Worked example: Determining the effect of temperature on thermodynamic favorability

WEBVTT
Kind: captions
Language: en

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- [Lecturer] Let's do a work example

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where we calculate the standard change

00:00:03.240 --> 00:00:06.690
in free energy, delta G naught,
for a chemical reaction.

00:00:06.690 --> 00:00:07.830
And for our reaction,

00:00:07.830 --> 00:00:10.620
let's look at the synthesis of ammonia gas

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from nitrogen gas and hydrogen gas.

00:00:13.560 --> 00:00:15.180
At 25 degrees Celsius,

00:00:15.180 --> 00:00:17.370
delta H naught for this reaction is equal

00:00:17.370 --> 00:00:21.870
to negative 92.2 kilojoules
per mole of reaction.

00:00:21.870 --> 00:00:23.560
Since delta H naught is negative,

00:00:23.560 --> 00:00:25.850
this reaction is exothermic

00:00:25.850 --> 00:00:27.640
and at 25 degrees Celsius,

00:00:27.640 --> 00:00:29.620
delta S naught for this reaction is equal

00:00:29.620 --> 00:00:34.330
to negative 198.7 joules
per kelvin mole of reaction.

00:00:34.330 --> 00:00:36.580
Our goal is to calculate delta G naught

00:00:36.580 --> 00:00:39.530
for this reaction at 25 degrees Celsius.

00:00:39.530 --> 00:00:41.480
Remember if delta G naught is negative,

00:00:41.480 --> 00:00:44.500
the forward reaction is
thermodynamically favorable,

00:00:44.500 --> 00:00:46.500
but if delta G naught is positive,

00:00:46.500 --> 00:00:50.550
the forward reaction is
thermodynamically unfavorable.

00:00:50.550 --> 00:00:52.640
Since the change in enthalpy is negative,

00:00:52.640 --> 00:00:54.810
and the change in entropy is negative,

00:00:54.810 --> 00:00:56.790
whether or not delta G naught is negative

00:00:56.790 --> 00:00:58.810
depends on the temperature.

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To calculate the value for delta G naught,

00:01:00.790 --> 00:01:02.970
we're going to use the following equation.

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Delta G naught is equal to delta H naught

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minus T delta S naught,

00:01:07.710 --> 00:01:10.320
where T is the temperature in kelvin.

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Next, we need to plug
everything into our equation.

00:01:12.760 --> 00:01:14.690
So delta H nought is equal

00:01:14.690 --> 00:01:19.690
to negative 92.2 kilojoules
per mole of reaction.

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The temperature is 25 degrees Celsius.

00:01:22.800 --> 00:01:27.800
So 25 plus 273 is equal to 298 kelvin.

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So we're gonna plug 298
kelvin into our equation.

00:01:32.290 --> 00:01:35.280
And for delta S naught,
notice we have this in joules,

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whereas for delta H naught,
it was in kilojoules.

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So we need to convert delta S naught

00:01:40.140 --> 00:01:43.130
into kilojoules per
kelvin mole of reaction.

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One way to do that is just to move

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the decimal place three to the left,

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which gives us negative 0.1987 kilojoules

00:01:51.350 --> 00:01:53.520
per kelvin mole of reaction.

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So we're gonna plug all of
this in for delta S naught.

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Here's our equation with
everything plugged in.

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Notice that kelvin will cancel out,

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which gives us kilojoules per
mole of reaction as our units.

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And when we do the math,

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we find that delta G naught is equal

00:02:09.970 --> 00:02:14.940
to negative 33.0 kilojoules
per mole of reaction.

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Because delta G naught is negative,

00:02:16.770 --> 00:02:19.850
that means the reaction is
thermodynamically favorable

00:02:19.850 --> 00:02:22.110
in the forward direction.

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And the superscript naught
means that both the reactants

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and products are in their standard states.

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So what our calculation tells
us is if we had a mixture

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of nitrogen, hydrogen, and ammonia gas

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at 25 degrees Celsius,

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and all three gases had a partial pressure

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of one atmosphere,

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the reaction is
thermodynamically favorable

00:02:44.400 --> 00:02:45.790
in the forward direction,

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meaning nitrogen gas and
hydrogen gas would come together

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to synthesize or make more ammonia.

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So at 298 kelvin, delta G naught

00:02:56.130 --> 00:02:58.870
for this reaction is negative.

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Let's do another calculation
for the same reaction

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at a different temperature,

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a temperature of 1,000 kelvin.

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Also, even though the
value for delta G naught

00:03:08.880 --> 00:03:10.940
is highly dependent on temperature,

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the values for delta H
naught and delta S naught

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are not as dependent on the temperature.

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Therefore, we're gonna
assume that these values

00:03:19.690 --> 00:03:23.190
for delta H naught and
delta S naught don't change,

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and we're gonna use the same ones

00:03:24.880 --> 00:03:28.140
at the higher temperature of 1,000 kelvin.

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Here's the equation with
everything plugged in

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for the calculation of delta G naught

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at the higher temperature of 1,000 kelvin.

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And notice that the values
we're using for delta H naught

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and delta S naught are
the same as the ones

00:03:40.970 --> 00:03:45.540
that we used at the lower
temperature of 298 kelvin.

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Once again, kelvin cancels out

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and gives us kilojoules
per mole of reaction

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as our units for our final answer.

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And after we do the math,

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we find that delta G naught is equal

00:03:55.700 --> 00:04:00.350
to positive 106.5 kilojoules
per mole of reaction.

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Since delta G naught is positive,

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that means the forward reaction

00:04:04.460 --> 00:04:06.810
is thermodynamically unfavorable,

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which means the reverse reaction

00:04:08.970 --> 00:04:11.000
is thermodynamically favorable.

00:04:11.000 --> 00:04:14.310
So if we had a mixture of
nitrogen gas, hydrogen gas,

00:04:14.310 --> 00:04:17.780
and ammonia gas at a
temperature of 1,000 kelvin,

00:04:17.780 --> 00:04:20.460
and all three gases had a partial pressure

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of one atmosphere,

00:04:22.140 --> 00:04:25.470
the ammonia gas would
turn into nitrogen gas

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and hydrogen gas.

00:04:26.950 --> 00:04:29.560
And while we have both
calculations on the screen,

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let's analyze why one calculation
gives a negative value

00:04:32.830 --> 00:04:33.860
for delta G naught,

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and the other one gives a positive value.

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For the calculation at 298 kelvin,

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the temperature was low enough

00:04:42.460 --> 00:04:45.570
that the entropy term didn't overwhelm

00:04:45.570 --> 00:04:47.970
the negative value for the enthalpy term,

00:04:47.970 --> 00:04:50.700
and we ended up with a
negative overall value

00:04:50.700 --> 00:04:52.600
for delta G naught.

00:04:52.600 --> 00:04:55.720
However, for the
calculation at 1,000 kelvin,

00:04:55.720 --> 00:04:58.230
this time the temperature was high enough

00:04:58.230 --> 00:05:01.290
where the entropy term outweighed

00:05:01.290 --> 00:05:04.050
the negative value for the enthalpy term,

00:05:04.050 --> 00:05:06.060
and therefore the overall value

00:05:06.060 --> 00:05:09.640
for delta G naught
ended up being positive.

00:05:09.640 --> 00:05:12.490
We've just seen that for
the synthesis of ammonia

00:05:12.490 --> 00:05:15.060
from nitrogen gas and hydrogen gas

00:05:15.060 --> 00:05:19.470
at relatively low
temperatures like 298 kelvin,

00:05:19.470 --> 00:05:22.260
delta G naught is less than zero,

00:05:22.260 --> 00:05:26.040
and at relatively high
temperatures like 1,000 kelvin,

00:05:26.040 --> 00:05:28.800
delta G naught is greater than zero.

00:05:28.800 --> 00:05:30.580
So there must be some temperature

00:05:30.580 --> 00:05:34.410
between 298 kelvin and 1,000 kelvin

00:05:34.410 --> 00:05:36.380
where the forward reaction goes

00:05:36.380 --> 00:05:38.760
from being thermodynamically favorable

00:05:38.760 --> 00:05:41.500
to thermodynamically unfavorable.

00:05:41.500 --> 00:05:45.385
And this crossover point
occurs when the enthalpy term

00:05:45.385 --> 00:05:49.220
and the entropy term
are perfectly balanced.

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Therefore, delta G naught is equal to zero

00:05:52.700 --> 00:05:54.870
at this crossover point.

00:05:54.870 --> 00:05:57.490
So we plug in zero for delta G naught

00:05:57.490 --> 00:06:00.890
and we solve for the temperature
at the crossover point.

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After some algebra,

00:06:02.160 --> 00:06:04.060
we find that the temperature is equal

00:06:04.060 --> 00:06:07.330
to delta H naught divided
by delta S naught,

00:06:07.330 --> 00:06:10.660
which is equal to negative 92.2 kilojoules

00:06:10.660 --> 00:06:15.660
per mole of reaction divided
by negative 0.1987 kilojoules

00:06:15.960 --> 00:06:18.380
per kelvin mole of reaction.

00:06:18.380 --> 00:06:21.230
Kilojoules per mole of
reaction cancels out

00:06:21.230 --> 00:06:26.230
and gives us 464 kelvin for
the crossover temperature.

00:06:26.690 --> 00:06:28.400
So for this particular reaction,

00:06:28.400 --> 00:06:32.890
delta G naught is equal
to zero at 464 kelvin.

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At temperatures less than 464 kelvin,

00:06:35.490 --> 00:06:39.020
the forward reaction is
thermodynamically favorable.

00:06:39.020 --> 00:06:42.620
However, at temperatures
higher than 464 kelvin,

00:06:42.620 --> 00:06:46.153
the forward reaction is
thermodynamically unfavorable.


Thermodynamic favorability and temperature

WEBVTT
Kind: captions
Language: en

00:00:00.230 --> 00:00:02.360
- [Instructor] The
thermodynamic favorability

00:00:02.360 --> 00:00:03.590
of a chemical reaction

00:00:03.590 --> 00:00:05.690
can be affected by the temperature.

00:00:05.690 --> 00:00:07.580
Let's say we have a generic reaction

00:00:07.580 --> 00:00:09.910
where the reactants
turn into the products.

00:00:09.910 --> 00:00:13.060
As a quick review, when the
standard change in free energy,

00:00:13.060 --> 00:00:14.070
delta G naught,

00:00:14.070 --> 00:00:15.690
is less than zero,

00:00:15.690 --> 00:00:19.740
the forward reaction is
thermodynamically favorable.

00:00:19.740 --> 00:00:22.090
Therefore, the net
reaction goes to the right

00:00:22.090 --> 00:00:24.110
to make more of the products.

00:00:24.110 --> 00:00:26.440
And when delta G naught
is greater than zero,

00:00:26.440 --> 00:00:30.230
the forward reaction is
thermodynamically unfavorable.

00:00:30.230 --> 00:00:32.930
That means that the reverse
reaction is favorable

00:00:32.930 --> 00:00:34.830
and the net reaction would go to the left

00:00:34.830 --> 00:00:37.150
to make more of the reactants.

00:00:37.150 --> 00:00:38.990
One way to calculate delta G naught

00:00:38.990 --> 00:00:40.890
is to use the following equation.

00:00:40.890 --> 00:00:43.410
Delta G naught is equal to delta H naught

00:00:43.410 --> 00:00:47.590
minus the absolute temperature
times delta S naught.

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And since delta H naught

00:00:48.870 --> 00:00:51.870
is talking about the
standard change enthalpy,

00:00:51.870 --> 00:00:53.050
for this equation

00:00:53.050 --> 00:00:56.240
delta H naught will be
called the enthalpy term.

00:00:56.240 --> 00:00:57.580
And because delta S naught

00:00:57.580 --> 00:01:00.340
is talking about the
standard change in entropy,

00:01:00.340 --> 00:01:02.780
the absolute temperature
times delta S naught

00:01:02.780 --> 00:01:06.590
will be referred to as the entropy term.

00:01:06.590 --> 00:01:08.050
So when we're trying to figure out

00:01:08.050 --> 00:01:09.610
if delta G naught is negative

00:01:09.610 --> 00:01:11.730
or if delta G naught is positive,

00:01:11.730 --> 00:01:15.210
we need to consider the
sign for delta H naught,

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the sign for delta S naught,

00:01:17.750 --> 00:01:20.020
and sometimes we also need to consider

00:01:20.020 --> 00:01:22.020
what the temperature is.

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There are four possible
combinations of signs

00:01:24.770 --> 00:01:27.190
for delta H naught and delta S naught

00:01:27.190 --> 00:01:28.970
as we can see in this table.

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For each of these four combinations,

00:01:30.750 --> 00:01:32.260
we're gonna think about the temperatures

00:01:32.260 --> 00:01:34.930
at which delta G naught is less than zero

00:01:34.930 --> 00:01:37.930
and we're gonna finish
filling out this table.

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To start with, we're gonna
think about the situation

00:01:40.180 --> 00:01:42.270
where delta H naught is negative

00:01:42.270 --> 00:01:45.060
and delta S naught is positive.

00:01:45.060 --> 00:01:47.710
So for this first possible combination,

00:01:47.710 --> 00:01:50.180
when delta H naught is negative,

00:01:50.180 --> 00:01:52.750
we say that's an exothermic reaction.

00:01:52.750 --> 00:01:55.620
And when delta S naught is positive,

00:01:55.620 --> 00:01:57.780
that's saying there's
an increase in entropy

00:01:57.780 --> 00:02:00.310
as reactants turn into products.

00:02:00.310 --> 00:02:02.660
So let's think about the
sign for delta G naught

00:02:02.660 --> 00:02:05.380
if we plug in some numbers
for delta H naught,

00:02:05.380 --> 00:02:08.070
temperature, and delta S naught.

00:02:08.070 --> 00:02:09.460
And when I plug in some numbers,

00:02:09.460 --> 00:02:11.940
I'm not really concerned
about a particular reaction.

00:02:11.940 --> 00:02:13.540
I'm not concerned about units.

00:02:13.540 --> 00:02:15.920
I'm not concerned about
significant figures.

00:02:15.920 --> 00:02:17.930
I'm just interested in
looking at the numbers

00:02:17.930 --> 00:02:22.210
and seeing how the math affects
the sign for delta G naught.

00:02:22.210 --> 00:02:24.760
So for delta H naught,
I'm saying that's -100.

00:02:26.508 --> 00:02:30.050
For T, I'm saying that's 100,

00:02:30.050 --> 00:02:32.423
and for delta S naught,
I'm saying that's +0.1.

00:02:36.039 --> 00:02:37.307
So this would be equal to -100 - 10,

00:02:40.996 --> 00:02:43.550
which is equal to -110.

00:02:43.550 --> 00:02:45.930
So delta G naught is negative.

00:02:45.930 --> 00:02:48.940
Notice how a negative entropy term

00:02:48.940 --> 00:02:53.180
favors getting a negative
value for delta G naught.

00:02:53.180 --> 00:02:55.910
And notice how a positive entropy term,

00:02:55.910 --> 00:02:59.080
since we're subtracting
that in our equation

00:02:59.080 --> 00:03:02.070
is also favorable for
getting a negative value

00:03:02.070 --> 00:03:03.770
for delta G naught.

00:03:03.770 --> 00:03:06.670
Also notice, no matter what
we put in for the temperature

00:03:06.670 --> 00:03:09.400
since delta S naught is positive,

00:03:09.400 --> 00:03:12.180
the entropy term will always be positive.

00:03:12.180 --> 00:03:15.010
And since we're subtracting
it in our equation,

00:03:15.010 --> 00:03:18.990
we would always get a negative
value for delta G naught.

00:03:18.990 --> 00:03:22.492
So whenever delta H naught is negative

00:03:22.492 --> 00:03:25.200
and when delta S naught is positive,

00:03:25.200 --> 00:03:26.750
delta G naught will be negative

00:03:26.750 --> 00:03:29.040
no matter what the temperature.

00:03:29.040 --> 00:03:30.630
So if we go back to our chart

00:03:30.630 --> 00:03:32.800
for our first possible combination,

00:03:32.800 --> 00:03:35.690
when delta H naught is
negative, that's favorable.

00:03:35.690 --> 00:03:39.510
And when delta S naught as
positive, that's also favorable.

00:03:39.510 --> 00:03:44.090
Therefore delta G naught is
less than 0 at all temperatures.

00:03:44.090 --> 00:03:45.367
So I'm gonna go ahead and write in here,

00:03:45.367 --> 00:03:47.930
'at all temperatures' in our chart.

00:03:47.930 --> 00:03:49.000
As an example of this,

00:03:49.000 --> 00:03:53.710
let's consider the conversion
of ozone gas into oxygen gas.

00:03:53.710 --> 00:03:55.360
Since energy is released,

00:03:55.360 --> 00:03:57.780
this is an exothermic reaction,

00:03:57.780 --> 00:04:00.470
and delta H naught is negative.

00:04:00.470 --> 00:04:02.600
And since we're going from 2 moles of gas

00:04:02.600 --> 00:04:03.700
on the reactant side

00:04:03.700 --> 00:04:06.450
to 3 moles of gas on the product side,

00:04:06.450 --> 00:04:08.460
that's an increase in entropy.

00:04:08.460 --> 00:04:10.830
So delta S naught is positive.

00:04:10.830 --> 00:04:12.640
Since delta H naught is negative

00:04:12.640 --> 00:04:14.720
and delta S naught is positive,

00:04:14.720 --> 00:04:16.320
delta G naught for this reaction

00:04:16.320 --> 00:04:19.080
is less than zero at all temperatures.

00:04:19.080 --> 00:04:21.580
That means the forward reaction

00:04:21.580 --> 00:04:24.150
is thermodynamically favorable,

00:04:24.150 --> 00:04:28.170
and ozone gas would turn into oxygen gas,

00:04:28.170 --> 00:04:31.160
and this reaction would
occur at all temperatures.

00:04:31.160 --> 00:04:34.030
The second possible
combination for the signs

00:04:34.030 --> 00:04:36.710
is when delta H naught is positive

00:04:36.710 --> 00:04:39.380
and when delta S naught is negative.

00:04:39.380 --> 00:04:41.240
When delta H naught is positive,

00:04:41.240 --> 00:04:43.470
that's an endothermic reaction.

00:04:43.470 --> 00:04:45.450
And when delta S naught is negative,

00:04:45.450 --> 00:04:47.210
there's a decrease in entropy

00:04:47.210 --> 00:04:49.610
going from reactants to products.

00:04:49.610 --> 00:04:51.710
Once again we're gonna
plug in some numbers

00:04:51.710 --> 00:04:53.780
and solve for delta G naught.

00:04:53.780 --> 00:04:55.290
So for delta H naught,

00:04:55.290 --> 00:04:58.070
let's say that's +100,

00:04:58.070 --> 00:05:01.760
for the temperature term,
let's say that's 100,

00:05:01.760 --> 00:05:05.133
and delta S naught, let's say that's -0.1.

00:05:06.638 --> 00:05:07.733
When we do the math,

00:05:10.162 --> 00:05:11.862
we get positive 100 - negative 10,

00:05:15.345 --> 00:05:18.030
which is equal to +110.

00:05:18.030 --> 00:05:20.520
So delta G naught is positive.

00:05:20.520 --> 00:05:23.370
Notice when the entropy term is positive,

00:05:23.370 --> 00:05:26.120
that's not favorable for
getting a negative value

00:05:26.120 --> 00:05:27.780
for delta G naught.

00:05:27.780 --> 00:05:31.000
And when the entropy term is negative,

00:05:31.000 --> 00:05:34.060
since we're subtracting this negative,

00:05:34.060 --> 00:05:37.110
we would actually be
adding the entropy term.

00:05:37.110 --> 00:05:38.990
And that would not favor

00:05:38.990 --> 00:05:41.840
a negative value for delta G naught.

00:05:41.840 --> 00:05:42.673
Also notice,

00:05:42.673 --> 00:05:45.030
no matter what you put in
for the temperature here,

00:05:45.030 --> 00:05:46.900
you're always gonna get a negative value

00:05:46.900 --> 00:05:48.640
for the entropy term.

00:05:48.640 --> 00:05:51.040
And if the enthalpy term is positive,

00:05:51.040 --> 00:05:54.260
since we would mathematically
be adding the entropy term,

00:05:54.260 --> 00:05:57.880
we would always get a positive
value for delta G naught.

00:05:57.880 --> 00:06:00.040
So whenever delta H naught is positive

00:06:00.040 --> 00:06:02.660
and delta S naught is negative,

00:06:02.660 --> 00:06:05.240
delta G naught will be positive

00:06:05.240 --> 00:06:07.330
no matter what the temperature.

00:06:07.330 --> 00:06:10.930
So going back to our charge
when delta H naught is positive,

00:06:10.930 --> 00:06:12.430
that was not favorable,

00:06:12.430 --> 00:06:14.820
and when delta S naught was negative,

00:06:14.820 --> 00:06:16.940
that was also not favorable.

00:06:16.940 --> 00:06:18.700
So it doesn't matter
what the temperature is.

00:06:18.700 --> 00:06:21.440
Delta G naught will not be less than zero.

00:06:21.440 --> 00:06:24.610
Therefore, on our chart
here we can write in,

00:06:24.610 --> 00:06:28.730
at no temperature will delta
G naught be less than zero.

00:06:28.730 --> 00:06:29.810
As an example of this,

00:06:29.810 --> 00:06:32.530
let's look at the reverse
of the reaction we saw

00:06:32.530 --> 00:06:34.820
for the first possible combination.

00:06:34.820 --> 00:06:37.700
Before we had ozone
turning into oxygen gas,

00:06:37.700 --> 00:06:41.170
and now we have oxygen
gas turning into ozone.

00:06:41.170 --> 00:06:43.010
Because heat is on the reactant side,

00:06:43.010 --> 00:06:44.980
we know this reaction is endothermic

00:06:44.980 --> 00:06:47.240
and delta H naught is positive.

00:06:47.240 --> 00:06:49.360
And since we're going from 3 moles of gas

00:06:49.360 --> 00:06:50.440
on the reactant side

00:06:50.440 --> 00:06:53.120
to 2 moles of gas on the product side,

00:06:53.120 --> 00:06:55.270
that's a decrease in entropy,

00:06:55.270 --> 00:06:58.190
and delta S naught is negative.

00:06:58.190 --> 00:07:00.180
Since delta H naught is positive

00:07:00.180 --> 00:07:02.640
and delta S naught is negative,

00:07:02.640 --> 00:07:05.150
delta G naught will
never be less than zero.

00:07:05.150 --> 00:07:07.470
It doesn't matter what the temperature is.

00:07:07.470 --> 00:07:09.150
And so the forward reaction

00:07:09.150 --> 00:07:12.770
is always thermodynamically unfavorable.

00:07:12.770 --> 00:07:14.880
The third possible combination of signs

00:07:14.880 --> 00:07:17.930
has both delta H naught and delta S naught

00:07:17.930 --> 00:07:19.430
as being negative.

00:07:19.430 --> 00:07:22.080
Once again we're gonna plug
some numbers into the equation

00:07:22.080 --> 00:07:24.750
and see what happens to
the sign of delta G naught.

00:07:24.750 --> 00:07:27.890
So let's say delta H naught is -100.

00:07:27.890 --> 00:07:31.810
The temperature is 100 and
delta S naught is -0.1.

00:07:32.870 --> 00:07:36.630
Doing the math, we get
negative 100 minus negative 10.

00:07:36.630 --> 00:07:38.640
So that's -100 + 10,

00:07:38.640 --> 00:07:40.127
which is equal to -90.

00:07:40.970 --> 00:07:43.260
Let's think about why
we got a negative value

00:07:43.260 --> 00:07:44.920
for delta G naught.

00:07:44.920 --> 00:07:46.720
The enthalpy term is negative,

00:07:46.720 --> 00:07:50.060
which favors a negative
value for delta G naught.

00:07:50.060 --> 00:07:52.590
The entropy term is also negative,

00:07:52.590 --> 00:07:54.870
but since we're subtracting
the entropy term,

00:07:54.870 --> 00:07:56.210
it actually gets added

00:07:56.210 --> 00:07:59.000
for the overall delta
G naught calculation.

00:07:59.000 --> 00:08:01.850
So the entropy term is not favorable.

00:08:01.850 --> 00:08:03.260
However, in this case,

00:08:03.260 --> 00:08:05.460
the favorable entropy term

00:08:05.460 --> 00:08:08.900
outweighs the unfavorable entropy term.

00:08:08.900 --> 00:08:12.050
And that's the reason why
we get a negative value

00:08:12.050 --> 00:08:15.300
for delta G naught at this temperature.

00:08:15.300 --> 00:08:18.110
Let's compare the
calculation that we just did

00:08:18.110 --> 00:08:19.900
at a temperature of 100,

00:08:19.900 --> 00:08:22.720
which gave us a negative
value for delta G naught,

00:08:22.720 --> 00:08:24.280
to a similar calculation,

00:08:24.280 --> 00:08:26.620
except this time we have
a much higher temperature.

00:08:26.620 --> 00:08:28.450
This time it's 2000.

00:08:28.450 --> 00:08:30.870
But notice how the
values for delta H naught

00:08:30.870 --> 00:08:33.960
and delta S naught are the same as before.

00:08:33.960 --> 00:08:35.950
The reason why we can use the same values

00:08:35.950 --> 00:08:37.340
at a different temperature

00:08:37.340 --> 00:08:39.620
is because delta H and delta S naught

00:08:39.620 --> 00:08:42.100
don't change much with temperature.

00:08:42.100 --> 00:08:45.610
However, delta G naught does
change a lot with temperature.

00:08:45.610 --> 00:08:48.470
Because at this higher
temperature, when we do the math,

00:08:48.470 --> 00:08:52.940
we see that now we get delta
G naught is equal to +100.

00:08:52.940 --> 00:08:54.540
So at this higher temperature,

00:08:54.540 --> 00:08:57.670
even though the enthalpy
term is favorable,

00:08:57.670 --> 00:09:00.470
now the unfavorable entropy term

00:09:00.470 --> 00:09:03.550
outweighs the favorable enthalpy term

00:09:03.550 --> 00:09:07.410
and gives us a positive
value for delta G naught.

00:09:07.410 --> 00:09:09.960
Therefore, at a relatively
lower temperature,

00:09:09.960 --> 00:09:12.190
delta G naught is negative.

00:09:12.190 --> 00:09:14.180
However, at a higher temperature,

00:09:14.180 --> 00:09:16.900
delta G naught is positive.

00:09:16.900 --> 00:09:20.200
So when delta H naught is
negative, that's favorable,

00:09:20.200 --> 00:09:24.010
but when delta S naught is
negative, that's not favorable.

00:09:24.010 --> 00:09:25.800
And we just saw from the calculations

00:09:25.800 --> 00:09:29.730
that delta G naught will be
negative at low temperatures.

00:09:29.730 --> 00:09:32.850
So let's write that in here in our chart.

00:09:32.850 --> 00:09:35.130
So far we've been talking
about delta G naught

00:09:35.130 --> 00:09:37.840
relative to a chemical process.

00:09:37.840 --> 00:09:42.120
However, delta G naught also
applies to a physical process.

00:09:42.120 --> 00:09:42.953
In this case

00:09:42.953 --> 00:09:46.070
we're looking at liquid water
turning into solid water.

00:09:46.070 --> 00:09:48.740
So this is the process of freezing.

00:09:48.740 --> 00:09:51.320
The process of freezing gives off energy.

00:09:51.320 --> 00:09:53.320
So this process is exothermic,

00:09:53.320 --> 00:09:55.670
and delta H naught will be negative.

00:09:55.670 --> 00:09:59.780
Going from a liquid to a solid
is a decrease in entropy,

00:09:59.780 --> 00:10:02.500
therefore, delta S naught is negative.

00:10:02.500 --> 00:10:06.800
For this combination of signs,
delta G naught is negative

00:10:06.800 --> 00:10:09.500
only at relatively low temperatures.

00:10:09.500 --> 00:10:12.860
Therefore, the forward process of freezing

00:10:12.860 --> 00:10:16.340
is thermodynamically
favorable at low temperatures.

00:10:16.340 --> 00:10:18.790
And this matches with what
we already know about water,

00:10:18.790 --> 00:10:22.200
water freezes at a
relatively low temperature.

00:10:22.200 --> 00:10:25.320
Finally, let's look at our
fourth combination of signs,

00:10:25.320 --> 00:10:27.530
and that's when both delta H naught

00:10:27.530 --> 00:10:30.010
and delta S naught are positive.

00:10:30.010 --> 00:10:33.360
So let's say delta H
naught is equal to +100,

00:10:33.360 --> 00:10:35.990
delta S naught is equal able to +0.1,

00:10:35.990 --> 00:10:39.800
and first we'll look at the
lower temperature of 100.

00:10:39.800 --> 00:10:42.600
Doing the math, we get positive 100 - 10,

00:10:42.600 --> 00:10:46.080
which gives us a positive
value for delta G naught.

00:10:46.080 --> 00:10:47.400
And if we do the calculation

00:10:47.400 --> 00:10:49.580
at a higher temperature of 2000,

00:10:49.580 --> 00:10:52.180
notice that delta H naught is still +100

00:10:52.180 --> 00:10:54.043
and delta S not is still +0.1.

00:10:55.340 --> 00:10:58.680
This time we get -100 for delta G naught.

00:10:58.680 --> 00:11:01.640
Notice how for this example
at the higher temperature,

00:11:01.640 --> 00:11:03.710
the entropy term is unfavorable

00:11:03.710 --> 00:11:06.540
for giving a negative
value for delta G naught.

00:11:06.540 --> 00:11:09.600
However, the entropy term is favorable.

00:11:09.600 --> 00:11:11.810
A positive value for delta S naught

00:11:11.810 --> 00:11:14.460
gives a positive value
for the entropy term.

00:11:14.460 --> 00:11:16.500
And since we are subtracting it,

00:11:16.500 --> 00:11:20.040
the entropy term outweighs
the enthalpy term,

00:11:20.040 --> 00:11:23.020
giving a negative value
for delta G naught.

00:11:23.020 --> 00:11:26.080
However, for the first example
at the lower temperature,

00:11:26.080 --> 00:11:27.780
since the temperature is lower,

00:11:27.780 --> 00:11:30.450
the entropy term is a smaller value,

00:11:30.450 --> 00:11:35.450
and is unable to overcome the
unfavorable enthalpy term.

00:11:35.890 --> 00:11:37.800
And that's the reason why delta G naught

00:11:37.800 --> 00:11:41.110
ends up being positive
at the lower temperature.

00:11:41.110 --> 00:11:43.020
So for this combination of signs,

00:11:43.020 --> 00:11:46.560
delta G naught is negative
at higher temperatures.

00:11:46.560 --> 00:11:48.400
So for our fourth combination,

00:11:48.400 --> 00:11:50.560
when delta H naught is positive,

00:11:50.560 --> 00:11:52.250
that is unfavorable.

00:11:52.250 --> 00:11:56.460
However, when delta S naught
is positive, that is favorable.

00:11:56.460 --> 00:12:00.840
And we just saw that delta
G naught is less than zero

00:12:00.840 --> 00:12:02.170
at high temperatures.

00:12:02.170 --> 00:12:05.217
So I'll go ahead and write
in on our table here,

00:12:05.217 --> 00:12:07.420
'at high temperatures.'

00:12:07.420 --> 00:12:08.570
As an example of this,

00:12:08.570 --> 00:12:12.490
let's consider solid water
turning into liquid water.

00:12:12.490 --> 00:12:16.280
So the forward process is
the process of melting.

00:12:16.280 --> 00:12:17.170
It takes energy

00:12:17.170 --> 00:12:20.030
to disrupt the intermolecular
forces of the solid,

00:12:20.030 --> 00:12:22.140
therefore, heat is on the reactant side

00:12:22.140 --> 00:12:24.830
and delta H naught is positive.

00:12:24.830 --> 00:12:27.820
Melting is an endothermic process.

00:12:27.820 --> 00:12:32.250
Going from a solid to a liquid
is an increase in entropy.

00:12:32.250 --> 00:12:35.300
Therefore, delta S naught is positive.

00:12:35.300 --> 00:12:37.130
And when both signs are positive,

00:12:37.130 --> 00:12:38.930
our calculations showed us

00:12:38.930 --> 00:12:43.630
that delta G naught is less
than zero at high temperatures.

00:12:43.630 --> 00:12:46.870
Therefore, the forward process of melting

00:12:46.870 --> 00:12:49.100
is thermodynamically favorable

00:12:49.100 --> 00:12:51.380
at relatively high temperatures.

00:12:51.380 --> 00:12:53.780
And this matches what we
already know from experience.

00:12:53.780 --> 00:12:57.050
At relatively low temperatures
ice does not melt.

00:12:57.050 --> 00:13:00.940
However, at relatively high
temperatures ice does melt.

00:13:00.940 --> 00:13:02.490
Now that we have our chart filled out,

00:13:02.490 --> 00:13:04.070
let's summarize what we've learned

00:13:04.070 --> 00:13:06.490
about these four possible combinations.

00:13:06.490 --> 00:13:10.350
When delta H naught and delta
S naught are both favorable,

00:13:10.350 --> 00:13:13.660
the forward process is
thermodynamically favorable

00:13:13.660 --> 00:13:15.450
at all temperatures.

00:13:15.450 --> 00:13:19.670
When delta H naught and delta
S naught are both unfavorable,

00:13:19.670 --> 00:13:23.170
the forward process is
thermodynamically favorable

00:13:23.170 --> 00:13:24.760
at no temperatures,

00:13:24.760 --> 00:13:26.750
or you could say the forward process

00:13:26.750 --> 00:13:30.690
is thermodynamically
unfavorable at all temperatures.

00:13:30.690 --> 00:13:32.980
When delta H naught is favorable

00:13:32.980 --> 00:13:35.920
and delta S naught is unfavorable,

00:13:35.920 --> 00:13:39.970
the forward process is only
thermodynamically favorable

00:13:39.970 --> 00:13:42.500
at low temperatures.

00:13:42.500 --> 00:13:45.540
And finally, if delta H
naught is unfavorable,

00:13:45.540 --> 00:13:47.930
but delta S naught is favorable,

00:13:47.930 --> 00:13:51.370
the forward process is
thermodynamically favorable

00:13:51.370 --> 00:13:53.553
only at high temperatures.


Introduction to Gibbs free energy

WEBVTT
Kind: captions
Language: en

00:00:00.510 --> 00:00:03.340
- [Instructor] Gibbs Free
Energy is symbolized by G

00:00:03.340 --> 00:00:08.020
and change in Gibbs Free Energy
is symbolized by delta G.

00:00:08.020 --> 00:00:09.970
And the change in free energy, delta G,

00:00:09.970 --> 00:00:12.450
is equal to the change
in enthalpy, delta H,

00:00:12.450 --> 00:00:14.470
minus the temperature in Kelvin,

00:00:14.470 --> 00:00:18.120
times the change in entropy, delta S.

00:00:18.120 --> 00:00:20.140
When delta G is less than zero,

00:00:20.140 --> 00:00:22.880
a chemical or physical process is favored

00:00:22.880 --> 00:00:24.780
in the forward direction.

00:00:24.780 --> 00:00:27.520
Therefore, we say that the forward process

00:00:27.520 --> 00:00:30.390
is thermodynamically favored.

00:00:30.390 --> 00:00:32.740
As an example, if we look at a reaction

00:00:32.740 --> 00:00:34.940
where reactants turn into products,

00:00:34.940 --> 00:00:37.270
if delta G is less than zero,

00:00:37.270 --> 00:00:40.600
the forward reaction as
thermodynamically favored,

00:00:40.600 --> 00:00:42.960
meaning the reaction will go to the right

00:00:42.960 --> 00:00:44.780
to make more products.

00:00:44.780 --> 00:00:47.250
Textbooks will often use
the word spontaneous.

00:00:47.250 --> 00:00:50.170
So, when delta G is less than zero,

00:00:50.170 --> 00:00:54.220
the reaction would be spontaneous
in the forward direction.

00:00:54.220 --> 00:00:56.620
When delta G is greater than zero,

00:00:56.620 --> 00:00:58.550
the chemical or physical process

00:00:58.550 --> 00:01:01.130
is favored in the reverse direction.

00:01:01.130 --> 00:01:03.410
Therefore, the forward process

00:01:03.410 --> 00:01:06.490
is not thermodynamically favored.

00:01:06.490 --> 00:01:09.460
Going back to our reaction, as an example,

00:01:09.460 --> 00:01:12.390
if delta G is greater than zero,

00:01:12.390 --> 00:01:15.500
that means the reverse
reaction is favored,

00:01:15.500 --> 00:01:18.880
which favors the formation
of the reactants.

00:01:18.880 --> 00:01:19.760
For this example,

00:01:19.760 --> 00:01:22.210
textbooks will often say that the reaction

00:01:22.210 --> 00:01:25.430
is non-spontaneous in
the forward direction,

00:01:25.430 --> 00:01:27.470
which means the reaction is spontaneous

00:01:27.470 --> 00:01:29.620
in the reverse direction.

00:01:29.620 --> 00:01:31.520
And when delta G is equal to zero,

00:01:31.520 --> 00:01:35.400
the chemical or physical
process is at equilibrium.

00:01:35.400 --> 00:01:39.370
So, for our chemical reaction,
if delta G is equal to zero,

00:01:39.370 --> 00:01:42.150
the reaction is at equilibrium
and the concentration

00:01:42.150 --> 00:01:45.660
of reactants and products
will remain constant.

00:01:45.660 --> 00:01:48.210
When we see delta G with
a superscript naught,

00:01:48.210 --> 00:01:51.650
we're talking about the
change in Gibbs Free Energy

00:01:51.650 --> 00:01:54.770
when substances are in
their standard states.

00:01:54.770 --> 00:01:58.630
By convention, the standard
state of a solid or liquid

00:01:58.630 --> 00:02:01.960
is referring to the pure
solid or the pure liquid

00:02:01.960 --> 00:02:04.810
under a pressure of one atmosphere.

00:02:04.810 --> 00:02:08.010
The standard state of a gas
is referring to the pure gas

00:02:08.010 --> 00:02:10.000
at a pressure of one atmosphere.

00:02:10.000 --> 00:02:11.920
And the standard state of a solution

00:02:11.920 --> 00:02:14.990
is talking about a one
Mueller concentration.

00:02:14.990 --> 00:02:17.010
If our substances are
in the standard state,

00:02:17.010 --> 00:02:20.970
we can add a superscript to the
equation that we saw before.

00:02:20.970 --> 00:02:23.140
So, we could calculate delta G naught,

00:02:23.140 --> 00:02:25.360
the standard change in free energy

00:02:25.360 --> 00:02:28.830
by getting the standard change
in enthalpy and from that,

00:02:28.830 --> 00:02:31.980
subtracting the absolute
temperature in Kelvin,

00:02:31.980 --> 00:02:35.270
times the standard change in entropy.

00:02:35.270 --> 00:02:38.270
When the substances are
in their standard states,

00:02:38.270 --> 00:02:41.830
delta G naught is equal to delta G.

00:02:41.830 --> 00:02:44.720
Therefore, we can say
that if delta G naught

00:02:44.720 --> 00:02:46.340
is less than zero,

00:02:46.340 --> 00:02:47.910
if we're talking about a reaction,

00:02:47.910 --> 00:02:50.140
the reaction is thermodynamically favored

00:02:50.140 --> 00:02:52.330
in the forward direction.

00:02:52.330 --> 00:02:55.100
And if delta G naught
is greater than zero,

00:02:55.100 --> 00:02:58.580
we could say the reaction is
not thermodynamically favored

00:02:58.580 --> 00:03:00.450
in the forward direction.

00:03:00.450 --> 00:03:02.540
Next, let's calculate delta G naught

00:03:02.540 --> 00:03:04.260
for a chemical reaction.

00:03:04.260 --> 00:03:05.410
And for our reaction,

00:03:05.410 --> 00:03:08.300
let's look at the synthesis
of hydrogen fluoride gas

00:03:08.300 --> 00:03:11.480
from hydrogen gas and fluorine gas.

00:03:11.480 --> 00:03:13.730
Our goal is to calculate delta G naught

00:03:13.730 --> 00:03:17.170
for this reaction at 25 degrees Celsius.

00:03:17.170 --> 00:03:20.640
Delta H naught for this
reaction at 25 degrees Celsius

00:03:20.640 --> 00:03:25.640
is equal to negative 537.2
kilojoules per mole of reaction.

00:03:26.470 --> 00:03:30.000
And delta S naught for this
reaction at 25 degrees Celsius

00:03:30.000 --> 00:03:34.680
is equal to 13.7 joules per
Kelvin mole of reaction.

00:03:34.680 --> 00:03:37.430
The next step is to plug
everything into our equation.

00:03:37.430 --> 00:03:40.020
So, to calculate delta
G naught of reaction,

00:03:40.020 --> 00:03:42.610
we need to plug in delta
H naught of reaction,

00:03:42.610 --> 00:03:44.180
delta S naught of reaction,

00:03:44.180 --> 00:03:46.940
and also the temperature in Kelvin.

00:03:46.940 --> 00:03:49.900
So, we can plug in delta
H naught of reaction

00:03:49.900 --> 00:03:51.410
into our equations.

00:03:51.410 --> 00:03:56.230
That's negative 537.2
kilojoules per mole of reaction.

00:03:56.230 --> 00:03:58.670
Next, we think about the temperature.

00:03:58.670 --> 00:04:01.340
The temperature is 25 degrees Celsius

00:04:01.340 --> 00:04:03.640
and we need to convert that into Kelvin.

00:04:03.640 --> 00:04:08.640
So, 25 plus 273 is equal to 298 Kelvin.

00:04:08.670 --> 00:04:10.870
Next we think about delta S naught,

00:04:10.870 --> 00:04:12.820
and here we have to be careful with units

00:04:12.820 --> 00:04:15.360
because delta H naught was in kilojoules

00:04:15.360 --> 00:04:18.860
and delta S naught was
given to us in joules.

00:04:18.860 --> 00:04:19.900
So, one approach is

00:04:19.900 --> 00:04:23.020
to convert delta S naught into kilojoules

00:04:23.020 --> 00:04:25.010
per Kelvin mole of reaction.

00:04:25.010 --> 00:04:27.260
So, we could divide this number by 1,000,

00:04:27.260 --> 00:04:31.450
or we could move the decimal
place three to the left.

00:04:31.450 --> 00:04:35.300
So, 13.7 joules per
Kelvin mole of reaction

00:04:35.300 --> 00:04:40.300
is equal to 0.0137 kilojoules
per Kelvin mole of reaction.

00:04:41.770 --> 00:04:44.660
Looking at our units,
Kelvin will cancel out

00:04:44.660 --> 00:04:47.950
and that gives us kilojoules
per mole of reaction.

00:04:47.950 --> 00:04:50.850
So, when we do the math, delta
G naught for this reaction

00:04:50.850 --> 00:04:55.850
is equal to negative 541.3
kilojoules per mole of reaction.

00:04:57.350 --> 00:04:59.970
Since delta G naught for
this reaction is negative,

00:04:59.970 --> 00:05:01.830
that means the forward reaction

00:05:01.830 --> 00:05:03.760
is thermodynamically favored.

00:05:03.760 --> 00:05:06.550
So, we can think about the
reactants coming together

00:05:06.550 --> 00:05:08.340
to make the products.

00:05:08.340 --> 00:05:10.860
And since we calculated delta G naught,

00:05:10.860 --> 00:05:14.170
the reactants and products
are in their standard states.

00:05:14.170 --> 00:05:17.690
So, what our calculation
means is if we had a mixture

00:05:17.690 --> 00:05:21.830
of hydrogen gas, fluorine gas,
and hydrogen fluoride gas,

00:05:21.830 --> 00:05:24.960
and we're at a temperature
of 25 degrees Celsius,

00:05:24.960 --> 00:05:29.000
and each gas had a partial
pressure of one atmosphere,

00:05:29.000 --> 00:05:32.380
the forward reaction is
thermodynamically favored,

00:05:32.380 --> 00:05:35.460
which means the hydrogen
gas and fluorine gas

00:05:35.460 --> 00:05:38.663
would come together to make
more hydrogen fluoride.


Absolute entropy and entropy change

WEBVTT
Kind: captions
Language: en

00:00:00.400 --> 00:00:02.810
- [Instructor] Entropy can be
measured on an absolute scale,

00:00:02.810 --> 00:00:06.020
which means there is a
point of zero entropy.

00:00:06.020 --> 00:00:09.110
And that point is reached for
a pure crystalline substance

00:00:09.110 --> 00:00:11.760
when the temperature
is equal to zero kelvin

00:00:11.760 --> 00:00:13.350
or absolute zero.

00:00:13.350 --> 00:00:14.500
At zero kelvin,

00:00:14.500 --> 00:00:17.880
the entropy of the pure
crystalline substance, S,

00:00:17.880 --> 00:00:19.960
is equal to zero.

00:00:19.960 --> 00:00:22.450
We can think about why the
entropy is equal to zero

00:00:22.450 --> 00:00:25.160
by looking at the equation
developed by Boltzmann,

00:00:25.160 --> 00:00:29.620
that relates entropy, S, to
number of microstates, W.

00:00:29.620 --> 00:00:32.790
A microstate refers to a
microscopic arrangement

00:00:32.790 --> 00:00:35.470
of all of the positions and energies

00:00:35.470 --> 00:00:37.170
of all of the particles.

00:00:37.170 --> 00:00:41.000
For our pure crystalline
substance at absolute zero,

00:00:41.000 --> 00:00:44.010
all of the particles are perfectly ordered

00:00:44.010 --> 00:00:46.000
in their lattice states.

00:00:46.000 --> 00:00:48.910
And at zero kelvin
there's no thermal motion.

00:00:48.910 --> 00:00:50.430
So all of the particles

00:00:50.430 --> 00:00:52.770
are perfectly ordered
in their lattice states

00:00:52.770 --> 00:00:54.250
with no thermal motion.

00:00:54.250 --> 00:00:57.310
That means there's only
one possible arrangement

00:00:57.310 --> 00:00:59.050
for these particles.

00:00:59.050 --> 00:01:02.040
And that means there's
only one microstate.

00:01:02.040 --> 00:01:03.820
So when we think about our equation,

00:01:03.820 --> 00:01:07.550
if we plug in the number of
microstates is equal to one,

00:01:07.550 --> 00:01:10.900
the natural log of one is equal to zero,

00:01:10.900 --> 00:01:15.130
which means that the entropy
is equal to zero at zero kelvin

00:01:15.130 --> 00:01:17.670
for this pure crystalline substance.

00:01:17.670 --> 00:01:20.900
This is called the third
law of thermodynamics.

00:01:20.900 --> 00:01:22.560
Next, let's think about what happens

00:01:22.560 --> 00:01:24.950
to our hypothetically perfect crystal

00:01:24.950 --> 00:01:27.050
if we increase the temperature.

00:01:27.050 --> 00:01:30.570
The increased temperature
means the particles gain energy

00:01:30.570 --> 00:01:33.380
and have motion around
their lattice states.

00:01:33.380 --> 00:01:34.213
Therefore,

00:01:34.213 --> 00:01:37.800
there's an increase in the
number of possible microstates.

00:01:37.800 --> 00:01:40.930
And if there's an increase
in the number of microstates,

00:01:40.930 --> 00:01:43.410
according to the equation
developed by Boltzmann,

00:01:43.410 --> 00:01:46.790
that also means an increase in entropy.

00:01:46.790 --> 00:01:50.240
Since we started with zero
entropy at zero kelvin,

00:01:50.240 --> 00:01:52.490
and the entropy increases,

00:01:52.490 --> 00:01:55.670
at all temperatures that are
greater than zero kelvin,

00:01:55.670 --> 00:01:58.380
the entropy must be greater than zero,

00:01:58.380 --> 00:02:01.030
or you can say the entropy is positive.

00:02:01.030 --> 00:02:02.780
We can get the units for entropy

00:02:02.780 --> 00:02:04.910
from the Boltzmann constant, K.

00:02:04.910 --> 00:02:07.000
K is equal to 1.38

00:02:07.000 --> 00:02:11.040
times 10 to the negative
23rd joules per kelvin.

00:02:11.040 --> 00:02:13.970
So we will use these units for entropy.

00:02:13.970 --> 00:02:16.680
Now that we understand the
concept of zero entropy,

00:02:16.680 --> 00:02:19.110
let's look at the entropy of a substance.

00:02:19.110 --> 00:02:22.510
And let's say that we
have one mole of carbon

00:02:22.510 --> 00:02:24.430
in the form of graphite.

00:02:24.430 --> 00:02:25.740
Standard entropy

00:02:25.740 --> 00:02:29.260
refers to the absolute
entropy of a substance

00:02:29.260 --> 00:02:31.620
at a pressure of one atmosphere

00:02:31.620 --> 00:02:33.750
and a specified temperature.

00:02:33.750 --> 00:02:37.690
Often that temperature
is 25 degrees Celsius.

00:02:37.690 --> 00:02:42.010
The standard entropy of
graphite at 25 degrees Celsius

00:02:42.010 --> 00:02:44.920
is equal to 5.7 joules per kelvin.

00:02:44.920 --> 00:02:47.920
And since we're dealing
with one mole of graphite,

00:02:47.920 --> 00:02:51.680
we could write the units
as joules per kelvin mole.

00:02:51.680 --> 00:02:55.510
And often this is called
standard molar entropy.

00:02:55.510 --> 00:02:59.150
The standard molar entropy
of graphite is positive

00:02:59.150 --> 00:03:00.550
because it's being compared

00:03:00.550 --> 00:03:03.330
to a hypothetically
perfect crystal of graphite

00:03:03.330 --> 00:03:04.960
at zero kelvin.

00:03:04.960 --> 00:03:07.093
So really it's a change in entropy

00:03:07.093 --> 00:03:10.270
and therefore it would be 5.7 minus zero

00:03:10.270 --> 00:03:13.910
with zero being the entropy
of a hypothetical crystal

00:03:13.910 --> 00:03:15.420
at zero kelvin.

00:03:15.420 --> 00:03:16.253
However,

00:03:16.253 --> 00:03:17.720
when we write standard molar entropies,

00:03:17.720 --> 00:03:19.700
we don't include the delta sign

00:03:19.700 --> 00:03:22.810
and we reserve the
delta sign for processes

00:03:22.810 --> 00:03:26.930
such as phase changes or
also chemical reactions.

00:03:26.930 --> 00:03:28.290
The superscript, not,

00:03:28.290 --> 00:03:32.110
refers to the standard
state of the substance.

00:03:32.110 --> 00:03:33.080
By convention,

00:03:33.080 --> 00:03:35.350
the standard state of a solid or liquid

00:03:35.350 --> 00:03:38.380
is referring to the pure
solid or pure liquid

00:03:38.380 --> 00:03:41.010
under a pressure of one atmosphere.

00:03:41.010 --> 00:03:44.330
For gases, standard state
is referring to the pure gas

00:03:44.330 --> 00:03:46.670
at a pressure of one atmosphere.

00:03:46.670 --> 00:03:49.860
And for solutions, we're
talking about a solution

00:03:49.860 --> 00:03:52.390
with a concentration of one molar.

00:03:52.390 --> 00:03:54.850
In our case, we're talking
about one mole of carbon

00:03:54.850 --> 00:03:57.720
in the form of graphite
in the solid state.

00:03:57.720 --> 00:03:58.553
Therefore,

00:03:58.553 --> 00:04:01.450
the standard molar entropy of graphite

00:04:01.450 --> 00:04:04.030
is referring to the entropy value

00:04:04.030 --> 00:04:07.020
for one mole of a pure solid

00:04:07.020 --> 00:04:09.950
under a pressure of one atmosphere.

00:04:09.950 --> 00:04:11.170
Next let's look at a table

00:04:11.170 --> 00:04:14.330
showing standard molar entropies
of different substances

00:04:14.330 --> 00:04:16.750
at 25 degrees Celsius.

00:04:16.750 --> 00:04:20.070
We just saw that the standard
molar entropy of carbon

00:04:20.070 --> 00:04:21.270
in the form of graphite

00:04:21.270 --> 00:04:25.130
is equal to 5.7 joules per kelvin mole.

00:04:25.130 --> 00:04:26.920
Let's compare that solid

00:04:26.920 --> 00:04:29.820
to two other standard molar entropies.

00:04:29.820 --> 00:04:30.653
For example,

00:04:30.653 --> 00:04:34.740
liquid water has a standard
molar entropy of 69.9,

00:04:34.740 --> 00:04:39.490
and methane gas has a standard
molar entropy of 186.3.

00:04:39.490 --> 00:04:41.200
Looking at these numbers, in general,

00:04:41.200 --> 00:04:45.300
gasses have higher standard
molar entropies than liquids,

00:04:45.300 --> 00:04:47.410
as we can see comparing the values,

00:04:47.410 --> 00:04:49.160
and liquids, in general,

00:04:49.160 --> 00:04:52.480
have higher standard molar
entropies than solids.

00:04:52.480 --> 00:04:54.430
The reason why this is generally true

00:04:54.430 --> 00:04:57.120
has to do with the number
of microstates available

00:04:57.120 --> 00:04:59.610
to solids, liquids, and gasses.

00:04:59.610 --> 00:05:02.420
Solids are held together
by either chemical bonds

00:05:02.420 --> 00:05:04.830
or by intermolecular forces.

00:05:04.830 --> 00:05:08.320
Liquids are held together
by intermolecular forces.

00:05:08.320 --> 00:05:10.920
And gases, if we assume ideal gases,

00:05:10.920 --> 00:05:14.920
there are no intermolecular
forces between the particles.

00:05:14.920 --> 00:05:19.170
So in general, as we go from
a solid to a liquid to a gas,

00:05:19.170 --> 00:05:20.150
there's an increase

00:05:20.150 --> 00:05:23.340
in the number of possible
arrangements of particles.

00:05:23.340 --> 00:05:25.640
There's an increased freedom of movement.

00:05:25.640 --> 00:05:28.770
And therefore going from
solid to liquid to gas means

00:05:28.770 --> 00:05:32.470
an increase in the number
of possible microstates.

00:05:32.470 --> 00:05:35.620
And an increase in the number
of possible microstates means

00:05:35.620 --> 00:05:38.660
an increase in entropy.

00:05:38.660 --> 00:05:39.493
Next,

00:05:39.493 --> 00:05:42.070
let's compare the standard
molar entropies of two gases.

00:05:42.070 --> 00:05:46.553
Methane is 186.3, whereas ethane is 229.6.

00:05:48.230 --> 00:05:49.510
Looking at their dots structures,

00:05:49.510 --> 00:05:52.820
methane is on the left and
ethane is on the right.

00:05:52.820 --> 00:05:55.110
They're both composed
of carbon and hydrogen.

00:05:55.110 --> 00:05:58.900
However, ethane has more
carbons and more hydrogens.

00:05:58.900 --> 00:06:01.210
Single bonds allow free rotation.

00:06:01.210 --> 00:06:03.830
And since ethane has
more bonds than methane,

00:06:03.830 --> 00:06:07.470
there's more ways for the
ethane molecule to rotate.

00:06:07.470 --> 00:06:10.600
Bonds can also stretch and
compress like a spring.

00:06:10.600 --> 00:06:13.330
So we could imagine
this carbon-carbon bond

00:06:13.330 --> 00:06:15.600
having some vibrational motion.

00:06:15.600 --> 00:06:18.830
And since ethane has
more bonds than methane,

00:06:18.830 --> 00:06:22.680
ethane has more ways to
vibrate than methane does.

00:06:22.680 --> 00:06:26.170
More ways to rotate and
more ways to vibrate mean

00:06:26.170 --> 00:06:27.810
that there are more possible ways

00:06:27.810 --> 00:06:31.350
to distribute the energy
in ethane than in methane.

00:06:31.350 --> 00:06:34.270
And an increased number of ways
of distributing energy means

00:06:34.270 --> 00:06:37.140
an increase in the number
of available microstates,

00:06:37.140 --> 00:06:39.970
which means an increase in the entropy.

00:06:39.970 --> 00:06:40.840
So in general,

00:06:40.840 --> 00:06:43.310
as there's an increase
in the number of atoms,

00:06:43.310 --> 00:06:45.200
or an increase in the molar mass,

00:06:45.200 --> 00:06:47.800
from say methane to ethane,

00:06:47.800 --> 00:06:50.840
there's also an increase in entropy.

00:06:50.840 --> 00:06:53.810
Next, let's calculate the
standard change in entropy

00:06:53.810 --> 00:06:55.740
for a chemical reaction.

00:06:55.740 --> 00:06:56.680
For our reaction,

00:06:56.680 --> 00:06:58.030
let's look at the decomposition

00:06:58.030 --> 00:07:00.370
of one mole of calcium carbonate

00:07:00.370 --> 00:07:02.850
turning into one mole of calcium oxide

00:07:02.850 --> 00:07:05.640
and one mole of carbon dioxide gas.

00:07:05.640 --> 00:07:08.520
To calculate the standard
change in entropy

00:07:08.520 --> 00:07:10.300
for this chemical reaction,

00:07:10.300 --> 00:07:13.460
we need to sum up the
standard molar entropies

00:07:13.460 --> 00:07:14.330
of the products.

00:07:14.330 --> 00:07:15.163
And from that,

00:07:15.163 --> 00:07:18.400
we subtract the sum of the
standard molar entropies

00:07:18.400 --> 00:07:19.870
of the reactants.

00:07:19.870 --> 00:07:22.600
So first we think about our products,

00:07:22.600 --> 00:07:26.670
which are calcium oxide
and carbon dioxide.

00:07:26.670 --> 00:07:29.430
So we need to sum the
standard molar entropies

00:07:29.430 --> 00:07:31.630
of these two substances.

00:07:31.630 --> 00:07:33.610
Next, we think about our reactants.

00:07:33.610 --> 00:07:37.010
And for this reaction, we
have only calcium carbonate.

00:07:37.010 --> 00:07:37.843
So therefore,

00:07:37.843 --> 00:07:40.500
we're gonna plug in the
standard molar entropy

00:07:40.500 --> 00:07:42.120
of calcium carbonate.

00:07:42.120 --> 00:07:45.530
Before we plug in our values
for standard molar entropies,

00:07:45.530 --> 00:07:48.930
let's predict the sign for
the standard entropy change

00:07:48.930 --> 00:07:50.290
for this reaction.

00:07:50.290 --> 00:07:51.130
In our reaction,

00:07:51.130 --> 00:07:52.714
we go from one mole of a solid,

00:07:52.714 --> 00:07:57.260
to one mole of another
solid and one mole of a gas.

00:07:57.260 --> 00:07:58.910
We already know that in general

00:07:58.910 --> 00:08:02.010
gases have higher values
for entropy than solids.

00:08:02.010 --> 00:08:05.430
And since we have zero moles
of gas on the reactant side,

00:08:05.430 --> 00:08:08.590
and one mole of gas on the product side,

00:08:08.590 --> 00:08:09.530
we would predict

00:08:09.530 --> 00:08:12.980
that the sum of the standard
molar entropies of the products

00:08:12.980 --> 00:08:16.520
would be greater than the sum
of the standard molar entropy

00:08:16.520 --> 00:08:17.460
of the reactants.

00:08:17.460 --> 00:08:19.370
In this case, only one reactant.

00:08:19.370 --> 00:08:21.970
And since we'd be
subtracting a smaller number

00:08:21.970 --> 00:08:23.740
from a larger number,

00:08:23.740 --> 00:08:27.850
we would predict that the
standard change in entropy

00:08:27.850 --> 00:08:32.610
for this reaction would be positive.

00:08:32.610 --> 00:08:33.990
Now that we've made our prediction,

00:08:33.990 --> 00:08:36.810
let's plug in our
standard molar entropies,

00:08:36.810 --> 00:08:39.110
and these are for a 25 degrees Celsius.

00:08:39.110 --> 00:08:41.550
And see if our prediction was correct.

00:08:41.550 --> 00:08:45.030
The standard molar
entropy of calcium oxide

00:08:45.030 --> 00:08:48.640
is equal to 39.8 joules per kelvin mole.

00:08:48.640 --> 00:08:49.870
In our balanced equation,

00:08:49.870 --> 00:08:52.980
we have one mole of calcium oxide.

00:08:52.980 --> 00:08:55.840
So we're gonna multiply
one mole of calcium oxide

00:08:55.840 --> 00:08:59.120
by the standard molar entropy
and moles will cancel.

00:08:59.120 --> 00:08:59.953
Next,

00:08:59.953 --> 00:09:03.280
we look up the standard molar
entropy of carbon dioxide,

00:09:03.280 --> 00:09:07.120
which is equal to 213.6
joules per kelvin mole

00:09:07.120 --> 00:09:09.170
and our balanced equation, once again,

00:09:09.170 --> 00:09:12.170
there's a one in front of carbon dioxide.

00:09:12.170 --> 00:09:14.280
So we're gonna multiply
the standard molar entropy

00:09:14.280 --> 00:09:16.930
by one mole and moles will cancel.

00:09:16.930 --> 00:09:19.200
So we're gonna add those two together.

00:09:19.200 --> 00:09:20.033
And from that,

00:09:20.033 --> 00:09:23.130
we're going to subtract
the standard molar entropy

00:09:23.130 --> 00:09:26.980
of our reactants, which
is calcium carbonate.

00:09:26.980 --> 00:09:29.280
So the standard molar
entropy of calcium carbonate

00:09:29.280 --> 00:09:31.970
is 92.9 joules per kelvin mole.

00:09:31.970 --> 00:09:33.560
And in our balanced equation,

00:09:33.560 --> 00:09:35.420
there's a one as a coefficient

00:09:35.420 --> 00:09:37.470
in front of calcium carbonate.

00:09:37.470 --> 00:09:38.990
So multiply that by one mole.

00:09:38.990 --> 00:09:41.500
And once again, the moles will cancel out.

00:09:41.500 --> 00:09:43.990
For this particular reaction,
in the balanced equation,

00:09:43.990 --> 00:09:46.430
all of the coefficients happen to be one.

00:09:46.430 --> 00:09:47.263
However,

00:09:47.263 --> 00:09:49.790
if one of our reactants or
products in a different reaction

00:09:49.790 --> 00:09:51.750
happened to have a coefficient of two,

00:09:51.750 --> 00:09:56.298
we would multiply the standard
molar entropy by two moles.

00:09:56.298 --> 00:09:57.910
After we do the math we find

00:09:57.910 --> 00:10:00.800
that the standard change in
entropy for this reaction

00:10:00.800 --> 00:10:05.260
is equal to positive
160.5 joules per kelvin.

00:10:05.260 --> 00:10:07.880
Therefore one mole of calcium carbonate

00:10:07.880 --> 00:10:10.750
decomposing to form one
mole of calcium oxide

00:10:10.750 --> 00:10:12.700
and one mole of carbon dioxide

00:10:12.700 --> 00:10:15.220
involves an increase in entropy

00:10:15.220 --> 00:10:18.370
just as we had predicted earlier on.

00:10:18.370 --> 00:10:20.740
Sometimes you'll see the
units as joules per kelvin,

00:10:20.740 --> 00:10:21.573
and sometimes,

00:10:21.573 --> 00:10:24.990
you might see joules per
kelvin mole of reaction.

00:10:24.990 --> 00:10:26.680
To understand how to get those units,

00:10:26.680 --> 00:10:27.513
let's just consider

00:10:27.513 --> 00:10:30.950
the standard molar
entropy of carbon dioxide.

00:10:30.950 --> 00:10:35.640
And that value is 213.6 joules per kelvin

00:10:35.640 --> 00:10:39.090
per one mole of carbon dioxide.

00:10:39.090 --> 00:10:41.650
And looking at the balanced equation,

00:10:41.650 --> 00:10:46.030
there's a one as a coefficient
in front of carbon dioxide.

00:10:46.030 --> 00:10:50.010
Therefore, there's one
mole of carbon dioxide

00:10:50.010 --> 00:10:52.760
per one mole of reaction.

00:10:52.760 --> 00:10:54.770
So molar reaction is just talking about,

00:10:54.770 --> 00:10:57.260
how the balanced equation is written.

00:10:57.260 --> 00:11:01.170
So the one mole of carbon
dioxide, it cancels out,

00:11:01.170 --> 00:11:06.170
and that leaves us with joules
per kelvin mole of reaction.

00:11:06.840 --> 00:11:08.990
So using these units is simply saying,

00:11:08.990 --> 00:11:12.950
for the decomposition of one
mole of calcium carbonate,

00:11:12.950 --> 00:11:14.410
there'll be a change in entropy

00:11:14.410 --> 00:11:18.490
of positive 160.5 joules per kelvin.

00:11:18.490 --> 00:11:19.323
Therefore,

00:11:19.323 --> 00:11:20.740
if we were talking about the decomposition

00:11:20.740 --> 00:11:22.950
of two moles of calcium carbonate,

00:11:22.950 --> 00:11:25.760
the change in entropy
would be twice this value

00:11:25.760 --> 00:11:27.353
that we already calculated.


Introduction to entropy

WEBVTT
Kind: captions
Language: en

00:00:00.600 --> 00:00:02.830
- [Instructor] The concept
of entropy is related

00:00:02.830 --> 00:00:05.620
to the idea of microstates.

00:00:05.620 --> 00:00:06.990
And to think about microstates,

00:00:06.990 --> 00:00:09.740
let's consider one mole of an ideal gas.

00:00:09.740 --> 00:00:13.940
So remember, n represents
moles at a specific pressure,

00:00:13.940 --> 00:00:16.710
volume, and temperature.

00:00:16.710 --> 00:00:19.990
If the system of gas
particles is at equilibrium,

00:00:19.990 --> 00:00:21.550
then the pressure, the volume,

00:00:21.550 --> 00:00:23.150
the number of moles, and the temperature

00:00:23.150 --> 00:00:24.500
all remain the same.

00:00:24.500 --> 00:00:27.050
So from a macroscopic point of view,

00:00:27.050 --> 00:00:29.350
nothing seems to be changing.

00:00:29.350 --> 00:00:31.530
However, from a microscopic point of view,

00:00:31.530 --> 00:00:33.370
things are changing all of the time.

00:00:33.370 --> 00:00:36.300
So looking at our gas particles
here in the first box,

00:00:36.300 --> 00:00:38.610
imagine these gas particles
slamming into the sides

00:00:38.610 --> 00:00:41.110
of the container and maybe
slamming into each other

00:00:41.110 --> 00:00:44.540
and transferring energy from
one particle to another.

00:00:44.540 --> 00:00:47.370
So if we think about the
particles in our system

00:00:47.370 --> 00:00:49.720
at one moment in time, in box 1,

00:00:49.720 --> 00:00:51.620
if we think about them at
a different moment in time,

00:00:51.620 --> 00:00:52.540
in box 2,

00:00:52.540 --> 00:00:55.490
the particles might be in
slightly different positions

00:00:55.490 --> 00:00:59.320
and the velocities might have changed.

00:00:59.320 --> 00:01:01.240
Remember that the speed
of a particle tells us

00:01:01.240 --> 00:01:03.040
how fast the particle is traveling.

00:01:03.040 --> 00:01:05.340
However, when we put an
arrow on each particle,

00:01:05.340 --> 00:01:07.340
that also gives us the direction.

00:01:07.340 --> 00:01:10.370
And the magnitude and the
direction give a velocity.

00:01:10.370 --> 00:01:14.060
So these arrows on these
particles are meant to represent

00:01:14.060 --> 00:01:16.640
the velocities of the particles.

00:01:16.640 --> 00:01:19.550
And the kinetic energy
of each particle is equal

00:01:19.550 --> 00:01:22.810
to 1/2 mv squared,

00:01:22.810 --> 00:01:27.480
where m is the mass of each
particle and v is the velocity.

00:01:27.480 --> 00:01:30.870
A microstate refers to a
microscopic arrangement

00:01:30.870 --> 00:01:34.900
of all of the positions and
energies of the gas particles.

00:01:34.900 --> 00:01:37.520
Since we're dealing
with an ideal gas here,

00:01:37.520 --> 00:01:40.420
by energies, we're referring
to the kinetic energies

00:01:40.420 --> 00:01:42.030
of the particles.

00:01:42.030 --> 00:01:46.280
So going back to our boxes,
box 1, box 2 and box 3,

00:01:46.280 --> 00:01:49.930
each box shows a different
microscopic arrangement

00:01:49.930 --> 00:01:53.710
of positions and energies
of the gas particles.

00:01:53.710 --> 00:01:57.640
Therefore, each box
represents one microstate.

00:01:57.640 --> 00:01:59.540
A good way to think about
a microstate would be

00:01:59.540 --> 00:02:03.220
like taking a picture of
our system of gas particles.

00:02:03.220 --> 00:02:05.130
So from a macroscopic point of view,

00:02:05.130 --> 00:02:06.690
nothing seems to change.

00:02:06.690 --> 00:02:10.140
But taking a picture at
the microscopic level,

00:02:10.140 --> 00:02:12.980
we see that the system is
moving from one microstate

00:02:12.980 --> 00:02:15.890
into another, into another, into another.

00:02:15.890 --> 00:02:19.340
So the number of microstates
available to this system

00:02:19.340 --> 00:02:22.710
of one mole of gas particles
is a number that's too high

00:02:22.710 --> 00:02:25.210
for us to even comprehend.

00:02:25.210 --> 00:02:27.560
Now that we understand the
concept of microstates,

00:02:27.560 --> 00:02:29.970
let's look at an equation
developed by Boltzmann

00:02:29.970 --> 00:02:33.530
that relates entropy to
the number of microstates.

00:02:33.530 --> 00:02:34.960
According to this equation,

00:02:34.960 --> 00:02:36.860
entropy, symbolized by S,

00:02:36.860 --> 00:02:39.510
is equal to Boltzmann's constant, k,

00:02:39.510 --> 00:02:42.050
times the natural log of W,

00:02:42.050 --> 00:02:45.960
and W represents the number
of microstates in a system.

00:02:45.960 --> 00:02:49.870
So if the number of microstates
of a system increases,

00:02:49.870 --> 00:02:53.220
that represents an increase in entropy,

00:02:53.220 --> 00:02:56.260
and if the number of
microstates decreases,

00:02:56.260 --> 00:03:00.650
that represents a decrease
in the entropy of the system.

00:03:00.650 --> 00:03:03.360
Sometimes, instead of
using the word microstates,

00:03:03.360 --> 00:03:05.930
people will describe
an increase in entropy

00:03:05.930 --> 00:03:08.160
as an increase in disorder

00:03:08.160 --> 00:03:13.160
or an increase in the dispersal
of either matter or energy.

00:03:13.560 --> 00:03:16.810
Sometimes, using these terms
helps us think about entropy.

00:03:16.810 --> 00:03:17.643
However,

00:03:17.643 --> 00:03:19.980
when we're using the equation
developed by Boltzmann,

00:03:19.980 --> 00:03:22.430
we should think about these
terms as meaning an increase

00:03:22.430 --> 00:03:23.860
in the number of microstates

00:03:23.860 --> 00:03:27.750
and therefore an increase in
the entropy of the system.

00:03:27.750 --> 00:03:30.460
And if we think about a
decrease in the disorder

00:03:30.460 --> 00:03:33.090
of the system or an increase in the order,

00:03:33.090 --> 00:03:37.610
or a decrease in the dispersal
of either matter or energy,

00:03:37.610 --> 00:03:39.380
that really relates to a decrease

00:03:39.380 --> 00:03:41.550
in the number of available microstates,

00:03:41.550 --> 00:03:44.970
which means a decrease in
the entropy of the system.

00:03:44.970 --> 00:03:47.200
Next, let's think about
the change in entropy

00:03:47.200 --> 00:03:49.720
for a number of different situations.

00:03:49.720 --> 00:03:50.840
In our first situation,

00:03:50.840 --> 00:03:53.310
we're starting off with
one mole of an ideal gas,

00:03:53.310 --> 00:03:56.520
so here are the gas
particles, in a container,

00:03:56.520 --> 00:03:59.970
and the container has a removable divider

00:03:59.970 --> 00:04:02.970
separating the container
into two compartments.

00:04:02.970 --> 00:04:05.230
Let's say we go ahead
and remove the divider.

00:04:05.230 --> 00:04:08.340
And now our gas particles
are free to travel around

00:04:08.340 --> 00:04:09.960
in a larger volume.

00:04:09.960 --> 00:04:12.150
So if the initial volume is V1,

00:04:12.150 --> 00:04:15.460
let's say we have twice the
volume for the final volume,

00:04:15.460 --> 00:04:17.360
therefore, V2, or the final volume,

00:04:17.360 --> 00:04:19.720
is equal to 2 times V1.

00:04:19.720 --> 00:04:21.190
The number of moles remains the same.

00:04:21.190 --> 00:04:24.360
So n1 is equal to one
mole of our ideal gas

00:04:24.360 --> 00:04:26.920
and n2 is also equal to one mole.

00:04:26.920 --> 00:04:29.170
So n2 is equal to n1.

00:04:29.170 --> 00:04:30.930
During the expansion of the gas,

00:04:30.930 --> 00:04:32.670
the temperature is kept constant.

00:04:32.670 --> 00:04:34.980
Therefore, the initial temperature, T1,

00:04:34.980 --> 00:04:37.670
is equal to the final temperature, T2.

00:04:37.670 --> 00:04:39.040
Next, let's think about what happens

00:04:39.040 --> 00:04:41.160
to the number of available microstates

00:04:41.160 --> 00:04:42.890
when the volume is doubled.

00:04:42.890 --> 00:04:43.900
When the volume is doubled,

00:04:43.900 --> 00:04:46.900
that increases the number
of possible positions

00:04:46.900 --> 00:04:48.420
for the gas particles.

00:04:48.420 --> 00:04:52.040
Therefore, the number of
microstates increases.

00:04:52.040 --> 00:04:53.530
And looking at our equation,

00:04:53.530 --> 00:04:56.560
if the number of microstates increases,

00:04:56.560 --> 00:04:59.220
then so does the entropy.

00:04:59.220 --> 00:05:04.220
Therefore, we can say that
S2 is greater than S1.

00:05:04.860 --> 00:05:08.530
And thinking about the change
in the entropy, delta S,

00:05:08.530 --> 00:05:11.130
if S2 is greater than S1,

00:05:11.130 --> 00:05:15.000
then S2 minus S1 would
be a positive value.

00:05:15.000 --> 00:05:15.833
Therefore,

00:05:15.833 --> 00:05:19.650
the change in entropy for
the free expansion of a gas

00:05:19.650 --> 00:05:23.110
when the temperature is
constant is positive.

00:05:23.110 --> 00:05:24.300
For our next situation,

00:05:24.300 --> 00:05:27.590
we're starting once again
with one mole of an ideal gas.

00:05:27.590 --> 00:05:30.300
However, this time the
volume will be held constant

00:05:30.300 --> 00:05:32.710
and the temperature will be increased.

00:05:32.710 --> 00:05:35.210
So we start with one mole of our ideal gas

00:05:35.210 --> 00:05:37.970
at a certain volume, so n1 and V1,

00:05:37.970 --> 00:05:39.890
and we are increasing the temperature,

00:05:39.890 --> 00:05:42.750
but we're not changing the
number of moles or the volume.

00:05:42.750 --> 00:05:44.770
Therefore, the final number of moles, n2,

00:05:44.770 --> 00:05:47.550
is equal to the initial
number, n1, which is one mole.

00:05:47.550 --> 00:05:49.110
And the final volume, V2,

00:05:49.110 --> 00:05:51.720
is equal to the initial volume, V1.

00:05:51.720 --> 00:05:53.530
Since we are increasing the temperature,

00:05:53.530 --> 00:05:55.210
the final temperature, T2,

00:05:55.210 --> 00:05:58.380
is greater than the
initial temperature, T1.

00:05:58.380 --> 00:06:02.040
Next, let's use a Maxwell-Boltzmann
distribution over here

00:06:02.040 --> 00:06:04.770
on the right to explain what we see

00:06:04.770 --> 00:06:07.910
in our particulate diagrams on the left.

00:06:07.910 --> 00:06:09.610
In the particulate diagram on the left,

00:06:09.610 --> 00:06:11.780
we can see that the particles are moving

00:06:11.780 --> 00:06:15.450
with different velocities or
different magnitudes or speeds.

00:06:15.450 --> 00:06:18.600
And we can see that by the
length of the arrows in the box.

00:06:18.600 --> 00:06:20.700
And that's what the
Maxwell-Boltzmann distribution

00:06:20.700 --> 00:06:21.533
tells us here.

00:06:21.533 --> 00:06:23.370
So for the one in blue,

00:06:23.370 --> 00:06:26.080
we know that there are a
range of speeds available

00:06:26.080 --> 00:06:26.980
to the particles.

00:06:26.980 --> 00:06:30.030
So the area under the curve
represents all of the particles

00:06:30.030 --> 00:06:31.400
in the gas sample.

00:06:31.400 --> 00:06:34.140
So some are moving at a slower speed.

00:06:34.140 --> 00:06:35.820
Some are moving at a higher speed,

00:06:35.820 --> 00:06:37.320
but most of them are moving at a speed

00:06:37.320 --> 00:06:39.830
close to the center of this peak.

00:06:39.830 --> 00:06:42.900
Increasing the temperature
means that, on average,

00:06:42.900 --> 00:06:45.240
the particles are moving faster.

00:06:45.240 --> 00:06:46.690
So we can see that, on average,

00:06:46.690 --> 00:06:49.410
the length of these arrows is longer

00:06:49.410 --> 00:06:51.920
than the length of the arrows
in the particulate diagram

00:06:51.920 --> 00:06:53.270
on the left.

00:06:53.270 --> 00:06:54.103
But it also means,

00:06:54.103 --> 00:06:55.990
if we look at the
Maxwell-Boltzmann distribution,

00:06:55.990 --> 00:06:58.670
there are greater range
of speeds available

00:06:58.670 --> 00:07:00.300
to the particles.

00:07:00.300 --> 00:07:02.020
And if there's a greater range of speeds

00:07:02.020 --> 00:07:03.520
or greater range of velocities,

00:07:03.520 --> 00:07:06.600
that means there's a greater
range of kinetic energies,

00:07:06.600 --> 00:07:09.770
which means there are more
possible microstates available

00:07:09.770 --> 00:07:11.980
to the system of gas particles.

00:07:11.980 --> 00:07:15.330
Therefore, increasing the
temperature causes an increase

00:07:15.330 --> 00:07:18.290
in the number of possible microstates.

00:07:18.290 --> 00:07:21.310
And an increase in the
number of microstates means

00:07:21.310 --> 00:07:23.920
an increase in the entropy.

00:07:23.920 --> 00:07:26.360
Therefore, the final entropy, S2,

00:07:26.360 --> 00:07:30.200
is greater than the initial entropy, S1.

00:07:30.200 --> 00:07:32.240
So when we think about
the change in entropy,

00:07:32.240 --> 00:07:35.470
delta S for the system,
if S2 is greater than S1,

00:07:35.470 --> 00:07:38.390
the change in entropy is positive.

00:07:38.390 --> 00:07:40.260
So for a system of gas particles

00:07:40.260 --> 00:07:42.910
with constant volume and
constant number of moles,

00:07:42.910 --> 00:07:44.570
if we increased the temperature,

00:07:44.570 --> 00:07:47.610
there's an increase in the entropy.

00:07:47.610 --> 00:07:48.690
For our next situation,

00:07:48.690 --> 00:07:51.980
we're once again starting
with one mole of an ideal gas

00:07:51.980 --> 00:07:56.450
at a certain temperature,
T1, and a certain volume, V1.

00:07:56.450 --> 00:07:57.283
This time,

00:07:57.283 --> 00:07:58.980
we're going to increase
the number of moles

00:07:58.980 --> 00:08:02.500
from one mole of an ideal gas
to two moles of an ideal gas.

00:08:02.500 --> 00:08:05.180
So if we had four particles on the left,

00:08:05.180 --> 00:08:07.140
now we have eight particles

00:08:08.583 --> 00:08:11.050
in the particulate diagram on the right.

00:08:11.050 --> 00:08:12.930
So the final number of moles, n2,

00:08:12.930 --> 00:08:15.930
is greater than the initial number, n1.

00:08:15.930 --> 00:08:17.600
But we're gonna keep the
temperature the same.

00:08:17.600 --> 00:08:18.980
So the final temperature is equal

00:08:18.980 --> 00:08:20.130
to the initial temperature,

00:08:20.130 --> 00:08:22.190
and we're gonna keep
the volume constant too.

00:08:22.190 --> 00:08:24.000
So V2, or the final volume,

00:08:24.000 --> 00:08:26.670
is equal to the initial volume, V1.

00:08:26.670 --> 00:08:29.030
Because we have increased
the number of particles,

00:08:29.030 --> 00:08:31.690
there are more possible
arrangements of particles

00:08:31.690 --> 00:08:34.600
and also more ways to
distribute the energy.

00:08:34.600 --> 00:08:37.010
Therefore, when we increase
the number of moles,

00:08:37.010 --> 00:08:40.100
there's an increase in the
number of possible microstates

00:08:41.620 --> 00:08:44.200
and increasing the number
of microstates in a system

00:08:44.200 --> 00:08:46.460
increases the entropy.

00:08:46.460 --> 00:08:48.630
Therefore, the final entropy, S2,

00:08:48.630 --> 00:08:52.030
is greater than the initial entropy, S1.

00:08:52.030 --> 00:08:54.410
And once again, the change in entropy,

00:08:54.410 --> 00:08:57.200
if S2 greater than S1, would be positive.

00:08:57.200 --> 00:09:01.080
So increasing the number
of moles of gas particles

00:09:01.080 --> 00:09:03.910
increases the entropy of the system.

00:09:03.910 --> 00:09:04.990
For our next situation,

00:09:04.990 --> 00:09:07.920
let's consider the
evaporation of liquid water

00:09:07.920 --> 00:09:09.620
into gaseous water.

00:09:09.620 --> 00:09:11.950
Water molecules in the liquid
state are held together

00:09:11.950 --> 00:09:14.590
by intermolecular forces,

00:09:14.590 --> 00:09:16.870
with hydrogen bonding
being the most important

00:09:16.870 --> 00:09:19.960
intermolecular force holding
the particles together.

00:09:19.960 --> 00:09:22.870
Water molecules in the liquid
state are still free to move.

00:09:22.870 --> 00:09:25.740
However, when water
molecules in the liquid state

00:09:25.740 --> 00:09:28.430
get converted into water
molecules in the gaseous state,

00:09:28.430 --> 00:09:31.390
we assume there are no
more intermolecular forces

00:09:31.390 --> 00:09:33.720
between the gas particles.

00:09:33.720 --> 00:09:35.810
So we're assuming this is an ideal gas.

00:09:35.810 --> 00:09:38.100
And if there are no intermolecular forces

00:09:38.100 --> 00:09:39.750
between the gas particles,

00:09:39.750 --> 00:09:41.710
we've increased the freedom of movement

00:09:41.710 --> 00:09:42.870
of the water molecules.

00:09:42.870 --> 00:09:45.600
We've increased the number
of possible positions,

00:09:45.600 --> 00:09:46.490
and, therefore,

00:09:46.490 --> 00:09:50.380
we've increased the number
of microstates available.

00:09:50.380 --> 00:09:52.210
And if we increase the
number of microstates,

00:09:52.210 --> 00:09:54.500
we increase the entropy.

00:09:54.500 --> 00:09:57.450
Therefore, thinking
about the entropy change,

00:09:57.450 --> 00:09:59.560
this would be the final entropy, S2,

00:09:59.560 --> 00:10:02.070
minus the initial entropy, S1.

00:10:02.070 --> 00:10:04.050
And since the entropy increased,

00:10:04.050 --> 00:10:06.680
going from liquid water to gaseous water,

00:10:06.680 --> 00:10:09.870
the change in entropy would be positive.

00:10:09.870 --> 00:10:12.480
This is also one example
where thinking about entropy

00:10:12.480 --> 00:10:14.900
in terms of disorder can be helpful

00:10:14.900 --> 00:10:17.850
because gasses are more
disordered than liquids,

00:10:17.850 --> 00:10:19.460
you could say, and, therefore,

00:10:19.460 --> 00:10:21.990
going from a liquid to a
gas increases the amount

00:10:21.990 --> 00:10:24.630
of disorder, which increases the entropy.

00:10:24.630 --> 00:10:25.520
But remember, really,

00:10:25.520 --> 00:10:27.630
disorder is really just a way to describe

00:10:27.630 --> 00:10:30.420
an increased number of
available microstates.

00:10:30.420 --> 00:10:31.790
And for our final situation,

00:10:31.790 --> 00:10:35.270
let's look at a reaction
that involves only gases.

00:10:35.270 --> 00:10:36.240
On the left side,

00:10:36.240 --> 00:10:39.770
we can see there's two moles
of SO2 and one mole of O2.

00:10:39.770 --> 00:10:43.600
So there are three moles of
gas on the reactant side.

00:10:43.600 --> 00:10:48.500
And on the product side,
there are two moles of gas.

00:10:48.500 --> 00:10:50.450
Next, let's think about
the change in entropy

00:10:50.450 --> 00:10:52.730
for this reaction, delta S.

00:10:52.730 --> 00:10:53.563
That'd be equal

00:10:53.563 --> 00:10:57.190
to the final entropy
minus the initial entropy.

00:10:57.190 --> 00:10:59.810
So thinking about the initial entropy, S1,

00:10:59.810 --> 00:11:02.130
and the final entropy, S2,

00:11:02.130 --> 00:11:05.500
we went from three moles
of gas to two moles of gas.

00:11:05.500 --> 00:11:08.590
Remember that a decrease in
the number of moles of gas

00:11:08.590 --> 00:11:12.350
means a decrease in the number
of possible microstates.

00:11:12.350 --> 00:11:15.050
That means a decrease in the entropy.

00:11:15.050 --> 00:11:18.810
Therefore, S2 is less than S1,

00:11:18.810 --> 00:11:22.710
or we could say that
S1 is greater than S2.

00:11:22.710 --> 00:11:25.020
Therefore, for the change in the entropy,

00:11:25.020 --> 00:11:27.720
if S1 is greater than S2,

00:11:27.720 --> 00:11:30.870
we're subtracting a larger
number from a smaller number,

00:11:30.870 --> 00:11:33.400
which means the change in
entropy for this reaction,

00:11:33.400 --> 00:11:35.840
delta S, will be negative.

00:11:35.840 --> 00:11:36.870
So as a quick summary,

00:11:36.870 --> 00:11:40.080
when trying to figure out
the change in entropy,

00:11:40.080 --> 00:11:43.020
we need to consider the number
of available microstates.

00:11:43.020 --> 00:11:45.670
If the number of available
microstates increases,

00:11:45.670 --> 00:11:47.840
then the change in entropy is positive.

00:11:47.840 --> 00:11:50.520
If the number of available
microstates decreases,

00:11:50.520 --> 00:11:52.943
then the change in entropy is negative.


Solving unit price problem

WEBVTT
Kind: captions
Language: en

00:00:00.210 --> 00:00:02.930
- [Instructor] We're told
that Nierria earns $75

00:00:02.930 --> 00:00:04.960
for 4 hours of tutoring.

00:00:04.960 --> 00:00:08.540
How much does Nierria earn
for 1 hour of tutoring?

00:00:08.540 --> 00:00:11.190
Pause this video, and see
if you can figure that out.

00:00:12.420 --> 00:00:16.810
Well, the key here is $75
for 4 hours of tutoring.

00:00:16.810 --> 00:00:18.480
And there's a couple of
ways you can think about it.

00:00:18.480 --> 00:00:22.600
The ratio here is $75 for every 4 hours.

00:00:22.600 --> 00:00:25.110
So if you wanted one hour of tutoring,

00:00:25.110 --> 00:00:28.220
well, it's just going
to be 1/4 of the $75,

00:00:28.220 --> 00:00:30.960
or you could just take 75 and divide by 4.

00:00:30.960 --> 00:00:32.320
Another way you could think about it,

00:00:32.320 --> 00:00:37.270
you could say Nierria gets $75

00:00:37.270 --> 00:00:40.563
for every 4 hours,

00:00:41.430 --> 00:00:43.640
which is another way of saying

00:00:43.640 --> 00:00:45.810
that Nierria is going to earn

00:00:45.810 --> 00:00:50.810
75 divided by 4 dollars per hour.

00:00:51.920 --> 00:00:53.200
So really we just have to figure out

00:00:53.200 --> 00:00:55.260
what 75 divided by 4 is.

00:00:55.260 --> 00:00:58.800
So let's do a little bit of long division.

00:00:58.800 --> 00:01:03.660
So 4 goes into 7 one time, 1 times 4 is 4.

00:01:03.660 --> 00:01:06.940
We're going to subtract. 7 minus 4 is 3.

00:01:06.940 --> 00:01:08.540
Bring down that 5.

00:01:08.540 --> 00:01:11.290
4 goes into 35 eight times.

00:01:11.290 --> 00:01:13.680
8 times 4 is 32.

00:01:13.680 --> 00:01:16.710
I will subtract again. I get a 3.

00:01:16.710 --> 00:01:18.930
Well now I'm going to
get some decimals here,

00:01:18.930 --> 00:01:22.620
so let me bring down a zero.

00:01:22.620 --> 00:01:25.270
4 goes into 30 seven times.

00:01:25.270 --> 00:01:27.580
7 times 4 is 28.

00:01:27.580 --> 00:01:32.080
Subtract, and then I get a
2. Bring down another zero.

00:01:32.080 --> 00:01:35.190
4 goes into 20 five times.

00:01:35.190 --> 00:01:38.260
5 times 4 is 20, and I am done,

00:01:38.260 --> 00:01:40.060
no more remainder here.

00:01:40.060 --> 00:01:44.770
So we get that Nierria
earns 18.75 dollars,

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dollars per hour,

00:01:50.480 --> 00:01:55.053
or you can just say that
she makes $18.75 per hour.


Sine equation algebraic solution set

WEBVTT
Kind: captions
Language: en

00:00:00.120 --> 00:00:01.260
- [Instructor] The goal of this video

00:00:01.260 --> 00:00:05.230
is to find the solution set
for the following equation.

00:00:05.230 --> 00:00:07.940
So all of the X values, and we're dealing

00:00:07.940 --> 00:00:12.090
with radians, that will
satisfy this equation.

00:00:12.090 --> 00:00:14.820
So I encourage you, like
always, pause this video

00:00:14.820 --> 00:00:16.480
and see if you can work
through this on your own

00:00:16.480 --> 00:00:18.930
before we worked through it together.

00:00:18.930 --> 00:00:21.240
All right, now let's
work through it together.

00:00:21.240 --> 00:00:23.700
Now your intuition, which
would be correct, might be,

00:00:23.700 --> 00:00:25.590
let's see if we can isolate the sin

00:00:25.590 --> 00:00:27.970
of X over four algebraically.

00:00:27.970 --> 00:00:30.420
And the first step I
would do is subtract 11

00:00:30.420 --> 00:00:31.290
from both sides.

00:00:31.290 --> 00:00:36.060
And if you do that, you
would get eight sin of X

00:00:36.060 --> 00:00:40.950
over four is equal to three,
just subtracted 11, both sides.

00:00:40.950 --> 00:00:44.850
Now to isolate the sin, I would
divide both sides by eight

00:00:44.850 --> 00:00:49.850
and I would get sin of X
over four is equal to 3/8.

00:00:51.790 --> 00:00:54.760
Now, before I go further,
let's think about

00:00:54.760 --> 00:00:57.700
whether this is the most
general solution here

00:00:57.700 --> 00:01:01.220
or whether we're going to find
all of the solution set here.

00:01:01.220 --> 00:01:02.700
Well, we have to remind ourselves,

00:01:02.700 --> 00:01:06.510
let me actually draw a little
bit of a unit circle here.

00:01:06.510 --> 00:01:08.800
So that's my X axis.

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That's my Y axis.

00:01:12.018 --> 00:01:13.350
Then a circle.

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And if we have some angle
theta right over here.

00:01:18.910 --> 00:01:21.920
So that's theta, we know that
the sin of theta is equal

00:01:21.920 --> 00:01:23.350
to the Y coordinate

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of where this radius
intersects the unit circle.

00:01:26.760 --> 00:01:27.750
And we also know

00:01:27.750 --> 00:01:30.720
that if we add an arbitrary
number of two pies here

00:01:30.720 --> 00:01:32.830
or if we subtract an
arbitrary number of two pis,

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we go all the way around the unit circle,

00:01:34.240 --> 00:01:35.420
back to where we began

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and so the sin of theta would be the same.

00:01:37.100 --> 00:01:42.100
So we know that sin of theta
plus any integer multiple

00:01:42.500 --> 00:01:47.487
of two pi, that's going to
be equal to the sin of theta.

00:01:49.150 --> 00:01:51.290
And so we can generalize
this a little bit.

00:01:51.290 --> 00:01:53.530
We can instead of just
saying, sin of X over four

00:01:53.530 --> 00:01:58.530
is equal to 3/8, we could
write that sin of X over four

00:01:58.570 --> 00:02:02.530
plus any integer multiple of
two pi is going to be equal

00:02:02.530 --> 00:02:05.300
to 3/8 where N is any integer.

00:02:05.300 --> 00:02:06.980
It could even be a negative
one, a negative two

00:02:06.980 --> 00:02:09.440
or of course it could be
zero, one, two, three,

00:02:09.440 --> 00:02:11.560
so on and so forth.

00:02:11.560 --> 00:02:15.420
So is this, if we solve now for X,

00:02:15.420 --> 00:02:19.410
is this going to give us the
most general solution set?

00:02:19.410 --> 00:02:24.190
Well, we can also remind ourselves
that if I have theta here

00:02:24.190 --> 00:02:25.830
and sin if theta gets there.

00:02:25.830 --> 00:02:27.660
There's one other point on the unit circle

00:02:27.660 --> 00:02:29.800
where I get the same sin.

00:02:29.800 --> 00:02:31.520
It would be right over here.

00:02:31.520 --> 00:02:33.810
The Y coordinate would be the same.

00:02:33.810 --> 00:02:35.940
And one way to think about it is

00:02:35.940 --> 00:02:38.810
if we start at pi radians,
which would be right over there

00:02:38.810 --> 00:02:41.210
and we were to subtract theta,

00:02:41.210 --> 00:02:43.030
we're going to get the same thing.

00:02:43.030 --> 00:02:45.160
So this angle right over here,

00:02:45.160 --> 00:02:49.000
you could view as pi minus theta.

00:02:49.000 --> 00:02:51.360
And you could keep trying
it out for any theta,

00:02:51.360 --> 00:02:53.300
even the theta that put
you in the second quadrant,

00:02:53.300 --> 00:02:56.840
third quadrant or fourth quadrant,
if you do pi minus theta,

00:02:56.840 --> 00:02:58.190
sin of pi minus theta,

00:02:58.190 --> 00:03:00.320
you're going to get the same sin value.

00:03:00.320 --> 00:03:04.990
So we also know that sin of pi minus theta

00:03:04.990 --> 00:03:08.920
is equal to sin of theta.

00:03:08.920 --> 00:03:12.670
And so let me write another
expression over here.

00:03:12.670 --> 00:03:14.560
So it's not just sin of X over four

00:03:14.560 --> 00:03:15.720
is equal to 3/8.

00:03:15.720 --> 00:03:20.560
We could also write that
sin of pi minus X over four,

00:03:20.560 --> 00:03:22.680
'cause X over four is the theta here.

00:03:22.680 --> 00:03:27.680
Sin of pi minus X over
four is equal to 3/8.

00:03:28.680 --> 00:03:30.460
And of course we can also
use the other principle

00:03:30.460 --> 00:03:33.570
that we can add 2 pi or
subtract two pi from this,

00:03:33.570 --> 00:03:34.910
an arbitrary number of times

00:03:34.910 --> 00:03:37.540
and the sin of that will
still be equal to 3/8.

00:03:37.540 --> 00:03:39.300
So I could write it like this.

00:03:39.300 --> 00:03:44.300
Sin of pi minus X over four
plus an integer multiple

00:03:45.530 --> 00:03:49.890
of two pi, that is going
to be equal to 3/8.

00:03:49.890 --> 00:03:52.140
And if I solve both of these,

00:03:52.140 --> 00:03:54.980
the combination of them, the union of them

00:03:54.980 --> 00:03:56.990
would give me the broadest solution set.

00:03:56.990 --> 00:03:58.420
So let's do that.

00:03:58.420 --> 00:04:01.450
So over here, let me take the
inverse sin of both sides.

00:04:01.450 --> 00:04:06.450
I get X over four plus two pi N is equal

00:04:06.470 --> 00:04:11.050
to the inverse sign of 3/8.

00:04:11.050 --> 00:04:13.960
Now I could subtract two
pi N from both sides.

00:04:13.960 --> 00:04:16.530
I get X over four is equal

00:04:16.530 --> 00:04:21.530
to the inverse sin of 3/8, minus two pi N.

00:04:22.840 --> 00:04:25.330
And if you think about it,
because N can be any integer,

00:04:25.330 --> 00:04:27.050
this sin here in front of the two,

00:04:27.050 --> 00:04:28.420
this negative really doesn't matter.

00:04:28.420 --> 00:04:29.650
It could even be a positive.

00:04:29.650 --> 00:04:32.100
And now let's multiply both sides by four.

00:04:32.100 --> 00:04:36.720
We get X is equal to four
times the inverse sin

00:04:36.720 --> 00:04:41.720
of 3/8 minus eight pi N.

00:04:42.410 --> 00:04:45.430
And then if I work on this
blue part right over here,

00:04:45.430 --> 00:04:47.910
same idea, take the
inverse side of both sides.

00:04:47.910 --> 00:04:52.910
We get pi minus X over
four plus two pi N is equal

00:04:53.370 --> 00:04:57.093
to the inverse sin of 3/8.

00:04:58.100 --> 00:05:00.510
And then let's see, I can
subtract pi from both sides

00:05:00.510 --> 00:05:02.360
and subtract two pi N from both sides.

00:05:02.360 --> 00:05:06.150
And so I get negative X over four is equal

00:05:06.150 --> 00:05:11.150
to the inverse sin of 3/8
minus pi minus two pi N.

00:05:14.470 --> 00:05:16.910
And I multiply both
sides by negative four.

00:05:16.910 --> 00:05:21.910
I get X is equal to negative
four times the inverse sin

00:05:22.220 --> 00:05:27.220
of 3/8 plus four pi plus eight pi N.

00:05:29.040 --> 00:05:31.810
And as I mentioned, the
union of both of these

00:05:31.810 --> 00:05:33.700
give us the entire solution set

00:05:33.700 --> 00:05:35.570
to our original equation here.