If K is much greater than one,

00:01:20.200 --> 00:01:22.960
then we can assume that the
second reaction essentially goes

00:01:22.960 --> 00:01:23.890
to completion.

00:01:23.890 --> 00:01:27.460
So all the i and the D will
react together to form E and F.

00:01:27.460 --> 00:01:28.700
And at equilibrium,

00:01:28.700 --> 00:01:32.210
there are way more products
than there are reactants.

00:01:32.210 --> 00:01:36.640
Remember that our goal was to
make a lot of the product C,

00:01:36.640 --> 00:01:39.460
but we can't get that by just
running the first reaction

00:01:39.460 --> 00:01:40.720
by itself.

00:01:40.720 --> 00:01:43.410
However, by coupling the first reaction,

00:01:43.410 --> 00:01:45.730
which is thermodynamically unfavorable,

00:01:45.730 --> 00:01:47.110
to the second reaction,

00:01:47.110 --> 00:01:49.300
which is thermodynamically favorable,

00:01:49.300 --> 00:01:52.760
we can make a significant
amount of product C.

00:01:52.760 --> 00:01:55.950
These two hypothetical reactions
can be coupled together

00:01:55.950 --> 00:01:59.320
because they share a
common intermediate, i.

00:01:59.320 --> 00:02:02.910
Even though reaction one is
thermodynamically unfavorable,

00:02:02.910 --> 00:02:05.740
even if it forms only a small amount of i,

00:02:05.740 --> 00:02:08.620
since reaction two is
thermodynamically favorable,

00:02:08.620 --> 00:02:12.350
the i from reaction one will
be used up in reaction two

00:02:12.350 --> 00:02:15.030
to form the products for reaction two.

00:02:15.030 --> 00:02:18.860
Removing the product i
causes the equilibrium

00:02:18.860 --> 00:02:21.390
to shift to the right
in the first reaction,

00:02:21.390 --> 00:02:25.390
therefore producing more
of our desired product C.

00:02:25.390 --> 00:02:27.930
So the thermodynamically
favorable reaction

00:02:27.930 --> 00:02:31.260
is driving the thermodynamically
unfavorable reaction

00:02:31.260 --> 00:02:32.580
to the right.

00:02:32.580 --> 00:02:35.540
Since we are coupling these
two reactions together,

00:02:35.540 --> 00:02:38.340
if we add reaction one to reaction two,

00:02:38.340 --> 00:02:40.510
the intermediate would cancel out.

00:02:40.510 --> 00:02:41.910
And for the reactants,

00:02:41.910 --> 00:02:44.350
we would get A plus B plus D;

00:02:44.350 --> 00:02:48.910
and for the products, we
would get C plus E plus F.

00:02:48.910 --> 00:02:51.850
To find delta G naught
for this overall equation,

00:02:51.850 --> 00:02:54.200
we would need to add
together delta G naught

00:02:54.200 --> 00:02:58.460
for reaction one and delta
G naught for reaction two.

00:02:58.460 --> 00:03:01.080
So delta G naught for
this overall equation

00:03:01.080 --> 00:03:02.850
would be less than zero,

00:03:02.850 --> 00:03:05.310
which tells us this overall equation

00:03:05.310 --> 00:03:07.740
is thermodynamically favorable.

00:03:07.740 --> 00:03:11.230
And the equilibrium constant
is greater than one,

00:03:11.230 --> 00:03:14.380
which means at equilibrium,
we'll have a decent amount

00:03:14.380 --> 00:03:16.940
of our desired product C.

00:03:16.940 --> 00:03:19.110
Next, let's look at a practical example

00:03:19.110 --> 00:03:20.920
of coupled reactions.

00:03:20.920 --> 00:03:24.150
Let's say that our goal
is to extract solid copper

00:03:24.150 --> 00:03:26.370
from copper(I) sulfide.

00:03:26.370 --> 00:03:29.390
The other product of
this reaction is sulfur.

00:03:29.390 --> 00:03:31.820
However, delta G naught for this reaction

00:03:31.820 --> 00:03:36.390
is equal to positive 86.1
kilojoules per mole of reaction.

00:03:36.390 --> 00:03:38.300
Since delta G naught is positive,

00:03:38.300 --> 00:03:41.200
this reaction is
thermodynamically unfavorable.

00:03:41.200 --> 00:03:44.900
And at equilibrium, there'd
be very little of our product,

00:03:44.900 --> 00:03:47.380
so we wouldn't get very much copper.

00:03:47.380 --> 00:03:49.990
The solution is to couple
this first reaction

00:03:49.990 --> 00:03:52.950
to a thermodynamically favorable reaction.

00:03:52.950 --> 00:03:57.150
And that's the conversion of
sulfur into sulfur dioxide.

00:03:57.150 --> 00:04:00.830
So solid sulfur plus oxygen
gas give sulfur dioxide,

00:04:00.830 --> 00:04:02.760
and delta G naught for this reaction

00:04:02.760 --> 00:04:07.760
is equal to negative 300.4
kilojoules per mole of reaction.

00:04:07.780 --> 00:04:09.880
So adding our two reactions together,

00:04:09.880 --> 00:04:12.820
the common intermediate,
sulfur, cancels out,

00:04:12.820 --> 00:04:16.340
and that gives us copper(I)
sulfide plus oxygen

00:04:16.340 --> 00:04:21.340
for the reactants goes to
solid copper and sulfur dioxide

00:04:22.150 --> 00:04:23.550
for the products.

00:04:23.550 --> 00:04:26.810
And if we add the two delta
G naught values together,

00:04:26.810 --> 00:04:30.990
positive 86.1 plus negative 300.4

00:04:30.990 --> 00:04:34.290
gives a delta G naught
for this overall equation

00:04:34.290 --> 00:04:39.290
equal to negative 214.3
kilojoules per mole of reaction.

00:04:39.570 --> 00:04:42.280
Since delta G naught is negative,

00:04:42.280 --> 00:04:45.910
now, this overall reaction is
thermodynamically favorable

00:04:45.910 --> 00:04:49.130
with an equilibrium
constant greater than one.

00:04:49.130 --> 00:04:52.500
Therefore, we will now
produce a lot of copper.

00:04:52.500 --> 00:04:56.210
So burning copper(I) sulfide
in air gives us a feasible way

00:04:56.210 --> 00:05:00.100
of extracting copper
from copper(I) sulfide.

00:05:00.100 --> 00:05:04.010
Coupled reactions are also
very important in biochemistry.

00:05:04.010 --> 00:05:08.250
For example, two amino acids
are alanine and glycine.

00:05:08.250 --> 00:05:10.970
And if these two amino
acids come together,

00:05:10.970 --> 00:05:13.670
they form what's called a dipeptide.

00:05:13.670 --> 00:05:15.010
It's very important for the body

00:05:15.010 --> 00:05:16.840
to be able to make dipeptides

00:05:16.840 --> 00:05:20.010
because if we were to add
on some more amino acids,

00:05:20.010 --> 00:05:22.550
we would form what's called a polypeptide,

00:05:22.550 --> 00:05:25.750
which eventually leads to
the formation of a protein.

00:05:25.750 --> 00:05:28.030
However, the formation of dipeptides

00:05:28.030 --> 00:05:30.670
is thermodynamically unfavorable.

00:05:30.670 --> 00:05:32.250
Therefore, the body needs a way

00:05:32.250 --> 00:05:35.350
of making this reaction
thermodynamically favorable

00:05:35.350 --> 00:05:37.400
so the body can make proteins.

00:05:37.400 --> 00:05:39.940
To do this, the body uses ATP,

00:05:39.940 --> 00:05:42.343
which stands for adenosine triphosphate.

00:05:43.720 --> 00:05:47.900
And here's the dot structure
for ATP in the ionized form.

00:05:47.900 --> 00:05:51.100
ATP is often called a
high energy compound,

00:05:51.100 --> 00:05:53.370
and the hydrolysis of ATP

00:05:53.370 --> 00:05:56.890
is used to drive protein synthesis.

00:05:56.890 --> 00:05:59.530
Next, let's look at the
equation showing the hydrolysis

00:05:59.530 --> 00:06:02.040
of ATP to ADP,

00:06:02.040 --> 00:06:06.410
which is adenosine diphosphate
and inorganic phosphate.

00:06:06.410 --> 00:06:10.450
The standard change in free
energy for the hydrolysis of ATP

00:06:10.450 --> 00:06:14.340
is equal to negative 31
kilojoules per mole of reaction.

00:06:14.340 --> 00:06:16.420
Since delta G naught is negative,

00:06:16.420 --> 00:06:20.580
the hydrolysis of ATP is
thermodynamically favorable.

00:06:20.580 --> 00:06:22.550
Alanine and glycine combine

00:06:22.550 --> 00:06:26.150
to form the dipeptide
alanylglycine and water.

00:06:26.150 --> 00:06:28.650
Delta G naught for the
formation of this dipeptide

00:06:28.650 --> 00:06:32.450
is equal to positive 29
kilojoules per mole of reaction.

00:06:32.450 --> 00:06:34.650
Since delta G naught is positive,

00:06:34.650 --> 00:06:38.290
this reaction is
thermodynamically unfavorable.

00:06:38.290 --> 00:06:39.700
With the help of an enzyme,

00:06:39.700 --> 00:06:42.330
the unfavorable reaction is coupled

00:06:42.330 --> 00:06:44.500
to the favorable reaction.

00:06:44.500 --> 00:06:46.460
And if we add the two reactions together,

00:06:46.460 --> 00:06:50.140
water would cancel out and
give us alanine plus glycine

00:06:50.140 --> 00:06:52.900
plus ATP for the reactants,

00:06:52.900 --> 00:06:57.190
and alanylglycine and ADP
and inorganic phosphate

00:06:57.190 --> 00:06:58.690
for the products.

00:06:58.690 --> 00:07:01.680
Adding the two values of
delta G naught together,

00:07:01.680 --> 00:07:02.740
we get delta G naught

00:07:02.740 --> 00:07:06.750
is equal to negative two
kilojoules per mole of reaction.

00:07:06.750 --> 00:07:08.800
And since delta G naught is negative,

00:07:08.800 --> 00:07:10.490
it's now favorable for the body

00:07:10.490 --> 00:07:12.853
to synthesize this dipeptide.


Free energy and equilibrium

WEBVTT
Kind: captions
Language: en

00:00:00.360 --> 00:00:02.330
- [Instructor] Let's say
we have a generic reaction

00:00:02.330 --> 00:00:04.470
where reactants turn into products.

00:00:04.470 --> 00:00:07.380
And our goal is to think
about the relationship

00:00:07.380 --> 00:00:10.240
between free energy and this reaction

00:00:10.240 --> 00:00:12.650
when it comes to equilibrium.

00:00:12.650 --> 00:00:15.100
First, we need to consider the equation

00:00:15.100 --> 00:00:17.780
that allows us to calculate
non-standard changes

00:00:17.780 --> 00:00:19.720
in free energy delta G.

00:00:19.720 --> 00:00:21.180
We can think about delta G

00:00:21.180 --> 00:00:24.010
as the instantaneous
difference in free energy

00:00:24.010 --> 00:00:26.770
between reactants and products.

00:00:26.770 --> 00:00:29.130
And from the equation we see that delta G

00:00:29.130 --> 00:00:34.090
is equal to delta naught plus RT ln of Q.

00:00:34.090 --> 00:00:37.540
Delta G naught is the
standard change in free energy

00:00:37.540 --> 00:00:40.030
between reactants and products.

00:00:40.030 --> 00:00:42.930
This value refers to the
difference in free energy

00:00:42.930 --> 00:00:46.020
between reactants and products
in their standard states

00:00:46.020 --> 00:00:48.150
at a specified temperature.

00:00:48.150 --> 00:00:50.510
R is the ideal gas constant,

00:00:50.510 --> 00:00:52.870
T is the absolute temperature in Kelvin,

00:00:52.870 --> 00:00:55.920
and Q is the reaction quotient.

00:00:55.920 --> 00:00:59.280
At equilibrium, the instantaneous
difference in free energy

00:00:59.280 --> 00:01:02.000
between reactants and products is zero,

00:01:02.000 --> 00:01:05.780
which means there's no more
driving force for the reaction.

00:01:05.780 --> 00:01:09.430
And at equilibrium,
the reaction quotient Q

00:01:09.430 --> 00:01:13.420
is equal to the equilibrium constant K.

00:01:13.420 --> 00:01:17.140
So we can plug in zero for
delta G in our equation.

00:01:17.140 --> 00:01:19.890
And we can plug in K for Q,

00:01:19.890 --> 00:01:22.530
that gives us zero is
equal to delta G naught

00:01:22.530 --> 00:01:25.590
plus RT ln of K.

00:01:25.590 --> 00:01:29.000
If we subtract RT ln of K from both sides,

00:01:29.000 --> 00:01:30.250
we get delta G naught

00:01:30.250 --> 00:01:34.160
is equal to negative RT ln of K.

00:01:34.160 --> 00:01:36.590
This equation is extremely useful

00:01:36.590 --> 00:01:38.770
because now we have a relationship

00:01:38.770 --> 00:01:41.840
between free energy and equilibrium.

00:01:41.840 --> 00:01:44.250
If we know the standard
change in free energy

00:01:44.250 --> 00:01:47.640
for a particular reaction
at a particular temperature,

00:01:47.640 --> 00:01:52.640
we can calculate the equilibrium
constant for that reaction.

00:01:52.640 --> 00:01:54.480
Next, let's go ahead
and go through the math

00:01:54.480 --> 00:01:57.360
to solve for the equilibrium constant K.

00:01:57.360 --> 00:01:58.530
To solve for K,

00:01:58.530 --> 00:02:02.763
we need to first divide
both sides by negative RT.

00:02:04.050 --> 00:02:05.300
So on the right side,

00:02:05.300 --> 00:02:08.060
negative RT would cancel out.

00:02:08.060 --> 00:02:10.420
We could then rewrite it as ln of K

00:02:10.420 --> 00:02:15.070
is equal to negative delta
G naught divided by RT.

00:02:15.070 --> 00:02:16.770
And to get rid of the ln,

00:02:16.770 --> 00:02:21.370
we need to take e to both sides,

00:02:21.370 --> 00:02:23.140
which gives us the following equation.

00:02:23.140 --> 00:02:25.160
The equilibrium constant K

00:02:25.160 --> 00:02:30.160
is equal to e to the negative
delta G naught divided by RT.

00:02:30.430 --> 00:02:32.520
So if we know the value for delta G naught

00:02:32.520 --> 00:02:34.290
for a particular reaction,

00:02:34.290 --> 00:02:35.550
and we know that temperature,

00:02:35.550 --> 00:02:39.010
we can calculate the equilibrium
constant for that reaction

00:02:39.010 --> 00:02:41.370
at that particular temperature.

00:02:41.370 --> 00:02:43.850
Let's calculate the equilibrium constant K

00:02:43.850 --> 00:02:47.820
for the synthesis of ammonia
from nitrogen and hydrogen.

00:02:47.820 --> 00:02:51.290
And let's do this for three
different temperatures.

00:02:51.290 --> 00:02:53.540
Well, let's start with
a temperature of 298K.

00:02:54.830 --> 00:02:56.203
delta G naught at this temperature

00:02:56.203 --> 00:03:01.203
is equal to - 33.0 kilojoules
per mole of reaction.

00:03:01.570 --> 00:03:04.500
So to solve for the
equilibrium constant K,

00:03:04.500 --> 00:03:08.430
we need to plug in the
value for delta G naught.

00:03:08.430 --> 00:03:11.040
We also need to plug in
the temperature in Kelvin

00:03:11.040 --> 00:03:13.760
and the ideal gas constant R.

00:03:13.760 --> 00:03:15.160
The ideal gas constant

00:03:15.160 --> 00:03:19.720
is equal to 8.314 joules
per mole of reaction Kelvin.

00:03:19.720 --> 00:03:21.790
And since we're using joules in the units,

00:03:21.790 --> 00:03:25.470
we need to make sure to convert
the units for delta G naught

00:03:25.470 --> 00:03:27.530
into joules per mole of reaction.

00:03:27.530 --> 00:03:31.590
So - 33.0 kilojoules per mole of reaction

00:03:31.590 --> 00:03:35.610
is equal to - 33,000
joules per mole of reaction

00:03:35.610 --> 00:03:38.130
still with three significant digits.

00:03:38.130 --> 00:03:41.770
And once we plug in our
temperature of 298 Kelvin,

00:03:41.770 --> 00:03:44.140
notice how Kelvin cancels out

00:03:44.140 --> 00:03:47.940
and also joules per mole
of reaction cancels out.

00:03:47.940 --> 00:03:48.850
When don't we do the math,

00:03:48.850 --> 00:03:51.130
we find that the equilibrium constant K

00:03:51.130 --> 00:03:54.380
is equal to e to the 13.3 power,

00:03:54.380 --> 00:03:57.310
which is equal to 6 times 10 to the fifth.

00:03:57.310 --> 00:04:01.370
Notice how equilibrium
constants don't have units.

00:04:01.370 --> 00:04:03.370
When e is raised to a power,

00:04:03.370 --> 00:04:06.670
the result has the same
number of significant figures

00:04:06.670 --> 00:04:09.400
as there are decimal places in the power.

00:04:09.400 --> 00:04:12.300
Since we have only one
decimal place in this power,

00:04:12.300 --> 00:04:16.130
we have one significant
figure for our final answer.

00:04:16.130 --> 00:04:19.990
Since the value for K for
this reaction at 298 Kelvin

00:04:19.990 --> 00:04:21.750
is much greater than one,

00:04:21.750 --> 00:04:25.360
that tells us at equilibrium
there are more products

00:04:25.360 --> 00:04:27.050
than there are reactants.

00:04:27.050 --> 00:04:28.980
So there's a lot more ammonia

00:04:28.980 --> 00:04:31.430
than there is nitrogen or hydrogen,

00:04:31.430 --> 00:04:36.430
when this reaction reaches
equilibrium at 298 Kelvin.

00:04:36.530 --> 00:04:39.340
So whenever delta G naught is negative,

00:04:39.340 --> 00:04:42.170
or you could say delta G
naught is less than zero.

00:04:42.170 --> 00:04:44.770
The equilibrium constant
is greater than one,

00:04:44.770 --> 00:04:49.360
and products are favored over
reactants at equilibrium.

00:04:49.360 --> 00:04:52.220
For the same reaction for
the synthesis of ammonia,

00:04:52.220 --> 00:04:54.540
let's calculate the equilibrium constant K

00:04:54.540 --> 00:04:55.750
at a different temperature,

00:04:55.750 --> 00:04:57.950
this time it's 1000 Kelvin.

00:04:57.950 --> 00:05:01.160
And that 1000 Kelvin delta
G naught for this reaction

00:05:01.160 --> 00:05:06.160
is equal to + 106.5 kilojoules
per mole of reaction.

00:05:06.450 --> 00:05:07.820
Just like the previous example,

00:05:07.820 --> 00:05:09.990
we're gonna plug everything
into our equation

00:05:09.990 --> 00:05:11.570
to calculate K.

00:05:11.570 --> 00:05:13.780
So we need to plug in delta G naught,

00:05:13.780 --> 00:05:16.730
the temperature which
is 1000 Kelvin this time

00:05:16.730 --> 00:05:19.030
and the ideal gas constant.

00:05:19.030 --> 00:05:21.130
So we plug in the ideal gas constant,

00:05:21.130 --> 00:05:23.960
the temperature and delta G naught.

00:05:23.960 --> 00:05:25.490
Once again, we had to convert

00:05:25.490 --> 00:05:27.480
from kilojoules per mole reaction

00:05:27.480 --> 00:05:30.490
into joules per mole of reaction.

00:05:30.490 --> 00:05:32.850
So joules per mole of reaction cancels out

00:05:32.850 --> 00:05:35.390
and Kelvin cancels out.

00:05:35.390 --> 00:05:40.390
That gives us K is equal
to e to the - 12.81 power,

00:05:40.670 --> 00:05:44.670
which is equal to 2.7 times 10 to the - 6.

00:05:44.670 --> 00:05:47.630
Because we have two decimal
places in the power.

00:05:47.630 --> 00:05:51.450
The final answer is to
two significant figures.

00:05:51.450 --> 00:05:54.810
For this example, K is much less than one

00:05:54.810 --> 00:05:57.060
which tells us at equilibrium,

00:05:57.060 --> 00:06:00.540
there are far more reactants
than there are products.

00:06:00.540 --> 00:06:02.350
So at 1000 Kelvin,

00:06:02.350 --> 00:06:04.260
if this reaction at equilibrium

00:06:04.260 --> 00:06:06.140
we'd have a lot more of our reactants,

00:06:06.140 --> 00:06:07.460
nitrogen and hydrogen,

00:06:07.460 --> 00:06:11.210
and only a small amount
of our product ammonia.

00:06:11.210 --> 00:06:14.080
So whenever delta G naught is positive,

00:06:14.080 --> 00:06:17.230
or you could say delta G
naught is greater than zero.

00:06:17.230 --> 00:06:20.600
The equilibrium constant
K is less than one,

00:06:20.600 --> 00:06:22.180
which means that equilibrium,

00:06:22.180 --> 00:06:26.210
there are a lot more reactants
than there are products.

00:06:26.210 --> 00:06:27.330
And for our last example,

00:06:27.330 --> 00:06:29.690
let's calculate the equilibrium constant K

00:06:29.690 --> 00:06:31.340
for the same reaction.

00:06:31.340 --> 00:06:33.970
This time at 464 Kelvin,

00:06:33.970 --> 00:06:36.010
at that temperature delta G naught

00:06:36.010 --> 00:06:38.560
for this reaction is equal to zero.

00:06:38.560 --> 00:06:41.160
Calculating K is a lot
easier for this example,

00:06:41.160 --> 00:06:44.320
because if we plug in
zero for delta G naught,

00:06:44.320 --> 00:06:48.360
K is equal to each of the zero
power which is equal to one.

00:06:48.360 --> 00:06:51.150
And when the equilibrium
constant is equal to one,

00:06:51.150 --> 00:06:53.630
that means that equilibrium
there's significant

00:06:53.630 --> 00:06:56.410
amounts of both reactants and products.

00:06:56.410 --> 00:07:01.360
So if this reaction were to
reach equilibrium at 464 Kelvin,

00:07:01.360 --> 00:07:04.500
there'll be a significant
amount of both our reactants,

00:07:04.500 --> 00:07:06.270
which are nitrogen and hydrogen,

00:07:06.270 --> 00:07:08.600
and our product which is ammonia.

00:07:08.600 --> 00:07:10.850
Let's summarize what we've
learned about the relationship

00:07:10.850 --> 00:07:13.810
between free energy and equilibrium.

00:07:13.810 --> 00:07:16.070
The standard at change in free energy

00:07:16.070 --> 00:07:18.110
for a reaction delta G naught

00:07:18.110 --> 00:07:22.800
is related to the equilibrium
constant K by this equation.

00:07:22.800 --> 00:07:24.560
And if we are using this equation,

00:07:24.560 --> 00:07:27.310
we know the reaction is at equilibrium

00:07:27.310 --> 00:07:29.140
because we found that equation

00:07:29.140 --> 00:07:31.940
by setting the instantaneous
change in free energy

00:07:31.940 --> 00:07:35.690
delta G equal to zero.

00:07:35.690 --> 00:07:38.080
It's very easy to think that delta G

00:07:38.080 --> 00:07:39.830
without the superscript naught,

00:07:39.830 --> 00:07:42.630
and delta G naught are the same thing.

00:07:42.630 --> 00:07:45.760
However, they are very
different quantities.

00:07:45.760 --> 00:07:49.330
For all three of the situations
that we talked about,

00:07:49.330 --> 00:07:53.070
delta G without the
superscript was equal to zero,

00:07:53.070 --> 00:07:56.370
because the reaction was at equilibrium.

00:07:56.370 --> 00:07:57.980
However, because delta G naught

00:07:57.980 --> 00:07:59.470
is the difference in free energy

00:07:59.470 --> 00:08:02.240
between reactants and products
and their standard states,

00:08:02.240 --> 00:08:04.910
delta G naught does not
have to be equal to zero

00:08:04.910 --> 00:08:06.270
at equilibrium.

00:08:06.270 --> 00:08:09.740
delta G naught is a constant
at a particular temperature,

00:08:09.740 --> 00:08:12.340
just like the equilibrium constant K.

00:08:12.340 --> 00:08:15.660
And if delta G naught is less than zero,

00:08:15.660 --> 00:08:17.310
K is greater than one,

00:08:17.310 --> 00:08:19.040
which means at equilibrium,

00:08:19.040 --> 00:08:21.850
there are more products than reactants.

00:08:21.850 --> 00:08:24.060
If delta G naught at a
particular temperature

00:08:24.060 --> 00:08:25.570
is greater than zero.

00:08:25.570 --> 00:08:28.650
That means the equilibrium
constant is less than one,

00:08:28.650 --> 00:08:30.120
which means that equilibrium

00:08:30.120 --> 00:08:32.470
there are more reactants than products.

00:08:32.470 --> 00:08:36.220
And when delta G naught is
equal to zero K is equal to one,

00:08:36.220 --> 00:08:38.970
which means significant
amounts of both reactants

00:08:38.970 --> 00:08:40.753
and products at equilibrium.


Nonstandard free energy changes

WEBVTT
Kind: captions
Language: en

00:00:00.360 --> 00:00:01.690
- [Instructor] Understanding the concept

00:00:01.690 --> 00:00:04.250
of nonstandard free energy changes

00:00:04.250 --> 00:00:07.380
is really important when it
comes to a chemical reaction.

00:00:07.380 --> 00:00:09.140
For this generic chemical reaction,

00:00:09.140 --> 00:00:11.760
the reactants turn into the products.

00:00:11.760 --> 00:00:16.450
And nonstandard free energy
change is symbolized by delta G.

00:00:16.450 --> 00:00:18.000
And notice, this delta G

00:00:18.000 --> 00:00:20.830
doesn't have the naught superscript,

00:00:20.830 --> 00:00:24.970
and therefore, it's nonstandard
change in free energy.

00:00:24.970 --> 00:00:26.930
The reason why this
concept is so important

00:00:26.930 --> 00:00:29.510
for a chemical reaction is by calculating

00:00:29.510 --> 00:00:31.770
the nonstandard change in free energy,

00:00:31.770 --> 00:00:34.210
we can figure out which direction

00:00:34.210 --> 00:00:36.520
the net reaction will proceed.

00:00:36.520 --> 00:00:38.710
When delta G is less than zero,

00:00:38.710 --> 00:00:40.540
so when delta G is negative,

00:00:40.540 --> 00:00:44.680
the forward reaction is
thermodynamically favored.

00:00:44.680 --> 00:00:48.050
Therefore, the reactants
will turn into the products.

00:00:48.050 --> 00:00:50.080
And the amount of products will increase,

00:00:50.080 --> 00:00:52.750
and the amount of reactants will decrease.

00:00:52.750 --> 00:00:54.790
When delta G is greater than zero,

00:00:54.790 --> 00:00:56.860
so when delta G is positive,

00:00:56.860 --> 00:01:01.340
the forward reaction is not
thermodynamically favored,

00:01:01.340 --> 00:01:04.210
which means the reverse
reaction is favored.

00:01:04.210 --> 00:01:07.020
So the net reaction goes
to the left to increase

00:01:07.020 --> 00:01:10.500
the amount of reactants and
decrease the amount of products.

00:01:10.500 --> 00:01:12.530
As long as there's a
difference in free energy

00:01:12.530 --> 00:01:14.530
between the reactants and the products,

00:01:14.530 --> 00:01:16.720
the net reaction will
proceed either to the left

00:01:16.720 --> 00:01:17.780
or to the right.

00:01:17.780 --> 00:01:20.260
However, when there's no
difference in free energy

00:01:20.260 --> 00:01:22.220
between the reactants and the products,

00:01:22.220 --> 00:01:24.580
or delta G is equal to zero,

00:01:24.580 --> 00:01:27.160
the reaction is at equilibrium.

00:01:27.160 --> 00:01:29.040
And when the reaction is at equilibrium,

00:01:29.040 --> 00:01:30.870
the concentrations of reactants

00:01:30.870 --> 00:01:33.450
and products remain constant.

00:01:33.450 --> 00:01:36.040
So it's useful to think
about nonstandard change

00:01:36.040 --> 00:01:38.410
in free energy as the driving force

00:01:38.410 --> 00:01:39.970
for a chemical reaction.

00:01:39.970 --> 00:01:42.150
As long as there's a
difference in free energy

00:01:42.150 --> 00:01:43.840
between reactants and products,

00:01:43.840 --> 00:01:46.510
the net reaction will move
one direction or the other

00:01:46.510 --> 00:01:48.140
to the left or to the right.

00:01:48.140 --> 00:01:50.620
However, when there's no
difference in free energy

00:01:50.620 --> 00:01:51.960
between reactants and products,

00:01:51.960 --> 00:01:53.510
there's no more driving force

00:01:53.510 --> 00:01:56.120
and the reaction is at equilibrium.

00:01:56.120 --> 00:01:57.580
Next, let's look at the equation

00:01:57.580 --> 00:02:01.540
that relates nonstandard
change in free energy, delta G,

00:02:01.540 --> 00:02:04.840
to standard change in free
energy, delta G naught.

00:02:04.840 --> 00:02:07.100
Remember, the superscript naught refers

00:02:07.100 --> 00:02:10.340
to the substances being
in their standard states

00:02:10.340 --> 00:02:12.660
under a pressure of one atmosphere.

00:02:12.660 --> 00:02:16.180
So delta G, or the nonstandard
change in free energy,

00:02:16.180 --> 00:02:18.550
refers to the instantaneous difference

00:02:18.550 --> 00:02:21.760
in free energy between the
reactants and the products.

00:02:21.760 --> 00:02:24.610
So when that reaction moves
to the left or to the right,

00:02:24.610 --> 00:02:26.550
this value is always changing.

00:02:26.550 --> 00:02:29.530
It's the driving force for the reaction.

00:02:29.530 --> 00:02:32.800
Delta G naught is a different situation.

00:02:32.800 --> 00:02:36.440
Delta G naught is a constant
at a certain temperature.

00:02:36.440 --> 00:02:39.140
And that's because it's
referring to the difference

00:02:39.140 --> 00:02:41.570
in free energy between
reactants and products

00:02:41.570 --> 00:02:43.320
in their standard states.

00:02:43.320 --> 00:02:46.130
And so this value will remain the same

00:02:46.130 --> 00:02:49.760
as the net reaction moves
to the left or to the right.

00:02:49.760 --> 00:02:51.580
To calculate the nonstandard change

00:02:51.580 --> 00:02:53.530
in free energy, delta G,

00:02:53.530 --> 00:02:55.500
it's equal to the standard change

00:02:55.500 --> 00:03:00.270
in free energy, delta G naught,
plus RT natural log of Q,

00:03:00.270 --> 00:03:02.930
where R is the ideal gas constant,

00:03:02.930 --> 00:03:05.000
T is the temperature in Kelvin,

00:03:05.000 --> 00:03:08.250
and Q is the reaction quotient.

00:03:08.250 --> 00:03:11.660
Next, let's calculate delta
G for a chemical reaction.

00:03:11.660 --> 00:03:14.940
And this is the reaction for
the synthesis of ammonia gas

00:03:14.940 --> 00:03:18.170
from nitrogen gas and hydrogen gas.

00:03:18.170 --> 00:03:22.540
Our goal is to calculate
delta G at 25 degrees Celsius

00:03:22.540 --> 00:03:24.720
at the moment in time
when the partial pressures

00:03:24.720 --> 00:03:27.440
of all three gases are one atmosphere.

00:03:27.440 --> 00:03:29.910
So one atmosphere for
nitrogen, for hydrogen,

00:03:29.910 --> 00:03:31.920
and for ammonia.

00:03:31.920 --> 00:03:36.030
At 25 degrees Celsius, delta
G naught for this reaction

00:03:36.030 --> 00:03:41.030
is equal to negative 33.0
kilojoules per mole of reaction.

00:03:41.170 --> 00:03:43.140
So our goal is to calculate delta G.

00:03:43.140 --> 00:03:45.240
And we know delta G naught,

00:03:45.240 --> 00:03:48.640
but we need to figure out what
Q is at this moment in time,

00:03:48.640 --> 00:03:50.610
when the partial
pressures of all the gases

00:03:50.610 --> 00:03:52.430
are one atmosphere.

00:03:52.430 --> 00:03:55.240
We can get the expression
for the reaction quotient Q

00:03:55.240 --> 00:03:57.230
from the balanced equation.

00:03:57.230 --> 00:03:59.430
So it would be the partial
pressure of ammonia.

00:03:59.430 --> 00:04:02.000
And since there's a two,
it'd be the partial pressure

00:04:02.000 --> 00:04:04.600
of ammonia raised to the second power

00:04:04.600 --> 00:04:07.510
divided by the partial
pressure of nitrogen.

00:04:07.510 --> 00:04:09.400
And since there's a one as a coefficient

00:04:09.400 --> 00:04:10.460
in the balanced equation,

00:04:10.460 --> 00:04:12.310
it's the partial pressure
of nitrogen raised

00:04:12.310 --> 00:04:13.470
to the first power,

00:04:13.470 --> 00:04:15.710
times the partial pressure of hydrogen.

00:04:15.710 --> 00:04:17.040
And since we have a three here,

00:04:17.040 --> 00:04:19.210
it'd be the partial
pressure of hydrogen raised

00:04:19.210 --> 00:04:20.910
to the third power.

00:04:20.910 --> 00:04:23.210
Since all three gases
have a partial pressure

00:04:23.210 --> 00:04:26.980
of one atmosphere, when
we plug in one atmosphere,

00:04:26.980 --> 00:04:30.400
we find that Q is equal to one.

00:04:30.400 --> 00:04:34.400
And the natural log of
one is equal to zero.

00:04:34.400 --> 00:04:38.290
So this second term here in
our equation goes to zero.

00:04:38.290 --> 00:04:40.240
And at this moment in time,

00:04:40.240 --> 00:04:44.410
delta G is equal to delta G naught.

00:04:44.410 --> 00:04:46.000
And since delta G naught is equal

00:04:46.000 --> 00:04:50.680
to negative 33.0 kilojoules
per mole of reaction,

00:04:50.680 --> 00:04:53.945
since delta G is negative
at this moment in time,

00:04:53.945 --> 00:04:57.320
the reaction is thermodynamically favored

00:04:57.320 --> 00:04:59.600
in the forward direction.

00:04:59.600 --> 00:05:02.840
So the reactants will
react together to form

00:05:02.840 --> 00:05:04.260
more of the products.

00:05:04.260 --> 00:05:07.080
The reason why delta G is
equal to delta G naught

00:05:07.080 --> 00:05:10.520
at this moment in time is because
our reactants and products

00:05:10.520 --> 00:05:13.300
are all in their standard states.

00:05:13.300 --> 00:05:14.830
By convention for a gas,

00:05:14.830 --> 00:05:17.750
standard state refers to
the pure gas at a pressure

00:05:17.750 --> 00:05:19.700
of one atmosphere.

00:05:19.700 --> 00:05:21.870
And since all of the
partial pressures are equal

00:05:21.870 --> 00:05:25.940
to one atmosphere, that
gave us Q is equal to one,

00:05:25.940 --> 00:05:27.950
which made the second term equal to zero.

00:05:27.950 --> 00:05:31.700
So only when the substances
are in their standard states

00:05:31.700 --> 00:05:34.970
is delta G equal to delta G naught.

00:05:34.970 --> 00:05:37.700
So if we had a different
set of partial pressures,

00:05:37.700 --> 00:05:41.640
delta G wouldn't be
equal to delta G naught.

00:05:41.640 --> 00:05:44.680
Let's calculate delta
G for the same reaction

00:05:44.680 --> 00:05:46.730
at 25 degrees Celsius.

00:05:46.730 --> 00:05:49.180
However, this time, instead
of having partial pressures

00:05:49.180 --> 00:05:52.850
of one atmosphere, all three
gasses at this moment in time

00:05:52.850 --> 00:05:55.990
have a partial pressure
of two atmospheres.

00:05:55.990 --> 00:05:59.480
Let's start by calculating
the reaction quotient Q.

00:05:59.480 --> 00:06:01.460
So when we plug in two atmospheres

00:06:01.460 --> 00:06:03.560
for the partial pressures of our gasses,

00:06:03.560 --> 00:06:07.220
we find that Q is equal to .25.

00:06:07.220 --> 00:06:09.492
Next, to calculate delta G,
we need to plug everything

00:06:09.492 --> 00:06:11.530
into our equation.

00:06:11.530 --> 00:06:16.530
So delta G naught is still equal
to negative 33.0 kilojoules

00:06:16.540 --> 00:06:17.880
per mole of reaction.

00:06:17.880 --> 00:06:20.040
So we can see, we plugged that in here.

00:06:20.040 --> 00:06:24.390
The ideal gas constant R is 8.314 joules

00:06:24.390 --> 00:06:26.640
per Kelvin mole of reaction.

00:06:26.640 --> 00:06:29.080
But to keep our units
the same in kilojoules,

00:06:29.080 --> 00:06:31.910
you can see I've converted
the ideal gas constant

00:06:31.910 --> 00:06:34.460
into kilojoules per
Kelvin mole of reaction.

00:06:34.460 --> 00:06:35.293
So it's .008314.

00:06:38.420 --> 00:06:41.280
We said the temperature
was 25 degrees Celsius.

00:06:41.280 --> 00:06:46.280
So 25 plus 273 is equal to 298 Kelvin.

00:06:46.370 --> 00:06:49.880
And the reaction quotient
Q is equal to .25.

00:06:49.880 --> 00:06:53.200
So be the it'd natural log of .25.

00:06:53.200 --> 00:06:55.740
So Kelvin will cancel out.

00:06:55.740 --> 00:06:57.360
And when we do the calculation,

00:06:57.360 --> 00:06:59.210
we find that delta G is equal

00:06:59.210 --> 00:07:03.860
to negative 36.4 kilojoules
per mole of reaction.

00:07:03.860 --> 00:07:07.120
Since delta G is negative
at this moment in time,

00:07:07.120 --> 00:07:09.790
the reaction is
thermodynamically favorable

00:07:09.790 --> 00:07:12.010
in the forward direction.

00:07:12.010 --> 00:07:14.370
So the net reaction will move to the right

00:07:14.370 --> 00:07:15.820
to make more products,

00:07:15.820 --> 00:07:18.890
and to decrease the amount of reactants.

00:07:18.890 --> 00:07:21.690
An increase in the amount
of ammonia means an increase

00:07:21.690 --> 00:07:23.490
in the partial pressure of ammonia.

00:07:23.490 --> 00:07:25.740
And a decrease in the amount of reactants

00:07:25.740 --> 00:07:28.030
means a decrease in the partial pressures

00:07:28.030 --> 00:07:29.340
of the reactants.

00:07:29.340 --> 00:07:32.140
Therefore, when the net
reaction goes to the right,

00:07:32.140 --> 00:07:35.910
the reaction quotient Q increases.

00:07:35.910 --> 00:07:40.510
So we started with a relatively
small value for Q of .25.

00:07:40.510 --> 00:07:43.480
And we know that Q is going to increase.

00:07:43.480 --> 00:07:45.120
Let's just jump ahead to a point,

00:07:45.120 --> 00:07:49.640
and let's say that Q is equal
to 1.0 times 10 to the third,

00:07:49.640 --> 00:07:51.970
and let's plug that into our equation

00:07:51.970 --> 00:07:54.510
and see what happens to delta G.

00:07:54.510 --> 00:07:56.730
So delta G naught stays the same

00:07:56.730 --> 00:08:00.770
at negative 33.0 kilojoules
per mole of reaction.

00:08:00.770 --> 00:08:02.730
The ideal gas constant's the same.

00:08:02.730 --> 00:08:03.920
The temperature's the same.

00:08:03.920 --> 00:08:06.810
So all we've done is increase Q.

00:08:06.810 --> 00:08:08.790
Once again, Kelvin cancels out,

00:08:08.790 --> 00:08:10.870
and we find that delta G,

00:08:10.870 --> 00:08:13.470
the instantaneous
difference in free energy,

00:08:13.470 --> 00:08:17.440
is equal to negative 15.9
kilojoules per mole of reaction

00:08:17.440 --> 00:08:19.180
at this moment in time.

00:08:19.180 --> 00:08:21.800
Since delta G is negative, once again,

00:08:21.800 --> 00:08:24.360
the forward reaction is favored.

00:08:24.360 --> 00:08:27.380
So the reaction moves to the
right to increase products

00:08:27.380 --> 00:08:29.530
and to decrease reactants.

00:08:29.530 --> 00:08:33.690
So again, Q will continue to increase.

00:08:33.690 --> 00:08:35.950
And think about what's
happening to delta G.

00:08:35.950 --> 00:08:40.660
It was negative 36.4, and
now it's negative 15.9.

00:08:40.660 --> 00:08:44.950
And as Q keeps increasing,
delta G keeps getting closer

00:08:44.950 --> 00:08:47.320
and closer to zero.

00:08:47.320 --> 00:08:52.320
Q will keep increasing until
delta G is equal to zero

00:08:53.030 --> 00:08:56.140
and the reaction is at equilibrium.

00:08:56.140 --> 00:08:59.890
So as long as Q is not equal
to the equilibrium constant K

00:08:59.890 --> 00:09:04.040
for this reaction, the
reaction is not at equilibrium.

00:09:04.040 --> 00:09:06.500
But eventually, Q will be equal

00:09:06.500 --> 00:09:08.440
to the equilibrium constant K.

00:09:08.440 --> 00:09:12.030
And if we were to plug that
value into our equation,

00:09:12.030 --> 00:09:15.480
we would find that delta
G is equal to zero.

00:09:15.480 --> 00:09:19.020
And therefore, the
reaction is at equilibrium.

00:09:19.020 --> 00:09:21.060
Let's summarize the difference between

00:09:21.060 --> 00:09:24.180
the nonstandard change
in free energy, delta G,

00:09:24.180 --> 00:09:27.890
and the standard change in
free energy, delta G naught.

00:09:27.890 --> 00:09:30.310
Delta G naught is talking
about the difference

00:09:30.310 --> 00:09:32.900
in free energy between
reactants and products

00:09:32.900 --> 00:09:35.950
when reactants and products
are in their standard states.

00:09:35.950 --> 00:09:39.040
And since that value is a constant,

00:09:39.040 --> 00:09:40.910
if the temperature is constant,

00:09:40.910 --> 00:09:42.820
notice for all of our calculations,

00:09:42.820 --> 00:09:43.990
delta G naught was equal

00:09:43.990 --> 00:09:48.220
to negative 33.0 kilojoules
per mole of reaction.

00:09:48.220 --> 00:09:51.690
Delta G is talking about
the instantaneous difference

00:09:51.690 --> 00:09:54.750
in free energy between
reactants and products.

00:09:54.750 --> 00:09:58.020
And as long as there's a
difference in free energy,

00:09:58.020 --> 00:10:00.120
there's a driving force
for the net reaction

00:10:00.120 --> 00:10:02.420
to go to the left or to the right.

00:10:02.420 --> 00:10:05.420
And eventually, when
delta G is equal to zero,

00:10:05.420 --> 00:10:07.760
the reaction is at equilibrium.

00:10:07.760 --> 00:10:10.900
Delta G is only equal to delta G naught

00:10:10.900 --> 00:10:12.500
when the reactants and products

00:10:12.500 --> 00:10:14.723
are in their standard states.


Thermodynamics vs. kinetics

WEBVTT
Kind: captions
Language: en

00:00:00.480 --> 00:00:01.313
- [Tutor] In chemistry,

00:00:01.313 --> 00:00:02.370
it's important to distinguish

00:00:02.370 --> 00:00:05.270
between thermodynamics and kinetics.

00:00:05.270 --> 00:00:06.103
For example,

00:00:06.103 --> 00:00:08.370
if we think about the conversion of carbon

00:00:08.370 --> 00:00:10.430
as a solid in the diamond form

00:00:10.430 --> 00:00:13.290
to carbon as a solid in the graphite form,

00:00:13.290 --> 00:00:15.970
thermodynamics tells us what will happen.

00:00:15.970 --> 00:00:19.370
Will this conversion happen
at a specific temperature?

00:00:19.370 --> 00:00:23.410
Whereas kinetics tells us when
the conversion could happen.

00:00:23.410 --> 00:00:25.040
Let's start with thermodynamics.

00:00:25.040 --> 00:00:27.480
We can calculate if this
conversion will happen

00:00:27.480 --> 00:00:30.210
at room temperature of 25 degrees Celsius

00:00:30.210 --> 00:00:33.130
by calculating the standard
change in free energy

00:00:33.130 --> 00:00:36.130
for this reaction at that temperature.

00:00:36.130 --> 00:00:37.950
To calculate the standard change

00:00:37.950 --> 00:00:40.200
in free energy for this reaction

00:00:40.200 --> 00:00:42.870
we need to know the standard change

00:00:42.870 --> 00:00:45.660
in free energy of
formation of the products.

00:00:45.660 --> 00:00:48.720
And from that we subtract
the standard change

00:00:48.720 --> 00:00:52.120
in free energy of
formation of the reactants.

00:00:52.120 --> 00:00:53.130
For this conversion

00:00:53.130 --> 00:00:56.430
our product is carbon
in the graphite form.

00:00:56.430 --> 00:00:58.360
And because carbon in the graphite form

00:00:58.360 --> 00:01:01.800
is the most stable
elemental form of carbon

00:01:01.800 --> 00:01:03.820
at a pressure of one atmosphere,

00:01:03.820 --> 00:01:07.540
the standard change in
free energy of formation

00:01:07.540 --> 00:01:10.800
of carbon in the graphite
form is equal to zero.

00:01:10.800 --> 00:01:12.860
Next, we think about our reactant,

00:01:12.860 --> 00:01:15.740
which is carbon in the form of diamond.

00:01:15.740 --> 00:01:18.600
The standard change in free
energy of formation of carbon

00:01:18.600 --> 00:01:22.580
in the diamond form is
2.84 kilojoules per mole.

00:01:22.580 --> 00:01:25.010
Looking at our balanced
equation for this conversion,

00:01:25.010 --> 00:01:26.880
we're converting one mole of diamond

00:01:26.880 --> 00:01:28.870
into one mole of graphite.

00:01:28.870 --> 00:01:32.250
Since there's a one as a
coefficient in front of diamond,

00:01:32.250 --> 00:01:34.920
we multiply the standard
change in free energy

00:01:34.920 --> 00:01:38.550
of formation of diamond
by one mole of diamond.

00:01:38.550 --> 00:01:42.600
Moles cancel, and we get
negative 2.84 kilojoules

00:01:42.600 --> 00:01:46.030
as the standard change in
free energy for this reaction.

00:01:46.030 --> 00:01:47.130
Instead of kilojoules,

00:01:47.130 --> 00:01:48.500
the units could've been written

00:01:48.500 --> 00:01:51.340
as kilojoules per mole of reaction.

00:01:51.340 --> 00:01:53.870
Because there's one mole of
carbon in the diamond form

00:01:53.870 --> 00:01:55.950
for how the balanced equation is written,

00:01:55.950 --> 00:01:58.470
we could write a conversion
factor of one mole

00:01:58.470 --> 00:02:02.570
of carbon in the diamond
form per mole of reaction.

00:02:02.570 --> 00:02:04.770
And because the standard
change in free energy

00:02:04.770 --> 00:02:06.070
of formation of diamond

00:02:06.070 --> 00:02:10.010
is equal to 2.84 kilojoules
per mole of diamond,

00:02:10.010 --> 00:02:12.100
moles of diamond would cancel out

00:02:12.100 --> 00:02:16.930
and give kilojoules per mole
of reaction as the units.

00:02:16.930 --> 00:02:18.190
Because the standard change

00:02:18.190 --> 00:02:21.760
in free energy for this
reaction is negative.

00:02:21.760 --> 00:02:25.310
We know that the conversion
of diamond into graphite

00:02:25.310 --> 00:02:30.310
at 25 degrees Celsius is
thermodynamically favorable.

00:02:30.590 --> 00:02:34.040
Next, let's think about the
kinetics of this reaction.

00:02:34.040 --> 00:02:35.850
If we look at the structure of diamond,

00:02:35.850 --> 00:02:40.400
each carbon atom is covalently
bonded to four other carbons.

00:02:40.400 --> 00:02:41.233
For example,

00:02:41.233 --> 00:02:45.060
this carbon right here is bonded
to four other carbon atoms

00:02:45.060 --> 00:02:50.060
and the tetrahedral geometry
around the central carbon atom.

00:02:50.060 --> 00:02:52.160
Graphite has a very different structure.

00:02:52.160 --> 00:02:52.993
In graphite,

00:02:52.993 --> 00:02:57.000
carbon atoms form covalently
bonded layers or sheets.

00:02:57.000 --> 00:02:58.770
And these layers or sheets

00:02:58.770 --> 00:03:02.200
are held together by
London dispersion forces.

00:03:02.200 --> 00:03:05.680
So in order to convert
diamond into graphite

00:03:05.680 --> 00:03:09.280
we would have to break a
lot of carbon-carbon bonds.

00:03:09.280 --> 00:03:11.590
And that would take a lot of energy.

00:03:11.590 --> 00:03:12.980
Since it takes a lot of energy

00:03:12.980 --> 00:03:15.610
to break the carbon-carbon
bonds in diamond,

00:03:15.610 --> 00:03:18.600
if we look at the energy
profile for this reaction,

00:03:18.600 --> 00:03:21.860
this reaction would have a
very high activation energy,

00:03:21.860 --> 00:03:24.360
symbolized by EA.

00:03:24.360 --> 00:03:27.820
And the higher the activation
energy for the reaction

00:03:27.820 --> 00:03:31.720
the slower the rate of the reaction.

00:03:31.720 --> 00:03:32.750
At room temperature,

00:03:32.750 --> 00:03:36.100
it's estimated this reaction
could take billions of years.

00:03:36.100 --> 00:03:38.010
Therefore we would say this reaction

00:03:38.010 --> 00:03:40.990
is kinetically unfavorable.

00:03:40.990 --> 00:03:42.710
Finally, let's summarize
what we've learned

00:03:42.710 --> 00:03:46.410
about the conversion of
diamond into graphite.

00:03:46.410 --> 00:03:50.480
Thermodynamics answers the
question, will the reaction go?

00:03:50.480 --> 00:03:52.630
And the answer to that question is yes.

00:03:52.630 --> 00:03:54.060
At room temperature,

00:03:54.060 --> 00:03:56.910
diamond will convert into graphite

00:03:56.910 --> 00:04:00.340
because delta G naught for
the reaction was negative.

00:04:00.340 --> 00:04:03.970
Kinetics answers the question,
how long will it take?

00:04:03.970 --> 00:04:05.160
And the answer to that question

00:04:05.160 --> 00:04:07.740
is it would take billions
of years for diamond

00:04:07.740 --> 00:04:10.640
to turn into graphite at room temperature.

00:04:10.640 --> 00:04:12.710
The reason why the reaction is so slow

00:04:12.710 --> 00:04:15.910
is because of the extremely
high activation energy.

00:04:15.910 --> 00:04:18.940
So even though the reaction
is thermodynamically favorable

00:04:18.940 --> 00:04:22.830
at room temperature, because
the reaction is so slow,

00:04:22.830 --> 00:04:25.260
the reaction is kinetically unfavored.

00:04:25.260 --> 00:04:28.893
And for all practical purposes,
the reaction doesn't happen.


Free energy of dissolution

WEBVTT
Kind: captions
Language: en

00:00:00.220 --> 00:00:02.420
- [Instructor] The term
dissolution refers to

00:00:02.420 --> 00:00:06.100
the dissolving of one
substance in a solvent.

00:00:06.100 --> 00:00:09.160
The dissolve substance
is now called a salute

00:00:09.160 --> 00:00:13.270
and the solute plus the
solvent form a solution.

00:00:13.270 --> 00:00:15.930
If the standard change in
free energy, delta G naught

00:00:15.930 --> 00:00:17.450
is less than zero,

00:00:17.450 --> 00:00:21.010
the dissolution is
thermodynamically favorable.

00:00:21.010 --> 00:00:23.000
So, if we were to put this substance

00:00:23.000 --> 00:00:25.030
in a solvent like water,

00:00:25.030 --> 00:00:28.440
the substance would dissolve
and it would form a solution.

00:00:28.440 --> 00:00:30.850
However, if delta G naught
is greater than zero,

00:00:30.850 --> 00:00:34.690
the dissolution is
thermodynamically unfavorable.

00:00:34.690 --> 00:00:37.940
So, if we try to dissolve
the substance in water,

00:00:37.940 --> 00:00:39.050
it wouldn't dissolve

00:00:39.050 --> 00:00:40.750
and therefore we would just see it

00:00:40.750 --> 00:00:43.360
on the bottom of the beaker.

00:00:43.360 --> 00:00:46.240
Since we're talking about
potentially making a solution,

00:00:46.240 --> 00:00:49.870
sometimes you see a
subscript soln written next

00:00:49.870 --> 00:00:51.140
to delta G naught.

00:00:51.140 --> 00:00:54.240
So, this would be delta
G naught of solution.

00:00:54.240 --> 00:00:57.710
We could also call this the
free energy of dissolution.

00:00:57.710 --> 00:01:00.650
We can calculate delta
G naught of solution

00:01:00.650 --> 00:01:04.270
by taking the standard change
and enthalpy of the solution

00:01:04.270 --> 00:01:07.100
and from that subtracting
the absolute temperature,

00:01:07.100 --> 00:01:11.200
times the standard change
in entropy of the solution.

00:01:11.200 --> 00:01:13.390
So, let's look in more
detail at what determines

00:01:13.390 --> 00:01:17.350
the signs for delta H
naught and delta S naught.

00:01:17.350 --> 00:01:19.700
We can think about the
disillusion of a solid

00:01:19.700 --> 00:01:21.970
in three hypothetical steps.

00:01:21.970 --> 00:01:25.110
The first step involves
breaking up the solid.

00:01:25.110 --> 00:01:27.570
Let's think about the change in enthalpy,

00:01:27.570 --> 00:01:31.520
delta H one, for this first step.

00:01:31.520 --> 00:01:34.120
Solids are held together
by attractive forces.

00:01:34.120 --> 00:01:37.010
For example, if we had an ionic solid,

00:01:37.010 --> 00:01:38.500
it would be ionic bonds

00:01:38.500 --> 00:01:41.160
or electrostatic
interactions holding together

00:01:41.160 --> 00:01:42.920
the opposite charges.

00:01:42.920 --> 00:01:46.320
So, if our goal is to pull
apart or break apart the solid,

00:01:46.320 --> 00:01:49.790
it would take energy to overcome
these attractive forces.

00:01:49.790 --> 00:01:53.440
Therefore, delta H one would be positive.

00:01:53.440 --> 00:01:55.550
Next, let's think about
the change in entropy.

00:01:55.550 --> 00:01:58.090
Delta S one for the first step.

00:01:58.090 --> 00:01:59.570
When the solid is broken up,

00:01:59.570 --> 00:02:03.000
the particles have a greater
number of possible positions

00:02:03.000 --> 00:02:05.400
and an increased number of
possible positions means

00:02:05.400 --> 00:02:08.640
an increase in number
of possible microstates.

00:02:08.640 --> 00:02:10.660
An increase in the number of microstates

00:02:10.660 --> 00:02:13.050
means an increase in entropy.

00:02:13.050 --> 00:02:14.880
Therefore, breaking up the solid

00:02:14.880 --> 00:02:17.180
means an increase in entropy

00:02:17.180 --> 00:02:20.360
and delta S one will be positive.

00:02:20.360 --> 00:02:22.810
Step two is the preparation of the solvent

00:02:22.810 --> 00:02:25.260
to receive the salute.

00:02:25.260 --> 00:02:28.180
And let's think about
the change in enthalpy

00:02:28.180 --> 00:02:30.870
for this second step.

00:02:30.870 --> 00:02:33.380
The solvent particles are held together

00:02:33.380 --> 00:02:36.240
by intermolecular forces.

00:02:36.240 --> 00:02:37.370
For water molecules,

00:02:37.370 --> 00:02:41.610
the most important intermolecular
force is hydrogen bonding.

00:02:41.610 --> 00:02:43.710
The goal of this theoretical second step

00:02:43.710 --> 00:02:45.920
is to break apart the solvent particles

00:02:45.920 --> 00:02:48.250
and to move them far apart from each other

00:02:48.250 --> 00:02:51.390
so there's enough room to
fit in a solute particle.

00:02:51.390 --> 00:02:53.350
Since it takes energy to overcome

00:02:53.350 --> 00:02:55.280
the intermolecular forces holding

00:02:55.280 --> 00:02:57.180
the solvent particles together,

00:02:57.180 --> 00:03:00.000
delta H two is positive.

00:03:00.000 --> 00:03:02.010
Next, let's think about
the change in entropy,

00:03:02.010 --> 00:03:04.960
delta S two for this second step.

00:03:04.960 --> 00:03:07.370
When the water molecules are pulled apart,

00:03:07.370 --> 00:03:10.050
there's an increase in the
number of possible positions

00:03:10.050 --> 00:03:11.650
of the water molecules.

00:03:11.650 --> 00:03:14.040
And like the first step,
an increase in the number

00:03:14.040 --> 00:03:16.090
of possible positions means an increase

00:03:16.090 --> 00:03:17.550
in the number of microstates,

00:03:17.550 --> 00:03:20.110
which means an increase in entropy.

00:03:20.110 --> 00:03:23.690
Therefore, delta S two is positive.

00:03:23.690 --> 00:03:26.060
The third step is called solvation,

00:03:26.060 --> 00:03:28.140
which refers to the interaction

00:03:28.140 --> 00:03:30.200
of the solute and the solvent.

00:03:30.200 --> 00:03:33.520
So, imagine there's an attractive force

00:03:33.520 --> 00:03:36.480
between this white solute particle

00:03:36.480 --> 00:03:39.790
and these four blue solvent particles.

00:03:39.790 --> 00:03:43.700
If the solvent is water, this
process is called hydration.

00:03:43.700 --> 00:03:46.140
And if we think about,
say a positive cation,

00:03:46.140 --> 00:03:47.610
interacting with water,

00:03:47.610 --> 00:03:49.520
since opposite charges attract,

00:03:49.520 --> 00:03:53.510
the positive cation is
attracted to the negative end

00:03:53.510 --> 00:03:55.990
of the water molecule.

00:03:55.990 --> 00:03:58.970
Since we have a positively
charged ion interacting

00:03:58.970 --> 00:04:02.730
with water, which is a polar
molecule with a dipole moment,

00:04:02.730 --> 00:04:03.750
this type of interaction

00:04:03.750 --> 00:04:06.990
is called an ion dipole interaction.

00:04:06.990 --> 00:04:09.850
And let's think about
the change in enthalpy

00:04:09.850 --> 00:04:12.210
for this third step.

00:04:12.210 --> 00:04:14.720
When the solute comes
together with the solvent,

00:04:14.720 --> 00:04:16.110
energy is released,

00:04:16.110 --> 00:04:19.680
therefore delta H three is negative.

00:04:19.680 --> 00:04:21.610
One way to think about
why this is negative

00:04:21.610 --> 00:04:25.710
is to consider the ion dipole
interactions on the right.

00:04:25.710 --> 00:04:27.130
Since it would take energy

00:04:27.130 --> 00:04:29.950
to break these ion dipole interactions,

00:04:29.950 --> 00:04:32.470
when those ion dipole interactions form,

00:04:32.470 --> 00:04:34.730
energy must be given off.

00:04:34.730 --> 00:04:36.740
Next, let's think about
the change in entropy.

00:04:36.740 --> 00:04:39.430
Delta S three for the third step.

00:04:39.430 --> 00:04:42.210
Because the water molecules are attracted

00:04:42.210 --> 00:04:44.690
to the ions in solution,

00:04:44.690 --> 00:04:47.890
the water molecules have a
decreased freedom of movement

00:04:47.890 --> 00:04:50.620
and therefore there are
fewer positions possible

00:04:50.620 --> 00:04:52.360
for the water molecules.

00:04:52.360 --> 00:04:55.070
A decrease in the number
of possible positions

00:04:55.070 --> 00:04:58.300
means a decrease in the
number of microstates,

00:04:58.300 --> 00:05:00.760
which means a decrease in entropy.

00:05:00.760 --> 00:05:05.760
Therefore, for the third step,
delta S three is negative.

00:05:05.790 --> 00:05:07.100
Now that we've talked about the signs

00:05:07.100 --> 00:05:10.500
for delta H and delta S for
each of the three steps,

00:05:10.500 --> 00:05:13.510
let's think about how they
influence the overall changes

00:05:13.510 --> 00:05:15.500
in delta H naught of solution

00:05:15.500 --> 00:05:18.220
and delta S naught of solution.

00:05:18.220 --> 00:05:21.010
And let's start with delta
H naught of solution,

00:05:21.010 --> 00:05:23.490
which is equal to the sum of the changes

00:05:23.490 --> 00:05:26.130
in enthalpy for the three steps.

00:05:26.130 --> 00:05:30.530
So, delta H one, plus delta
H two, plus delta H three.

00:05:30.530 --> 00:05:33.710
We've already seen that
delta H one is positive,

00:05:33.710 --> 00:05:38.610
delta H two is positive, and
delta H three is negative.

00:05:38.610 --> 00:05:41.790
So, adding delta H one
and delta H two together,

00:05:41.790 --> 00:05:43.870
it gives us a positive value.

00:05:43.870 --> 00:05:46.600
And since we're adding a
negative in delta H three,

00:05:46.600 --> 00:05:48.930
it's the magnitude of delta H three

00:05:48.930 --> 00:05:52.480
that determines if the overall
delta H naught of solution

00:05:52.480 --> 00:05:54.780
is positive or negative.

00:05:54.780 --> 00:05:57.840
If the magnitude of delta
H one, plus delta H two

00:05:57.840 --> 00:06:00.900
is greater than the
magnitude of delta H three,

00:06:00.900 --> 00:06:04.380
delta H naught of
solution will be positive.

00:06:04.380 --> 00:06:06.930
However, if the magnitude of delta H three

00:06:06.930 --> 00:06:08.460
is greater than the magnitude

00:06:08.460 --> 00:06:11.050
of delta H one plus delta H two,

00:06:11.050 --> 00:06:14.200
delta H naught of
solution will be negative.

00:06:14.200 --> 00:06:17.290
Therefore, it's possible for
delta H naught of solution

00:06:17.290 --> 00:06:19.160
to be positive or negative,

00:06:19.160 --> 00:06:21.740
depending on the
substance being dissolved.

00:06:21.740 --> 00:06:24.640
An example of a substance
that has a positive value

00:06:24.640 --> 00:06:28.370
for delta H naught of
solution is sodium chloride.

00:06:28.370 --> 00:06:29.870
At 25 degrees Celsius,

00:06:29.870 --> 00:06:32.560
delta H naught of solution
for sodium chloride

00:06:32.560 --> 00:06:37.130
is positive 3.9 kilojoules
per mole of reaction.

00:06:37.130 --> 00:06:39.750
An example of a substance
that has a negative value

00:06:39.750 --> 00:06:43.520
for delta H naught of solution
is magnesium chloride.

00:06:43.520 --> 00:06:45.040
At 25 degrees Celsius,

00:06:45.040 --> 00:06:47.900
delta H naught of solution
for magnesium chloride

00:06:47.900 --> 00:06:52.900
is equal to negative 160.0
kilojoules per mole of reaction.

00:06:53.350 --> 00:06:55.470
The main reason why these two substances

00:06:55.470 --> 00:06:58.110
have such different enthalpies of solution

00:06:58.110 --> 00:07:00.430
has to do with the third step.

00:07:00.430 --> 00:07:03.290
So, the value of delta H three.

00:07:03.290 --> 00:07:05.320
So, let's take a look
at some diagrams showing

00:07:05.320 --> 00:07:08.480
the ion dipole interaction of the cation

00:07:08.480 --> 00:07:11.220
of both of these substances
in aqueous solution,

00:07:11.220 --> 00:07:13.580
interacting with water molecules.

00:07:13.580 --> 00:07:16.770
For sodium chloride, we
would have a sodium cation

00:07:16.770 --> 00:07:19.930
in solution with a one plus charge.

00:07:19.930 --> 00:07:21.790
And for magnesium chloride,

00:07:21.790 --> 00:07:25.090
it would be an Mg two
plus cation interacting

00:07:25.090 --> 00:07:27.120
with the water molecules.

00:07:27.120 --> 00:07:30.730
Because the Mg two plus
cation has a higher charge

00:07:30.730 --> 00:07:33.920
than the sodium cation
and because it's smaller,

00:07:33.920 --> 00:07:37.750
the Mg two plus cation exerts
a greater electric force

00:07:37.750 --> 00:07:39.700
on the water molecules,

00:07:39.700 --> 00:07:42.660
and because the magnesium
two plus cation exerts

00:07:42.660 --> 00:07:45.730
a greater electric force
on the water molecules,

00:07:45.730 --> 00:07:49.840
more energy is released when
the magnesium two plus cation

00:07:49.840 --> 00:07:52.210
is hydrated compared to

00:07:52.210 --> 00:07:56.600
when the sodium one
plus cation is hydrated.

00:07:56.600 --> 00:08:00.240
And since more energy is
given off in the third step

00:08:00.240 --> 00:08:04.310
for the hydration of the
magnesium two plus cation,

00:08:04.310 --> 00:08:05.240
that's enough to turn

00:08:05.240 --> 00:08:09.490
the overall delta H naught
of solution negative.

00:08:09.490 --> 00:08:12.310
Therefore, the dissolution
of magnesium chloride

00:08:12.310 --> 00:08:14.970
is an exothermic process.

00:08:14.970 --> 00:08:18.110
For sodium chloride, when the
sodium cation is hydrated,

00:08:18.110 --> 00:08:20.710
not as much energy is released.

00:08:20.710 --> 00:08:25.050
Therefore, the overall delta
H of solution is positive

00:08:25.050 --> 00:08:27.410
and the dissolution of sodium chloride

00:08:27.410 --> 00:08:30.030
is an endothermic process.

00:08:30.030 --> 00:08:33.410
Next, let's think about
delta S naught of solution.

00:08:33.410 --> 00:08:34.900
Delta S naught of solution

00:08:34.900 --> 00:08:37.640
is equal to delta S one, plus delta S two,

00:08:37.640 --> 00:08:39.580
plus delta S three.

00:08:39.580 --> 00:08:41.610
Delta S one, the change in entropy

00:08:41.610 --> 00:08:44.460
for the first step, we saw was positive.

00:08:44.460 --> 00:08:46.830
Delta S two was also positive,

00:08:46.830 --> 00:08:49.610
but delta S3 was negative.

00:08:49.610 --> 00:08:52.300
Therefore, just like we
saw for delta H naught,

00:08:52.300 --> 00:08:55.360
the sign for delta S naught
depends on the magnitude

00:08:55.360 --> 00:08:59.970
of delta S one, plus delta S
two, compared to delta S three.

00:08:59.970 --> 00:09:01.960
If delta S one, plus delta S two

00:09:01.960 --> 00:09:04.650
is greater in magnitude
than delta S three,

00:09:04.650 --> 00:09:07.970
delta S naught of
solution will be positive.

00:09:07.970 --> 00:09:10.260
However, if the magnitude of delta S three

00:09:10.260 --> 00:09:12.510
is greater than the
magnitude of delta S one,

00:09:12.510 --> 00:09:14.250
plus delta S two,

00:09:14.250 --> 00:09:17.300
delta S naught of
solution will be negative.

00:09:17.300 --> 00:09:19.950
An example of a substance
that has a positive value

00:09:19.950 --> 00:09:22.970
for delta S naught of
solution is sodium chloride.

00:09:22.970 --> 00:09:24.500
At 25 degrees Celsius,

00:09:24.500 --> 00:09:26.780
delta S naught of solution is equal

00:09:26.780 --> 00:09:31.780
to positive 43.4 joules per
Kelvin mole of reaction.

00:09:32.360 --> 00:09:34.920
An example of a substance
that has a negative value

00:09:34.920 --> 00:09:38.350
for delta S naught of solution
is magnesium chloride.

00:09:38.350 --> 00:09:39.860
At 25 degrees Celsius,

00:09:39.860 --> 00:09:41.840
delta S naught of solution is equal

00:09:41.840 --> 00:09:46.840
to negative 114.7 joules
per Kelvin mole of reaction.

00:09:47.250 --> 00:09:49.530
The main reason for the
difference in entropies

00:09:49.530 --> 00:09:53.090
for these two substances has
to do with the third step.

00:09:53.090 --> 00:09:56.000
So, the magnitude of delta S three.

00:09:56.000 --> 00:09:57.310
So, once again, let's take a look

00:09:57.310 --> 00:09:59.630
at the diagrams showing the hydration

00:09:59.630 --> 00:10:02.110
of the cations in aqueous solution.

00:10:02.110 --> 00:10:04.240
We've already talked about
the fact that the smaller

00:10:04.240 --> 00:10:06.700
and more positive
magnesium two-plus cation

00:10:06.700 --> 00:10:08.360
has a greater electric force

00:10:08.360 --> 00:10:10.460
on its surrounding water molecules

00:10:10.460 --> 00:10:13.560
than the sodium one plus cation does

00:10:13.560 --> 00:10:15.980
on its surrounding water molecules.

00:10:15.980 --> 00:10:19.300
The stronger electric force
means a decreased freedom

00:10:19.300 --> 00:10:21.340
of movement of water molecules

00:10:21.340 --> 00:10:24.330
around the magnesium two plus cation,

00:10:24.330 --> 00:10:26.690
and a decreased freedom of movement means

00:10:26.690 --> 00:10:29.540
a decreased number of
possible microstates,

00:10:29.540 --> 00:10:33.620
which means a greater decrease in entropy.

00:10:33.620 --> 00:10:36.940
So, the hydration of the
magnesium two plus cation leads

00:10:36.940 --> 00:10:41.140
to a greater decrease in
entropy for step three,

00:10:41.140 --> 00:10:44.930
which outweighs the positive
values for steps one and two.

00:10:44.930 --> 00:10:47.430
And that gives an overall negative value

00:10:47.430 --> 00:10:49.610
for delta S naught of solution,

00:10:49.610 --> 00:10:51.740
therefore dissolving magnesium chloride

00:10:51.740 --> 00:10:55.560
in water results in a decrease in entropy.

00:10:55.560 --> 00:10:57.010
Since the sodium cation

00:10:57.010 --> 00:11:00.870
has a weaker electrostatic
attraction for water molecules,

00:11:00.870 --> 00:11:03.480
the decrease in entropy is not as much.

00:11:03.480 --> 00:11:05.990
And therefore, the
magnitude of the third term

00:11:05.990 --> 00:11:09.440
does not overcome the magnitude
of the first two terms,

00:11:09.440 --> 00:11:12.550
which means the overall
delta S naught of solution

00:11:12.550 --> 00:11:15.530
is positive and dissolving sodium chloride

00:11:15.530 --> 00:11:19.910
in water results in an
increase in entropy.

00:11:19.910 --> 00:11:22.390
We've just looked at
delta H naught of solution

00:11:22.390 --> 00:11:25.290
and delta S naught of
solution for two salts,

00:11:25.290 --> 00:11:28.170
sodium chloride and magnesium chloride.

00:11:28.170 --> 00:11:30.720
Let's calculate delta G naught of solution

00:11:30.720 --> 00:11:34.700
for both salts at 25 degrees Celsius.

00:11:34.700 --> 00:11:36.090
Let's start with the dissolution

00:11:36.090 --> 00:11:38.070
of solid sodium chloride and water

00:11:38.070 --> 00:11:41.310
to form sodium cations
and chloride anions.

00:11:41.310 --> 00:11:44.860
25 degrees Celsius is 298 Kelvin

00:11:44.860 --> 00:11:47.200
so we can plug that into our equation.

00:11:47.200 --> 00:11:49.830
And we can also plug in the
values for delta H naught

00:11:49.830 --> 00:11:51.640
and delta S naught.

00:11:51.640 --> 00:11:54.360
So, both of them were positive values

00:11:54.360 --> 00:11:57.250
for the dissolution of sodium chloride.

00:11:57.250 --> 00:12:00.880
Kelvin cancels out and gives
us delta G naught of solution

00:12:00.880 --> 00:12:05.260
is equal to negative 9.0
kilojoules per mole of reaction.

00:12:05.260 --> 00:12:07.750
Since delta G naught is negative,

00:12:07.750 --> 00:12:10.550
the dissolution of solid sodium chloride

00:12:10.550 --> 00:12:13.500
is a thermodynamically favorable process,

00:12:13.500 --> 00:12:15.630
which means at 25 degrees Celsius,

00:12:15.630 --> 00:12:18.380
sodium chloride is soluble in water.

00:12:18.380 --> 00:12:19.810
Notice how in this case,

00:12:19.810 --> 00:12:23.890
the favorable positive
entropy term outweighs

00:12:23.890 --> 00:12:26.790
the unfavorable positive enthalpy term

00:12:26.790 --> 00:12:29.960
to give a negative value
for delta G naught.

00:12:29.960 --> 00:12:32.410
Therefore, for the dissolution
of sodium chloride,

00:12:32.410 --> 00:12:36.420
the increase in entropy
drives the dissolution.

00:12:36.420 --> 00:12:38.390
Next, let's do the same calculation

00:12:38.390 --> 00:12:41.600
for the dissolution of solid
magnesium chloride and water

00:12:41.600 --> 00:12:43.760
to form the magnesium two plus cation

00:12:43.760 --> 00:12:45.710
and two chloride anions.

00:12:45.710 --> 00:12:48.300
We've already seen how the
increased positive charge

00:12:48.300 --> 00:12:50.920
of the Mg two plus cation in solution

00:12:50.920 --> 00:12:53.540
gave us a negative
value for delta H naught

00:12:53.540 --> 00:12:56.060
and a negative value for delta S naught.

00:12:56.060 --> 00:12:57.610
So, plugging in all our numbers

00:12:57.610 --> 00:12:59.940
and a temperature of 298 Kelvin,

00:12:59.940 --> 00:13:03.030
Kelvin cancels out and gives
us delta G naught of solution

00:13:03.030 --> 00:13:08.030
is equal to negative 125.8
kilojoules per mole of reaction.

00:13:08.120 --> 00:13:10.980
Since, delta G naught
of solution is negative,

00:13:10.980 --> 00:13:13.860
the dissolution of magnesium chloride

00:13:13.860 --> 00:13:17.190
is a thermodynamically favorable process.

00:13:17.190 --> 00:13:20.320
So, at room temperature
of 25 degrees Celsius,

00:13:20.320 --> 00:13:23.060
magnesium chloride is soluble in water.

00:13:23.060 --> 00:13:26.280
However, this time it's the
favorable negative value

00:13:26.280 --> 00:13:28.500
for the enthalpy term that outweighs

00:13:28.500 --> 00:13:30.520
the unfavorable negative value

00:13:30.520 --> 00:13:34.490
for the entropy term to give
a negative value overall

00:13:34.490 --> 00:13:36.590
for delta G naught of solution.

00:13:36.590 --> 00:13:39.650
So, this time it's the
decrease in enthalpy

00:13:39.650 --> 00:13:41.870
that drives the dissolution.

00:13:41.870 --> 00:13:45.130
Delta G naught has been negative
for both sodium chloride

00:13:45.130 --> 00:13:47.020
and for magnesium chloride.

00:13:47.020 --> 00:13:48.810
However, those are just the two examples

00:13:48.810 --> 00:13:49.970
that I chose for this video,

00:13:49.970 --> 00:13:52.150
it's certainly possible
to get a positive value

00:13:52.150 --> 00:13:54.040
for delta G naught of solution,

00:13:54.040 --> 00:13:56.630
which would mean an insoluble salt,

00:13:56.630 --> 00:13:59.350
and because it can be very
difficult to predict the signs

00:13:59.350 --> 00:14:01.710
for delta H naught and for delta S naught

00:14:01.710 --> 00:14:03.400
for a particular salt,

00:14:03.400 --> 00:14:06.160
it's even more difficult to
predict if delta G naught

00:14:06.160 --> 00:14:09.440
is positive or negative
for that particular salt.

00:14:09.440 --> 00:14:12.340
Therefore, it's often
necessary to do a calculation

00:14:12.340 --> 00:14:14.953
to see if delta G naught
is positive or negative.


Free energy of formation

WEBVTT
Kind: captions
Language: en

00:00:00.310 --> 00:00:02.960
- [Instructor] Free
energy is symbolized by G.

00:00:02.960 --> 00:00:05.620
And the change in free energy is delta G.

00:00:05.620 --> 00:00:07.160
When there's a subscript of F,

00:00:07.160 --> 00:00:08.570
this is talking about the change

00:00:08.570 --> 00:00:11.010
in the free energy of formation.

00:00:11.010 --> 00:00:13.360
Which refers to the
change in the free energy

00:00:13.360 --> 00:00:16.300
for the formation of
one mole of a substance

00:00:16.300 --> 00:00:19.440
from its elements in
their standard states.

00:00:19.440 --> 00:00:21.770
These superscript naught
refers to the fact

00:00:21.770 --> 00:00:25.260
that the substances are
in their standard states.

00:00:25.260 --> 00:00:27.410
By convention, the standard change

00:00:27.410 --> 00:00:29.920
in free energy of formation for an element

00:00:29.920 --> 00:00:32.910
and its standard state is equal to zero.

00:00:32.910 --> 00:00:36.900
Therefore, if we think about
forming one mole of oxygen gas,

00:00:36.900 --> 00:00:39.900
we would be making it
from elemental oxygen

00:00:39.900 --> 00:00:42.350
which is also oxygen gas.

00:00:42.350 --> 00:00:44.930
So forming one mole of
oxygen gas from one mole

00:00:44.930 --> 00:00:47.480
of oxygen gas means
there will be no change

00:00:47.480 --> 00:00:49.710
and therefore the standard change

00:00:49.710 --> 00:00:54.150
in free energy of
formation is equal to zero.

00:00:54.150 --> 00:00:57.220
Textbooks often tabulate
data for standard change

00:00:57.220 --> 00:00:59.430
in free energy of formation of substances

00:00:59.430 --> 00:01:01.110
at 25 degrees Celsius.

00:01:01.110 --> 00:01:03.760
However, it doesn't have
to be that temperature.

00:01:03.760 --> 00:01:06.730
The units are in kilojoules per mole.

00:01:06.730 --> 00:01:08.580
We've already seen that
the standard change

00:01:08.580 --> 00:01:11.670
in free energy of formation for an element

00:01:11.670 --> 00:01:13.690
is equal to zero.

00:01:13.690 --> 00:01:17.900
Therefore for elemental
oxygen, O2 delta G naught

00:01:17.900 --> 00:01:19.440
is equal to zero.

00:01:19.440 --> 00:01:23.550
And for carbon in the solid
state in the form of graphite,

00:01:23.550 --> 00:01:26.360
delta G naught is also equal to zero.

00:01:26.360 --> 00:01:29.670
The value for carbon monoxide
is naught equal to zero.

00:01:29.670 --> 00:01:30.920
Delta G naught is equal

00:01:30.920 --> 00:01:35.190
to negative 137.2 kilojoules per mole.

00:01:35.190 --> 00:01:38.670
So if we think about forming
one mole of carbon monoxide

00:01:38.670 --> 00:01:39.890
from its elements,

00:01:39.890 --> 00:01:43.360
carbon monoxide is composed
of carbon and oxygen.

00:01:43.360 --> 00:01:46.260
And the elemental form of oxygen is O2 gas

00:01:46.260 --> 00:01:48.070
and the elements of form of carbon

00:01:48.070 --> 00:01:50.690
is the solid form in graphite.

00:01:50.690 --> 00:01:52.960
So this value for the standard change

00:01:52.960 --> 00:01:57.050
in free energy of formation
of carbon monoxide is talking

00:01:57.050 --> 00:02:00.220
about the change in free
energy for the formation

00:02:00.220 --> 00:02:03.830
of one mole of carbon
monoxide from its elements

00:02:03.830 --> 00:02:05.860
in their standard states.

00:02:05.860 --> 00:02:09.210
Standard changes in free energy
of formation of substances

00:02:09.210 --> 00:02:12.100
are useful because they
can be used to calculate

00:02:12.100 --> 00:02:15.940
the standard change in free
energy for a chemical reaction.

00:02:15.940 --> 00:02:18.920
The standard change in free
energy of a chemical reaction

00:02:18.920 --> 00:02:22.440
is equal to the sum of the standard change

00:02:22.440 --> 00:02:25.440
in free energy of formation
of the products and from that

00:02:25.440 --> 00:02:27.670
you subtract the sum
of the standard change

00:02:27.670 --> 00:02:30.950
in free energy of
formation of the reactants.

00:02:30.950 --> 00:02:32.300
For our reaction,

00:02:32.300 --> 00:02:34.770
let's look at these
synthesis of ammonia gas

00:02:34.770 --> 00:02:37.500
from nitrogen gas and hydrogen gas.

00:02:37.500 --> 00:02:40.560
Our goal is to calculate the
standard change in free energy

00:02:40.560 --> 00:02:44.540
for this reaction at 25 degrees Celsius.

00:02:44.540 --> 00:02:46.580
First, we need to think
about our products.

00:02:46.580 --> 00:02:47.740
For this reaction,

00:02:47.740 --> 00:02:50.270
we have only one product
and that's ammonia.

00:02:50.270 --> 00:02:52.410
Notice how there's a two as a coefficient

00:02:52.410 --> 00:02:55.220
in front of ammonia and
the balanced equation.

00:02:55.220 --> 00:02:57.460
So we would need to look
up the standard change

00:02:57.460 --> 00:03:00.270
in free energy of formation of ammonia

00:03:00.270 --> 00:03:02.170
at 25 degrees Celsius.

00:03:02.170 --> 00:03:04.880
And because this value
is per mole of ammonia,

00:03:04.880 --> 00:03:06.970
we would need to multiply it by two

00:03:06.970 --> 00:03:10.260
because we have two moles
in our balanced equation.

00:03:10.260 --> 00:03:12.140
Next, we think about our reactants.

00:03:12.140 --> 00:03:15.220
And we have two reactants,
nitrogen and hydrogen.

00:03:15.220 --> 00:03:18.110
And we need the sum of
their standard changes

00:03:18.110 --> 00:03:20.320
in free energy of formation.

00:03:20.320 --> 00:03:22.030
So we would write the standard change

00:03:22.030 --> 00:03:24.240
in free energy of formation of nitrogen.

00:03:24.240 --> 00:03:26.460
And because there's a one
in our balanced equation,

00:03:26.460 --> 00:03:29.760
we multiply that value by
one and two that we add

00:03:29.760 --> 00:03:31.930
the standard change in
free energy of formation

00:03:31.930 --> 00:03:33.090
of hydrogen gas.

00:03:33.090 --> 00:03:35.150
And since we have a three
in our balanced equation,

00:03:35.150 --> 00:03:37.200
we would multiply that by three.

00:03:37.200 --> 00:03:38.730
So going back to ammonia,

00:03:38.730 --> 00:03:40.970
the standard change in
free energy of formation

00:03:40.970 --> 00:03:44.520
is equal to negative
16.5 kilojoules per mole,

00:03:44.520 --> 00:03:48.340
multiply that by two moles
and the moles will cancel out.

00:03:48.340 --> 00:03:50.080
For nitrogen and hydrogen,

00:03:50.080 --> 00:03:52.800
those are both elements
in their standard states

00:03:52.800 --> 00:03:55.900
and therefore their standard
free energy of formations

00:03:55.900 --> 00:03:57.180
are equal to zero.

00:03:57.180 --> 00:03:59.820
So this is just zero plus zero.

00:03:59.820 --> 00:04:03.450
So the standard change in
free energy for this reaction

00:04:03.450 --> 00:04:08.450
is equal to negative 33.0
kilojoules at 25 degrees Celsius.

00:04:08.980 --> 00:04:12.150
Standard state for a gas
refers to the pure gas

00:04:12.150 --> 00:04:14.250
at a pressure of one atmosphere.

00:04:14.250 --> 00:04:16.890
So we think about all three of our gases

00:04:16.890 --> 00:04:18.960
at a partial pressure of one atmosphere

00:04:18.960 --> 00:04:22.260
at a temperature of 25 degrees Celsius.

00:04:22.260 --> 00:04:25.260
The standard change and free
energy for this reaction

00:04:25.260 --> 00:04:28.190
is negative 33.0 kilojoules.

00:04:28.190 --> 00:04:32.140
Since the change in
free energy is negative,

00:04:32.140 --> 00:04:35.400
that tells us the reaction as
thermodynamically favorable

00:04:35.400 --> 00:04:37.760
in the forward direction.

00:04:37.760 --> 00:04:40.750
So nitrogen and hydrogen
will react together

00:04:40.750 --> 00:04:43.110
to form more ammonia.

00:04:43.110 --> 00:04:45.370
Textbooks often give
the units as kilojoules

00:04:45.370 --> 00:04:46.970
for the standard change and free energy

00:04:46.970 --> 00:04:48.480
of a chemical reaction.

00:04:48.480 --> 00:04:49.590
However, sometimes you

00:04:49.590 --> 00:04:53.530
also see kilojoules per mole of reaction.

00:04:53.530 --> 00:04:55.440
Per mole of reaction simply refers

00:04:55.440 --> 00:04:57.550
to how the balanced equation is written.

00:04:57.550 --> 00:05:00.880
And here we have the formation
of two moles of ammonia.

00:05:00.880 --> 00:05:03.660
So we can write a conversion
factor of two moles

00:05:03.660 --> 00:05:07.300
of ammonia per one mole of reaction.

00:05:07.300 --> 00:05:09.880
For standard free energies of formation,

00:05:09.880 --> 00:05:12.700
the units are kilojoules
per mole of the substance.

00:05:12.700 --> 00:05:13.533
So in this case,

00:05:13.533 --> 00:05:17.280
it's negative 16.5 kilojoules
per mole of ammonia.

00:05:17.280 --> 00:05:19.840
So moles of ammonia would cancel out,

00:05:19.840 --> 00:05:24.423
which gives kilojoules per
mole of reaction for the units.