Cosine equation algebraic solution set

WEBVTT
Kind: captions
Language: en

00:00:00.380 --> 00:00:01.470
- [Lecturer] The goal of this video

00:00:01.470 --> 00:00:04.740
is to find the solution set
for the following equation,

00:00:04.740 --> 00:00:06.760
negative six times the cosine

00:00:06.760 --> 00:00:10.640
of 8x plus four is equal to five.

00:00:10.640 --> 00:00:12.720
And like always, I encourage
you to pause this video

00:00:12.720 --> 00:00:14.380
and see if you can have a go at this

00:00:14.380 --> 00:00:15.370
before we do it together.

00:00:15.370 --> 00:00:18.680
And a reminder, we want
the entire solution set,

00:00:18.680 --> 00:00:20.123
not just one solution.

00:00:21.110 --> 00:00:23.210
All right, now let's work
through this together.

00:00:23.210 --> 00:00:24.780
Some of you might recognize

00:00:24.780 --> 00:00:29.030
that it would be valuable
to isolate the cosine of 8x,

00:00:29.030 --> 00:00:30.530
and a good way of doing that

00:00:30.530 --> 00:00:33.820
would be, first, to subtract
four from both sides,

00:00:33.820 --> 00:00:35.560
and then that would get us

00:00:35.560 --> 00:00:39.600
negative six times cosine of 8x,

00:00:39.600 --> 00:00:41.080
I subtracted four from the left,

00:00:41.080 --> 00:00:42.620
so that four is going to be gone,

00:00:42.620 --> 00:00:45.720
and then if I subtract four from the five,

00:00:45.720 --> 00:00:47.630
I am going to get a one there.

00:00:47.630 --> 00:00:50.700
And now I can multiply both sides

00:00:50.700 --> 00:00:55.700
of this equation by negative 1/6,

00:00:56.550 --> 00:00:59.040
I just wanna have a one
in front of the cosine,

00:00:59.040 --> 00:01:01.080
so negative 1/6.

00:01:01.080 --> 00:01:03.010
And so this is going to be one,

00:01:03.010 --> 00:01:06.390
so I'm just gonna have cosine of 8x

00:01:07.560 --> 00:01:10.660
is equal to negative 1/6.

00:01:10.660 --> 00:01:12.290
Now, if I just keep going,

00:01:12.290 --> 00:01:15.220
I could take the inverse
cosine of negative 1/6,

00:01:15.220 --> 00:01:17.160
and whatever that is divided by eight,

00:01:17.160 --> 00:01:19.010
I would get a solution,

00:01:19.010 --> 00:01:20.710
but this is a good time to pause

00:01:20.710 --> 00:01:24.370
and to make sure that we are
capturing all of the solutions.

00:01:24.370 --> 00:01:26.600
And I'll give us, or
I'll refresh our memories

00:01:26.600 --> 00:01:28.910
with some identities.

00:01:28.910 --> 00:01:30.130
And to help with these identities,

00:01:30.130 --> 00:01:33.023
I like to draw a quick unit circle.

00:01:34.270 --> 00:01:38.600
So this is our x-axis, this is our y-axis,

00:01:38.600 --> 00:01:43.310
and so my quick hand-drawn unit circle

00:01:45.250 --> 00:01:47.140
might look something like this, (laughing)

00:01:47.140 --> 00:01:49.020
it's not that nice looking,

00:01:49.020 --> 00:01:51.240
but we wanna think about all of the angles

00:01:51.240 --> 00:01:54.360
that when I take the cosine,
I get to negative 1/6.

00:01:54.360 --> 00:01:59.360
So negative 1/6 might be
something like right over here.

00:01:59.400 --> 00:02:02.760
And so you can see that
there might be an angle

00:02:02.760 --> 00:02:04.810
like this that would get us there,

00:02:04.810 --> 00:02:07.140
so let me draw that, draw the radius.

00:02:07.140 --> 00:02:11.610
We know the cosine of an
angle is the x-coordinate

00:02:11.610 --> 00:02:15.210
of where that radius that's
defined by that angle,

00:02:15.210 --> 00:02:17.710
where that radius
intersects the unit circle.

00:02:17.710 --> 00:02:19.790
But we also see there's another place,

00:02:19.790 --> 00:02:23.010
if we essentially take the
negative of that angle,

00:02:23.010 --> 00:02:25.420
we could go right over here

00:02:25.420 --> 00:02:27.490
and we would also get the same cosine.

00:02:27.490 --> 00:02:30.060
So we could go to the negative
of the angle, go that way.

00:02:30.060 --> 00:02:31.430
And that's where we get the identity

00:02:31.430 --> 00:02:34.040
that cosine of negative theta

00:02:34.040 --> 00:02:37.450
is equal to cosine of theta.

00:02:37.450 --> 00:02:39.480
And so if cosine of 8x

00:02:39.480 --> 00:02:41.400
is equal to negative 1/6,

00:02:41.400 --> 00:02:43.550
using this identity, we also know

00:02:43.550 --> 00:02:45.760
that cosine of the negative of this

00:02:45.760 --> 00:02:48.480
will also be equal to negative 1/6.

00:02:48.480 --> 00:02:49.530
So let me write that down,

00:02:49.530 --> 00:02:51.690
cosine of negative 8x

00:02:52.650 --> 00:02:55.723
is also going to be equal to negative 1/6.

00:02:56.700 --> 00:02:59.730
Now, already we have
expanded our solution set

00:02:59.730 --> 00:03:02.560
because this is going to
give us another x-value

00:03:02.560 --> 00:03:05.370
that's going to get us
the result that we want,

00:03:05.370 --> 00:03:06.720
but are we done?

00:03:06.720 --> 00:03:09.290
Well, the other thing to realize is,

00:03:09.290 --> 00:03:11.680
let's say I have some angle here,

00:03:11.680 --> 00:03:15.770
where if I take the cosine,
I get to negative 1/6,

00:03:15.770 --> 00:03:17.750
but then if I had two pi again,

00:03:17.750 --> 00:03:19.000
I'm gonna get to the same place,

00:03:19.000 --> 00:03:21.480
and the cosine is, once again,
going to be negative 1/6,

00:03:21.480 --> 00:03:23.260
and I could add two pi again,

00:03:23.260 --> 00:03:24.940
I could essentially add two pi

00:03:24.940 --> 00:03:27.820
an arbitrary integer number of times.

00:03:27.820 --> 00:03:32.340
So I could rewrite this
right over here as cosine,

00:03:32.340 --> 00:03:33.530
instead of just 8x,

00:03:33.530 --> 00:03:38.120
it's 8x plus an integer
multiple of two pi,

00:03:38.120 --> 00:03:41.270
that's also going to be
equal to negative 1/6.

00:03:41.270 --> 00:03:43.820
And similarly for negative 8x,

00:03:43.820 --> 00:03:48.060
I could say cosine of negative 8x

00:03:48.060 --> 00:03:50.670
plus an integer multiple of two pi,

00:03:50.670 --> 00:03:54.620
and is going to be some integer
in both of these situations,

00:03:54.620 --> 00:03:57.550
that's also going to
get us to negative 1/6.

00:03:57.550 --> 00:03:59.130
And so now we can feel pretty good

00:03:59.130 --> 00:04:00.820
that we're capturing all of the solutions

00:04:00.820 --> 00:04:02.380
when we solve for x.

00:04:02.380 --> 00:04:03.213
So in both of these,

00:04:03.213 --> 00:04:05.630
now let's take the inverse
cosine of negative 1/6

00:04:05.630 --> 00:04:08.420
in order to solve for x here.

00:04:08.420 --> 00:04:12.090
So if we were to take the
inverse cosine of both sides,

00:04:12.090 --> 00:04:15.370
we could get that 8x plus two pi

00:04:15.370 --> 00:04:17.750
times some arbitrary integer n

00:04:17.750 --> 00:04:22.553
is equal to the inverse
cosine of negative 1/6.

00:04:23.640 --> 00:04:25.560
And then now let's solve for x,

00:04:25.560 --> 00:04:28.770
we can subtract two pi n from both sides.

00:04:28.770 --> 00:04:33.770
So we could get 8x is
equal to the inverse cosine

00:04:33.920 --> 00:04:38.920
of negative 1/6 minus two pi n.

00:04:39.270 --> 00:04:40.330
Now, it's interesting to note

00:04:40.330 --> 00:04:42.300
that the sign on this two pi n term

00:04:42.300 --> 00:04:43.730
actually doesn't matter so much,

00:04:43.730 --> 00:04:45.560
'cause n could be a negative integer,

00:04:45.560 --> 00:04:48.420
but I'll just stick with
this negative two pi n.

00:04:48.420 --> 00:04:50.390
And so if we wanted to solve for x,

00:04:50.390 --> 00:04:52.410
we'd just divide both sides by eight,

00:04:52.410 --> 00:04:55.650
we get x is equal to 1/8

00:04:55.650 --> 00:04:59.890
times the inverse cosine of negative 1/6

00:04:59.890 --> 00:05:04.890
minus pi over four n.

00:05:05.080 --> 00:05:07.060
And now we can do the exact same thing

00:05:07.060 --> 00:05:10.690
in the other scenario, I'll
call this the yellow scenario,

00:05:10.690 --> 00:05:12.070
where if I take the inverse cosine,

00:05:12.070 --> 00:05:15.940
I get negative 8x plus two pi,

00:05:15.940 --> 00:05:20.073
n is equal to the inverse
cosine of negative 1/6.

00:05:21.060 --> 00:05:24.320
And now I can subtract
two pi n from both sides,

00:05:24.320 --> 00:05:25.570
so I get negative 8x

00:05:26.590 --> 00:05:31.550
is equal to inverse cosine of negative 1/6

00:05:31.550 --> 00:05:34.330
minus two pi n.

00:05:34.330 --> 00:05:36.550
Now I can multiply both
sides by negative 1/8,

00:05:36.550 --> 00:05:38.560
or divide both sides by negative eight,

00:05:38.560 --> 00:05:42.010
and I get x is equal to negative 1/8

00:05:42.010 --> 00:05:45.380
times the inverse cosine of negative 1/6

00:05:46.740 --> 00:05:50.303
plus pi over four n.

00:05:51.410 --> 00:05:53.660
So I will stop here for this video,

00:05:53.660 --> 00:05:56.610
where at least algebraically
we know the solution set,

00:05:56.610 --> 00:05:58.460
and this is the complete solution set

00:06:01.447 --> 00:06:04.220
if you take the combination
of both of these expressions.

00:06:04.220 --> 00:06:07.600
In a future video, we'll
evaluate this with a calculator,

00:06:07.600 --> 00:06:09.790
and we'll think about the solutions

00:06:09.790 --> 00:06:12.083
that fit within a given interval.


Proof of the tangent angle sum and difference identities

WEBVTT
Kind: captions
Language: en

00:00:00.150 --> 00:00:01.300
- [Instructor] In this
video I'm going to assume

00:00:01.300 --> 00:00:02.550
that you already know a few things

00:00:02.550 --> 00:00:03.383
and we've covered this.

00:00:03.383 --> 00:00:04.690
We've proved this in other videos

00:00:04.690 --> 00:00:08.810
that sine of x plus y is
equal to sine of x cosine y

00:00:08.810 --> 00:00:11.020
plus and then you swap
the cosines and the sines,

00:00:11.020 --> 00:00:13.430
cosine of x sine y,

00:00:13.430 --> 00:00:18.430
and then cosine of x plus y
is equal to cosine x cosine y

00:00:18.620 --> 00:00:20.770
minus sine x sin y.

00:00:20.770 --> 00:00:23.150
Once again, we've proven
this in this in other videos

00:00:23.150 --> 00:00:25.820
and then there's some other
properties we know of cosine

00:00:25.820 --> 00:00:27.610
and sine that we have
looked at another video's.

00:00:27.610 --> 00:00:30.972
Cosine of -x is equal to cosine of x

00:00:30.972 --> 00:00:35.450
and that sine of negative
x is equal to -sine of x.

00:00:35.450 --> 00:00:36.380
And that of course,

00:00:36.380 --> 00:00:38.540
the tangent of something is defined

00:00:38.540 --> 00:00:42.320
as a sine over cosine of that something.

00:00:42.320 --> 00:00:43.510
Now with that out of the way,

00:00:43.510 --> 00:00:44.980
I wanna come up with a formula

00:00:44.980 --> 00:00:48.380
for tangent of x plus y expressed just

00:00:48.380 --> 00:00:52.560
in terms of tangent of x and tangent of y.

00:00:52.560 --> 00:00:54.180
You can view it as the antilog

00:00:54.180 --> 00:00:56.663
for what we did up here
for sine and cosine.

00:00:57.690 --> 00:00:59.850
Well, the immediate thing
that you might recognize

00:00:59.850 --> 00:01:03.060
is that tangent of x plus
y based on the definition

00:01:03.060 --> 00:01:08.060
of tangent is the same
thing as sine of x plus y

00:01:08.210 --> 00:01:12.630
over cosine of x plus y.

00:01:12.630 --> 00:01:15.130
And what's that going to be equal to?

00:01:15.130 --> 00:01:17.350
Well, we know that sine of x plus y

00:01:17.350 --> 00:01:19.610
can be expressed this way.

00:01:19.610 --> 00:01:21.540
So let me write that down.

00:01:21.540 --> 00:01:26.447
So that's going to be sine of x cosine y

00:01:27.880 --> 00:01:32.120
plus cosine of x sine of y.

00:01:34.510 --> 00:01:36.230
And then, and actually,

00:01:36.230 --> 00:01:37.970
so that we can save a
little bit of writing,

00:01:37.970 --> 00:01:40.240
I'm gonna awkwardly write,

00:01:40.240 --> 00:01:42.400
make the line down here,

00:01:42.400 --> 00:01:44.990
because we're gonna put
something here in a second,

00:01:44.990 --> 00:01:46.100
but I think you'll get the idea.

00:01:46.100 --> 00:01:49.140
So there's going to be that
over cosine of x plus y

00:01:49.140 --> 00:01:52.060
which is this expression when
you just express it in terms

00:01:52.060 --> 00:01:55.620
of cosines of x and
cosines of y and sines of x

00:01:55.620 --> 00:01:56.550
and sines of y.

00:01:56.550 --> 00:01:57.720
So let me write it here.

00:01:57.720 --> 00:02:02.720
So you're gonna have cosine
of x cosine y minus sine of x,

00:02:07.100 --> 00:02:10.780
and then sine y.

00:02:10.780 --> 00:02:14.100
Now we wanna express
everything in terms of tangents

00:02:14.100 --> 00:02:16.080
of xs and ys.

00:02:16.080 --> 00:02:18.530
And so it might make sense here to say,

00:02:18.530 --> 00:02:22.410
all right, well, we know
tangent is sine over cosine.

00:02:22.410 --> 00:02:24.990
So what if we were to
divide both the numerator

00:02:24.990 --> 00:02:27.970
and the denominator by some
expression that can start

00:02:27.970 --> 00:02:30.110
to make the numerator
and denominator express

00:02:30.110 --> 00:02:31.700
in terms of tangents.

00:02:31.700 --> 00:02:33.950
And I will cut a little
bit to the chase here.

00:02:33.950 --> 00:02:35.400
So in the numerator,

00:02:35.400 --> 00:02:37.450
what I can do is,

00:02:37.450 --> 00:02:38.990
and I'm gonna do this
just in the numerator,

00:02:38.990 --> 00:02:40.920
and then I'm gonna do it
in the denominator as well.

00:02:40.920 --> 00:02:44.213
I'm gonna divide the numerator
by cosine of x cosine y.

00:02:47.780 --> 00:02:49.990
And of course, I can't
just divide the numerator

00:02:49.990 --> 00:02:51.570
by cosine of x cosine of y

00:02:51.570 --> 00:02:53.330
that would change the value of the,

00:02:53.330 --> 00:02:54.515
this rational expression.

00:02:54.515 --> 00:02:56.400
I have to do that to
the denominator as well.

00:02:56.400 --> 00:02:58.830
So I know this is a very
complex looking fraction here

00:02:58.830 --> 00:03:00.510
but it's going to simplify in a second.

00:03:00.510 --> 00:03:02.020
So I'm also going to
divide the denominator

00:03:02.020 --> 00:03:06.180
by cosine of x cosine of y.

00:03:06.180 --> 00:03:08.550
And now let's see if we can simplify this

00:03:08.550 --> 00:03:10.520
in certain ways.

00:03:10.520 --> 00:03:13.860
In the numerator, we can see
that this cosine y cancels

00:03:13.860 --> 00:03:15.340
with this cosine y.

00:03:15.340 --> 00:03:17.230
And so that first term becomes slightly

00:03:17.230 --> 00:03:18.220
in another color here.

00:03:18.220 --> 00:03:21.810
So this sine of x over cosine of x.

00:03:21.810 --> 00:03:23.200
And so the numerator,

00:03:23.200 --> 00:03:26.140
I can say this is going
to be equal to sine of x

00:03:26.140 --> 00:03:30.200
over cosine of x is tangent of x.

00:03:30.200 --> 00:03:32.490
And then the second term here,

00:03:32.490 --> 00:03:35.540
we can see that this cosine of x cancels

00:03:35.540 --> 00:03:36.960
with this cosine of x .

00:03:36.960 --> 00:03:41.120
So we're left with sine
of y over cosine of y,

00:03:41.120 --> 00:03:42.130
which is of course,

00:03:42.130 --> 00:03:43.620
tangent of y.

00:03:43.620 --> 00:03:47.500
So plus tangent of y

00:03:47.500 --> 00:03:49.860
and then all of that is going to be over,

00:03:49.860 --> 00:03:52.680
now we can look at the denominator.

00:03:52.680 --> 00:03:55.570
So this first term here,

00:03:55.570 --> 00:03:59.610
we can see the cosine of x
cancels with the cosine of x

00:03:59.610 --> 00:04:03.380
and the cosine of y cancels
out with the cosine of y.

00:04:03.380 --> 00:04:05.920
So you could view this first
term here when you divide

00:04:05.920 --> 00:04:08.130
by this cosine of x cosine y,

00:04:08.130 --> 00:04:10.400
it just becomes one

00:04:10.400 --> 00:04:13.240
and then we're going to have the minus.

00:04:13.240 --> 00:04:15.850
And now this second term is interesting.

00:04:15.850 --> 00:04:19.090
We have sine of x over cosine of x,

00:04:19.090 --> 00:04:20.523
sine of y over cosine of y.

00:04:21.569 --> 00:04:24.800
So sine of x over cosine of x

00:04:24.800 --> 00:04:26.910
that over there is tangent of x,

00:04:26.910 --> 00:04:31.850
and then sine of y over
cosine of y its tangent of y.

00:04:31.850 --> 00:04:36.850
So this is going to be tangent
of x times tangent of y.

00:04:38.850 --> 00:04:39.710
And just like that,

00:04:39.710 --> 00:04:41.190
we have come up with an expression

00:04:41.190 --> 00:04:44.550
for tangent of x plus y that just deals

00:04:44.550 --> 00:04:48.400
with tangent of xs and tangent of ys.

00:04:48.400 --> 00:04:49.840
Now the next question you might say,

00:04:49.840 --> 00:04:52.950
well, all right, that's
great for tangent of x plus y

00:04:52.950 --> 00:04:57.950
but what about tangent of x minus y?

00:04:58.250 --> 00:05:00.830
Well, here we just have
to recognize a little bit

00:05:00.830 --> 00:05:02.690
of what we've seen before.

00:05:02.690 --> 00:05:03.730
Let me write it over here.

00:05:03.730 --> 00:05:08.670
Tangent of -x is equal to sine of -x

00:05:10.790 --> 00:05:13.830
over cosine of negative x

00:05:13.830 --> 00:05:15.990
and what's that going to be equal to?

00:05:15.990 --> 00:05:17.600
And I know I'm running out of space.

00:05:17.600 --> 00:05:20.120
This is going to be equal to
sine of -x is the same thing

00:05:20.120 --> 00:05:24.360
as -sine of x, -sine of x,

00:05:24.360 --> 00:05:27.163
and then cosine of -x is just cosine of x.

00:05:28.490 --> 00:05:32.350
Well, this is just the
negative of the tangent of x.

00:05:32.350 --> 00:05:37.340
So this is negative tangent of x.

00:05:37.340 --> 00:05:40.460
And the reason why that is useful

00:05:40.460 --> 00:05:43.020
is I can rewrite this as being,

00:05:43.020 --> 00:05:43.853
write here.

00:05:43.853 --> 00:05:48.710
This is the same thing
as tangent of x plus -y.

00:05:50.950 --> 00:05:52.740
So everywhere we saw a y here,

00:05:52.740 --> 00:05:54.410
we can replace it with a -y.

00:05:54.410 --> 00:05:59.407
So this is going to be
equal to tangent of x

00:06:00.500 --> 00:06:05.500
plus the tangent of -y,

00:06:05.510 --> 00:06:10.510
all of that over 1 minus the tangent of x

00:06:11.900 --> 00:06:14.313
times the tangent of -y.

00:06:15.520 --> 00:06:16.900
Well, we know the tangent of -y

00:06:16.900 --> 00:06:21.900
is the same thing as the
negative tangent of y.

00:06:21.950 --> 00:06:24.070
And we know that over here as well.

00:06:24.070 --> 00:06:25.070
So this could just be,

00:06:25.070 --> 00:06:26.980
we could write the tangent of y here,

00:06:26.980 --> 00:06:30.010
and then the negative would
turn this into a plus.

00:06:30.010 --> 00:06:33.290
And so just to write everything neatly,

00:06:33.290 --> 00:06:38.290
we know that also the tangent of x minus y

00:06:38.300 --> 00:06:43.300
can be rewritten as tangent
of x minus tangent of y

00:06:46.010 --> 00:06:51.010
all of that over 1 plus
tangent of x and tangent of y.


Interpreting statements about vectors

WEBVTT
Kind: captions
Language: en

00:00:00.240 --> 00:00:01.770
- [Instructor] We're told
that particles A and B

00:00:01.770 --> 00:00:03.780
are moving along a plane.

00:00:03.780 --> 00:00:06.750
Their velocities are
represented by the vectors,

00:00:06.750 --> 00:00:09.490
vector a and vector b respectively.

00:00:09.490 --> 00:00:12.080
Which option best describes the meaning

00:00:12.080 --> 00:00:13.830
of the following statement.

00:00:13.830 --> 00:00:14.820
Choose one answer.

00:00:14.820 --> 00:00:17.340
So pause this video and try to
work through this on your own

00:00:17.340 --> 00:00:19.190
before we work through this together.

00:00:20.090 --> 00:00:22.810
All right, now let's work
through this together.

00:00:22.810 --> 00:00:26.190
So this is saying that
the magnitude of vector a

00:00:26.190 --> 00:00:30.180
is equal to the magnitude of vector b.

00:00:30.180 --> 00:00:31.730
So we know that a vector is specified

00:00:31.730 --> 00:00:34.090
by both a magnitude and a direction.

00:00:34.090 --> 00:00:36.870
And this is just saying that
the magnitudes are the same.

00:00:36.870 --> 00:00:41.870
So for example, vector
a could look like this,

00:00:42.100 --> 00:00:44.850
and vector b could look like this.

00:00:44.850 --> 00:00:48.490
It could do something like that

00:00:48.490 --> 00:00:52.070
where it has the same magnitude
and the same direction.

00:00:52.070 --> 00:00:55.700
Or vector b might be in a
completely different direction.

00:00:55.700 --> 00:00:57.213
The magnitudes being
equivalent just tells us

00:00:57.213 --> 00:00:59.920
that the length of these
arrows are the same,

00:00:59.920 --> 00:01:02.320
but we could have
different directions here.

00:01:02.320 --> 00:01:05.960
So what this tells me is
that we have the same speed

00:01:05.960 --> 00:01:07.830
which is the magnitude of velocity,

00:01:07.830 --> 00:01:10.670
but not necessarily the same direction.

00:01:10.670 --> 00:01:12.200
Now let's look at the choices here.

00:01:12.200 --> 00:01:14.540
The first choice is
that two particles move

00:01:14.540 --> 00:01:18.410
at the same speed and
in the same direction.

00:01:18.410 --> 00:01:21.430
So we've already said that
that's not necessarily the case.

00:01:21.430 --> 00:01:23.530
In order for choice A to be correct,

00:01:23.530 --> 00:01:26.330
they would essentially have
to be equivalent vectors.

00:01:26.330 --> 00:01:29.130
Choice A would be the case
if we said that vector a

00:01:29.130 --> 00:01:31.140
is equal to vector b,

00:01:31.140 --> 00:01:33.390
then they would have to
have the same magnitude

00:01:33.390 --> 00:01:34.550
and the same direction,

00:01:34.550 --> 00:01:37.270
the same magnitude and the same direction.

00:01:37.270 --> 00:01:38.310
But that's not what they told us.

00:01:38.310 --> 00:01:40.520
They just told us that the
magnitudes are the same,

00:01:40.520 --> 00:01:42.420
not necessarily the directions.

00:01:42.420 --> 00:01:44.130
So I'll rule that one out.

00:01:44.130 --> 00:01:45.950
The two particles move at the same speed,

00:01:45.950 --> 00:01:48.980
but not necessarily in the same direction.

00:01:48.980 --> 00:01:50.850
Yes, that's what we just talked about.

00:01:50.850 --> 00:01:51.990
They have the same speed,

00:01:51.990 --> 00:01:54.070
which is the magnitude of velocity,

00:01:54.070 --> 00:01:55.990
but they didn't tell us
anything about the direction,

00:01:55.990 --> 00:01:57.300
just the magnitudes.

00:01:57.300 --> 00:02:00.250
So I like this choice, but
let's look at choice C.

00:02:00.250 --> 00:02:03.040
The two particles move
in the same direction,

00:02:03.040 --> 00:02:06.240
but not necessarily at the same speed.

00:02:06.240 --> 00:02:08.230
Well, we know they move at the same speed.

00:02:08.230 --> 00:02:09.910
That's what this is telling us.

00:02:09.910 --> 00:02:11.610
The magnitudes are the same.

00:02:11.610 --> 00:02:13.620
We just don't know anything
about the direction.

00:02:13.620 --> 00:02:16.610
So I would rule this one out as well.

00:02:16.610 --> 00:02:18.890
In order for choice C to be the case,

00:02:18.890 --> 00:02:21.120
you would see something like this,

00:02:21.120 --> 00:02:24.160
maybe that is vector a right here,

00:02:24.160 --> 00:02:27.610
and then vector b would
move in the same direction,

00:02:27.610 --> 00:02:29.880
but it would have a different magnitude.

00:02:29.880 --> 00:02:31.810
And here you would visualize the magnitude

00:02:31.810 --> 00:02:33.140
as the length of the arrow.

00:02:33.140 --> 00:02:34.560
But that's not what they told us.

00:02:34.560 --> 00:02:37.003
They told us this right over there.


Representing quantities with vectors

WEBVTT
Kind: captions
Language: en

00:00:00.380 --> 00:00:01.860
- [Instructor] We're
told a powerful magnet

00:00:01.860 --> 00:00:05.410
is attracting a metal
ball on a flat surface.

00:00:05.410 --> 00:00:07.530
The magnet is pulling the ball

00:00:07.530 --> 00:00:10.120
at a force of 15 Newtons,

00:00:10.120 --> 00:00:13.890
and the magnet is 20 degrees to the south

00:00:13.890 --> 00:00:18.280
from the eastward direction
relative to the ball.

00:00:18.280 --> 00:00:21.500
Here are a few vectors where
the magnitude of vector A

00:00:21.500 --> 00:00:23.510
is equal to the magnitude of vector C

00:00:23.510 --> 00:00:25.230
is equal to 15 Newtons,

00:00:25.230 --> 00:00:27.180
and the magnitude of vector B is equal to

00:00:27.180 --> 00:00:30.530
the magnitude of vector D
which is equal to 20 Newtons.

00:00:30.530 --> 00:00:35.130
Which vectors can represent
the force of the team's pull?

00:00:35.130 --> 00:00:36.580
All right, pause this video and see

00:00:36.580 --> 00:00:38.300
if you can think about that on your own

00:00:38.300 --> 00:00:40.370
before we do it together.

00:00:40.370 --> 00:00:41.810
All right, now let's do it together.

00:00:41.810 --> 00:00:43.180
So before I even look at this,

00:00:43.180 --> 00:00:44.740
I'm just gonna look at the description.

00:00:44.740 --> 00:00:47.520
It has a magnitude of 15 Newtons.

00:00:47.520 --> 00:00:48.660
If we're talking about a force,

00:00:48.660 --> 00:00:51.800
you can view it as a
strength of 15 Newtons,

00:00:51.800 --> 00:00:54.760
and the magnet which
is pulling on the ball

00:00:54.760 --> 00:00:56.890
is 20 degrees to the south

00:00:56.890 --> 00:00:59.730
from the eastward direction
relative to the ball.

00:00:59.730 --> 00:01:01.800
So if this is the ball right over here,

00:01:01.800 --> 00:01:04.203
and if this is the eastward direction,

00:01:05.170 --> 00:01:07.570
it says that the magnet
is 20 degrees to the south

00:01:07.570 --> 00:01:10.010
from the eastward direction
relative to the ball.

00:01:10.010 --> 00:01:12.893
So the magnet would be in this direction,

00:01:13.780 --> 00:01:17.030
and this angle right
over here is 20 degrees,

00:01:17.030 --> 00:01:19.130
and the magnet is pulling on the ball,

00:01:19.130 --> 00:01:22.280
so the vector would go in that
direction towards the magnet,

00:01:22.280 --> 00:01:25.270
and we know it has a force of 15 Newtons,

00:01:25.270 --> 00:01:26.330
that's the magnitude.

00:01:26.330 --> 00:01:30.550
So it has to be a 15 Newton
magnitude right over here.

00:01:30.550 --> 00:01:32.770
So when we look at the choices,

00:01:32.770 --> 00:01:34.810
choice A is interesting,

00:01:34.810 --> 00:01:36.590
at least the direction looks right.

00:01:36.590 --> 00:01:39.520
It looks like it's 20
degrees south of due east,

00:01:39.520 --> 00:01:43.360
and they also tell us that the
magnitude of A is 15 Newtons.

00:01:43.360 --> 00:01:46.650
So I am liking A, now let's look at B.

00:01:46.650 --> 00:01:50.160
Well B looks 15 degrees south of due east,

00:01:50.160 --> 00:01:51.540
not 20 degrees south,

00:01:51.540 --> 00:01:52.860
so I will rule that out.

00:01:52.860 --> 00:01:56.330
And also B's magnitude is
wrong, it's 20 Newtons.

00:01:56.330 --> 00:01:59.880
C, the magnitude is
right, it's 15 Newtons,

00:01:59.880 --> 00:02:03.450
but the direction looks like
20 degrees north of due east.

00:02:03.450 --> 00:02:04.920
So I'll rule that one out.

00:02:04.920 --> 00:02:06.810
And last but not least, D,

00:02:06.810 --> 00:02:08.560
the direction is clearly wrong,

00:02:08.560 --> 00:02:10.850
it looks like 15 degrees
north of due east,

00:02:10.850 --> 00:02:14.150
and its magnitude is 20
Newtons, not 15 Newtons,

00:02:14.150 --> 00:02:15.453
so I'd rule that one out.

00:02:16.450 --> 00:02:19.290
Now to be clear a vector is only defined

00:02:19.290 --> 00:02:21.040
by its magnitude and its direction,

00:02:21.040 --> 00:02:22.710
not by its starting point.

00:02:22.710 --> 00:02:24.120
So if I had some other vector

00:02:24.120 --> 00:02:26.550
that looked like this right over here,

00:02:26.550 --> 00:02:28.690
that had the same magnitude and direction

00:02:28.690 --> 00:02:31.920
if this was right over
here, a 20 degree angle

00:02:31.920 --> 00:02:33.910
and it had a magnitude of 50 Newtons,

00:02:33.910 --> 00:02:36.420
then I would have
selected this one as well.

00:02:36.420 --> 00:02:38.830
You can shift a vector around like this

00:02:38.830 --> 00:02:41.060
as long as it has the same magnitude

00:02:41.060 --> 00:02:42.560
and it has the same direction,

00:02:42.560 --> 00:02:44.433
it is an equivalent vector.


Introduction to vector components

WEBVTT
Kind: captions
Language: en

00:00:00.630 --> 00:00:02.090
- [Instructor] In other
videos, we have talked

00:00:02.090 --> 00:00:04.100
about how a vector can
be completely defined

00:00:04.100 --> 00:00:06.880
by a magnitude and a
direction, you need both.

00:00:06.880 --> 00:00:08.260
And here we have done that.

00:00:08.260 --> 00:00:09.880
We have said that the magnitude

00:00:09.880 --> 00:00:12.570
of vector a is equal to three units,

00:00:12.570 --> 00:00:15.190
these parallel lines here on both sides,

00:00:15.190 --> 00:00:17.170
it looks like a double absolute value.

00:00:17.170 --> 00:00:19.090
That means the magnitude of vector a.

00:00:19.090 --> 00:00:23.150
And you can also specify
that visually by making sure

00:00:23.150 --> 00:00:26.200
that the length of this vector
arrow is three units long.

00:00:26.200 --> 00:00:27.560
And we also have its direction.

00:00:27.560 --> 00:00:29.610
We see the direction of
vector a is 30 degrees

00:00:29.610 --> 00:00:32.270
counter-clockwise of due East.

00:00:32.270 --> 00:00:34.860
Now in this video, we're
gonna talk about other ways

00:00:34.860 --> 00:00:38.220
or another way to specify
or to define a vector.

00:00:38.220 --> 00:00:41.050
And that's by using components.

00:00:41.050 --> 00:00:42.530
And the way that we're gonna do it is,

00:00:42.530 --> 00:00:44.100
we're gonna think about the tail

00:00:44.100 --> 00:00:47.300
of this vector and the
head of this vector.

00:00:47.300 --> 00:00:50.450
And think about as we go
from the tail to the head,

00:00:50.450 --> 00:00:53.990
what is our change in x?

00:00:53.990 --> 00:00:55.180
And we could see our change

00:00:55.180 --> 00:00:58.340
in x would be that right over there.

00:00:58.340 --> 00:01:00.980
We're going from this x
value to this x value.

00:01:00.980 --> 00:01:05.370
And then what is going
to be our change in y.

00:01:05.370 --> 00:01:07.980
And if we're going from
down here to up here,

00:01:07.980 --> 00:01:12.310
our change in y, we can
also specify like that.

00:01:12.310 --> 00:01:13.500
So let me label these.

00:01:13.500 --> 00:01:18.500
This is my change in x, and
then this is my change in y.

00:01:19.060 --> 00:01:19.920
And if you think about it,

00:01:19.920 --> 00:01:22.780
if someone told you your
change in x and change in y,

00:01:22.780 --> 00:01:25.390
you could reconstruct this
vector right over here

00:01:25.390 --> 00:01:27.490
by starting here, having that change in x,

00:01:27.490 --> 00:01:31.200
then having the change in y
and then defining where the tip

00:01:31.200 --> 00:01:34.740
of the vector would be
relative to the tail.

00:01:34.740 --> 00:01:38.800
The notation for this is
we would say that vector a

00:01:38.800 --> 00:01:42.870
is equal to, and we'll have parenthesis,

00:01:42.870 --> 00:01:46.290
and we'll have our change
in x comma, change in y.

00:01:46.290 --> 00:01:47.780
And so if we wanted to get tangible

00:01:47.780 --> 00:01:50.340
for this particular
vector right over here,

00:01:50.340 --> 00:01:53.550
we know the length of
this vector is three.

00:01:53.550 --> 00:01:55.540
Its magnitude is three.

00:01:55.540 --> 00:01:58.350
We know that this is, since
this is going due horizontally

00:01:58.350 --> 00:02:00.290
and then this is going
straight up and down.

00:02:00.290 --> 00:02:02.420
This is a right triangle.

00:02:02.420 --> 00:02:05.170
And so we can use a little
bit of geometry from the past.

00:02:05.170 --> 00:02:08.020
Don't worry if you need a little
bit of a refresher on this,

00:02:08.020 --> 00:02:09.620
but we could use a little bit of geometry,

00:02:09.620 --> 00:02:11.490
or a little bit of
trigonometry to establish,

00:02:11.490 --> 00:02:13.610
if we know this angle,
if we know the length

00:02:13.610 --> 00:02:17.210
of this hypotenuse, that
this side that's opposite

00:02:17.210 --> 00:02:20.180
the 30 degree angle is gonna
be half the hypotenuse,

00:02:20.180 --> 00:02:22.020
so it's going to be 3/2.

00:02:22.020 --> 00:02:24.200
And that the change in x is going to be

00:02:24.200 --> 00:02:26.960
the square root of three times the 3/2.

00:02:26.960 --> 00:02:31.080
So it's going to be three,
square roots of three over two.

00:02:31.080 --> 00:02:33.980
And so up here, we would
write our x component

00:02:33.980 --> 00:02:37.680
is three times the square
root of three over two.

00:02:37.680 --> 00:02:42.420
And we would write that
the y component is 3/2.

00:02:42.420 --> 00:02:43.820
Now I know a lot of you might be thinking

00:02:43.820 --> 00:02:47.260
this looks a lot like coordinates
in the coordinate plane,

00:02:47.260 --> 00:02:48.580
where this would be the x coordinate

00:02:48.580 --> 00:02:50.300
and this would be the y coordinate.

00:02:50.300 --> 00:02:51.970
But when you're dealing with vectors,

00:02:51.970 --> 00:02:54.610
that's not exactly the interpretation.

00:02:54.610 --> 00:02:57.000
It is the case that if the vector's tail

00:02:57.000 --> 00:03:00.860
were at the origin right
over here, then its head

00:03:00.860 --> 00:03:04.670
would be at these coordinates
on the coordinate plane.

00:03:04.670 --> 00:03:07.470
But we know that a vector is not defined

00:03:07.470 --> 00:03:10.180
by its position, by the
position of the tail.

00:03:10.180 --> 00:03:12.200
I could shift this vector around wherever

00:03:12.200 --> 00:03:13.840
and it would still be the same vector.

00:03:13.840 --> 00:03:15.590
It can start wherever.

00:03:15.590 --> 00:03:19.000
So when you use this
notation in a vector context,

00:03:19.000 --> 00:03:21.440
these aren't x coordinates
and y coordinates.

00:03:21.440 --> 00:03:26.440
This is our change in x,
and this is our change in y.

00:03:27.070 --> 00:03:28.480
Let me do one more example to show

00:03:28.480 --> 00:03:30.880
that we can actually go the other way.

00:03:30.880 --> 00:03:34.790
So let's say I defined some vector b,

00:03:34.790 --> 00:03:39.200
and let's say that its x
component is square root of two.

00:03:39.200 --> 00:03:43.520
And let's say that its y
component is square root of two.

00:03:43.520 --> 00:03:46.260
So let's think about what
that vector would look like.

00:03:46.260 --> 00:03:49.380
So it would, if this is its tail,

00:03:49.380 --> 00:03:51.410
and its x component which is its change

00:03:51.410 --> 00:03:53.030
in x is square root of two.

00:03:53.030 --> 00:03:55.460
So it might look something like this.

00:03:55.460 --> 00:04:00.460
So that would be change in x
is equal to square root of two.

00:04:00.800 --> 00:04:03.980
And then its y component would
also be square root of two.

00:04:03.980 --> 00:04:07.230
So I could write our change in y over here

00:04:07.230 --> 00:04:08.970
is square root of two.

00:04:08.970 --> 00:04:12.850
And so the vector would
look something like this.

00:04:12.850 --> 00:04:17.850
It would start here and
then it would go over here,

00:04:18.580 --> 00:04:20.590
and we can use a little bit of geometry

00:04:20.590 --> 00:04:21.980
to figure out the magnitude

00:04:21.980 --> 00:04:24.260
and the direction of this vector.

00:04:24.260 --> 00:04:26.760
You can use the Pythagorean
theorem to establish

00:04:26.760 --> 00:04:28.760
that this squared plus this squared

00:04:28.760 --> 00:04:30.410
is gonna be equal to that squared.

00:04:30.410 --> 00:04:32.380
And if you do that,
you're going to get this

00:04:32.380 --> 00:04:34.510
having a length of two, which tells you

00:04:34.510 --> 00:04:39.370
that the magnitude of
vector b is equal to two.

00:04:39.370 --> 00:04:42.420
And if you wanted to figure
out this angle right over here,

00:04:42.420 --> 00:04:43.870
you could do a little bit of trigonometry

00:04:43.870 --> 00:04:46.110
or even a little bit
of geometry recognizing

00:04:46.110 --> 00:04:49.500
that this is going to be a
right angle right over here,

00:04:49.500 --> 00:04:52.130
and that this side and that
side have the same length.

00:04:52.130 --> 00:04:53.410
So these are gonna be the same angles

00:04:53.410 --> 00:04:55.600
which are gonna be 45 degree angles.

00:04:55.600 --> 00:04:58.690
And so just like that, you could
also specify the direction,

00:04:58.690 --> 00:05:02.770
45 degrees counter-clockwise of due East.

00:05:02.770 --> 00:05:05.360
So hopefully you appreciate
that these are equivalent ways

00:05:05.360 --> 00:05:06.540
of representing a vector.

00:05:06.540 --> 00:05:08.950
You either can have a
magnitude and a direction,

00:05:08.950 --> 00:05:10.200
or you can have your components

00:05:10.200 --> 00:05:12.350
and you can go back and
forth between the two.

00:05:12.350 --> 00:05:15.283
And we'll get more practice
of that in future videos.


Personal Pronouns

WEBVTT
Kind: captions
Language: en

00:00:00.660 --> 00:00:02.190
- [Instructor] Hello, grammarians,

00:00:02.190 --> 00:00:04.140
let's talk about personal pronouns,

00:00:04.140 --> 00:00:07.040
but first let me lay
some sentences on you.

00:00:07.040 --> 00:00:10.050
Jake and I baked a loaf of bread.

00:00:10.050 --> 00:00:12.450
We baked a loaf of bread.

00:00:12.450 --> 00:00:14.053
You can learn anything.

00:00:14.940 --> 00:00:17.140
My friends are cool.

00:00:17.140 --> 00:00:18.890
They are cool.

00:00:18.890 --> 00:00:20.650
Now I'm gonna circle a few of these words,

00:00:20.650 --> 00:00:23.120
so the ones I wrote in yellow,
and point them out to you.

00:00:23.120 --> 00:00:24.650
I, we,

00:00:24.650 --> 00:00:27.540
you, my, they,

00:00:27.540 --> 00:00:29.130
these are personal pronouns.

00:00:29.130 --> 00:00:30.500
They're pronouns that change

00:00:30.500 --> 00:00:32.210
depending on how you're using them.

00:00:32.210 --> 00:00:34.780
On whether you're using them
as the object of a sentence,

00:00:34.780 --> 00:00:37.580
as the subject to show
ownership, and so on.

00:00:37.580 --> 00:00:39.220
But we'll get to those in later videos.

00:00:39.220 --> 00:00:40.730
For now, I'm gonna talk about

00:00:40.730 --> 00:00:44.290
the three basic types of English pronoun.

00:00:44.290 --> 00:00:47.100
Broadly speaking, there
are pronouns about me,

00:00:47.100 --> 00:00:48.840
pronouns about you,

00:00:48.840 --> 00:00:52.190
and pronouns about
something or someone else.

00:00:52.190 --> 00:00:54.610
This is an idea called grammatical person.

00:00:54.610 --> 00:00:56.950
Pronouns about me are first person,

00:00:56.950 --> 00:00:59.230
pronouns about you are second person,

00:00:59.230 --> 00:01:01.630
and pronouns about
something or someone else

00:01:01.630 --> 00:01:03.410
are third person.

00:01:03.410 --> 00:01:06.750
So, when I say I love my dog, Phryne.

00:01:06.750 --> 00:01:09.450
I is a first person pronoun.

00:01:09.450 --> 00:01:10.283
In the sentence,

00:01:10.283 --> 00:01:13.720
she is an excellent dog,
where she subs out for Phryne,

00:01:13.720 --> 00:01:16.860
she is a third person pronoun.

00:01:16.860 --> 00:01:18.420
Here is a picture of Phryne,

00:01:18.420 --> 00:01:21.040
I think we can all agree, she is perfect.

00:01:21.040 --> 00:01:21.873
Thank you.

00:01:21.873 --> 00:01:22.940
Okay. So what I want to do here

00:01:22.940 --> 00:01:24.330
is fill out this table

00:01:24.330 --> 00:01:25.970
with some of the basic pronouns

00:01:25.970 --> 00:01:27.730
we use to talk about ourselves

00:01:27.730 --> 00:01:30.560
divided between singular,
that is one person,

00:01:30.560 --> 00:01:33.600
and plural, or more than one person.

00:01:33.600 --> 00:01:37.990
Some first person pronouns
are I, me, my, and mine.

00:01:37.990 --> 00:01:39.780
But what if there's more than one of me?

00:01:39.780 --> 00:01:40.920
What if I'm part of a group

00:01:40.920 --> 00:01:43.280
and I wanna refer to that group?

00:01:43.280 --> 00:01:44.840
Well, then I'd use a plural pronoun

00:01:44.840 --> 00:01:48.290
like we, us, our, or ours.

00:01:48.290 --> 00:01:49.960
Second and third person are interesting

00:01:49.960 --> 00:01:52.750
because they have pronouns
that pull double duty.

00:01:52.750 --> 00:01:53.850
In second person,

00:01:53.850 --> 00:01:56.950
the singular and plural are identical.

00:01:56.950 --> 00:02:00.940
Singular second person
is you, your, and yours.

00:02:00.940 --> 00:02:02.760
And plural second person is the same,

00:02:02.760 --> 00:02:04.940
you, your, and yours.

00:02:04.940 --> 00:02:06.810
That is to say, it's
the same whether or not

00:02:06.810 --> 00:02:09.850
you're referring to one
person, here in singular,

00:02:09.850 --> 00:02:12.420
or many people, here in plural.

00:02:12.420 --> 00:02:14.570
Understanding whether you meet one you

00:02:14.570 --> 00:02:16.910
or a plural you depends on context.

00:02:16.910 --> 00:02:19.900
And it's usually very easy to figure out.

00:02:19.900 --> 00:02:22.870
Third person pronouns belong
to the most crowded category

00:02:22.870 --> 00:02:25.460
because the world is
full of things and people

00:02:25.460 --> 00:02:27.290
who are neither me nor you.

00:02:27.290 --> 00:02:30.750
For singular pronouns we
have she, her, and hers,

00:02:30.750 --> 00:02:33.080
he, him, and his,

00:02:33.080 --> 00:02:35.210
they, them, their and theirs

00:02:35.210 --> 00:02:37.660
and it, and its.

00:02:37.660 --> 00:02:41.100
They, like you, can refer
to both a single person

00:02:41.100 --> 00:02:42.140
or multiple people,

00:02:42.140 --> 00:02:45.570
but it doesn't specify a
gender like she and he do.

00:02:45.570 --> 00:02:47.190
This is extremely useful.

00:02:47.190 --> 00:02:49.010
I'd take note that the word it

00:02:49.010 --> 00:02:50.950
only refers to inanimate objects

00:02:50.950 --> 00:02:52.820
and sometimes non-human animals,

00:02:52.820 --> 00:02:54.370
but never to people.

00:02:54.370 --> 00:02:55.690
A robot? Yes.

00:02:55.690 --> 00:02:57.303
A person? Not so much.

00:02:58.180 --> 00:02:59.940
Now, in the plural category, much simpler,

00:02:59.940 --> 00:03:03.930
we have they, them, their, and theirs.

00:03:03.930 --> 00:03:06.430
Similar to singular versus plural you

00:03:06.430 --> 00:03:09.220
singular versus plural
they depends on context

00:03:09.220 --> 00:03:11.400
and it'll be obvious
from the words around it

00:03:11.400 --> 00:03:12.900
which one you mean.

00:03:12.900 --> 00:03:15.280
This is a lot of information to swallow.

00:03:15.280 --> 00:03:17.410
Pause this if you need to practice saying

00:03:17.410 --> 00:03:19.210
the different pronouns aloud,

00:03:19.210 --> 00:03:22.420
do the exercises on the
Khan Academy site or app.

00:03:22.420 --> 00:03:24.380
This is not the only time I'll be talking

00:03:24.380 --> 00:03:26.330
about personal pronouns in this course.

00:03:26.330 --> 00:03:28.480
So we have an opportunity to go deeper,

00:03:28.480 --> 00:03:30.660
especially if you want to
know more about the history

00:03:30.660 --> 00:03:33.623
of the singular use of
they, which is super cool.

00:03:34.490 --> 00:03:35.540
In the meantime,

00:03:35.540 --> 00:03:38.240
please enjoy this second
image of my dog, Phryne,

00:03:38.240 --> 00:03:41.930
who is, again, as I said, a perfect dog.

00:03:41.930 --> 00:03:43.540
You can learn anything.

00:03:43.540 --> 00:03:44.373
David out.


Cosine, sine and tangent of π/6 and π/3

WEBVTT
Kind: captions
Language: en

00:00:00.330 --> 00:00:01.290
- [Instructor] In this video,

00:00:01.290 --> 00:00:05.870
we're going to figure out what
the sine, cosine and tangent

00:00:05.870 --> 00:00:07.640
of two very important angles are.

00:00:07.640 --> 00:00:10.000
Angles that you will see a
lot in your trigonometric

00:00:10.000 --> 00:00:12.610
and just in general in your regular life.

00:00:12.610 --> 00:00:13.860
So these are the angles,

00:00:13.860 --> 00:00:17.720
pi over 3 radians and pi over 6 radians.

00:00:17.720 --> 00:00:21.540
And sometimes it's useful to
visualize them as degrees.

00:00:21.540 --> 00:00:25.030
pi over 3, you might remember
pi radians is 180 degrees,

00:00:25.030 --> 00:00:26.690
so you divide that by three,

00:00:26.690 --> 00:00:28.111
this is equivalent to 60 degrees.

00:00:28.111 --> 00:00:30.760
And once again, 180 degrees,

00:00:30.760 --> 00:00:32.990
which is the same thing as
pi radians divided by six

00:00:32.990 --> 00:00:35.500
is the same thing as 30 degrees.

00:00:35.500 --> 00:00:38.470
Now, I'm going to do it using
the unit circle definition

00:00:38.470 --> 00:00:40.430
of trig functions.

00:00:40.430 --> 00:00:41.480
But to help us there,

00:00:41.480 --> 00:00:43.160
I'm going to give us a
little bit of a reminder

00:00:43.160 --> 00:00:45.010
of what some of you might be familiar with

00:00:45.010 --> 00:00:46.980
as 30, 60, 90 triangles,

00:00:46.980 --> 00:00:48.660
which I guess we could
also call pi over six,

00:00:48.660 --> 00:00:51.250
pi over three, pi over two triangles.

00:00:51.250 --> 00:00:53.420
And so let me just draw one

00:00:53.420 --> 00:00:55.400
because this is going to be really helpful

00:00:55.400 --> 00:00:57.960
in establishing these trig functions

00:00:57.960 --> 00:01:00.520
using the unit circle definition.

00:01:00.520 --> 00:01:03.190
So let me draw a triangle here,

00:01:03.190 --> 00:01:04.180
it's hand drawn,

00:01:04.180 --> 00:01:06.190
so it's not as neat as it could be.

00:01:06.190 --> 00:01:08.840
So this right over here is a right angle,

00:01:08.840 --> 00:01:13.430
and let's say that this one
is pi over three radians

00:01:13.430 --> 00:01:15.640
which is the same thing as 60 degrees,

00:01:15.640 --> 00:01:18.890
and this one over here
is pi over six radians

00:01:18.890 --> 00:01:21.890
which is the same thing as 30 degrees.

00:01:21.890 --> 00:01:24.330
Now, let's also say that the longest side,

00:01:24.330 --> 00:01:26.453
the hypotenuse here has length one.

00:01:27.460 --> 00:01:29.850
Now, to help us think about
what the other two sides are,

00:01:29.850 --> 00:01:32.650
what I'm going to do is
flip this triangle over

00:01:32.650 --> 00:01:33.960
this side right over here,

00:01:33.960 --> 00:01:36.980
and essentially construct a mirror image.

00:01:36.980 --> 00:01:41.150
So because this right over
here is a mirror image,

00:01:41.150 --> 00:01:42.780
we immediately know a few things.

00:01:42.780 --> 00:01:44.600
We know that this length right over here

00:01:44.600 --> 00:01:48.040
is going to be congruent
to this length over here.

00:01:48.040 --> 00:01:51.530
And let me actually finish
drawing the entire triangle,

00:01:51.530 --> 00:01:53.920
it's going to look something like this.

00:01:53.920 --> 00:01:56.350
And since once again, it's a reflection,

00:01:56.350 --> 00:01:59.120
this length over here is
going to have length one,

00:01:59.120 --> 00:02:03.250
this is going to be pi over six radians,

00:02:03.250 --> 00:02:07.610
this is going to be pi over three radians.

00:02:07.610 --> 00:02:11.350
So what else do we know about
this larger triangle now?

00:02:11.350 --> 00:02:13.470
Well, we know it's an
equilateral triangle.

00:02:13.470 --> 00:02:15.610
All the angles, pi over three radians,

00:02:15.610 --> 00:02:16.443
pi over three radians

00:02:16.443 --> 00:02:18.080
and if you add two pi over sixes together,

00:02:18.080 --> 00:02:19.440
you're going to get pi over three as well,

00:02:19.440 --> 00:02:22.540
so it's a 60 degree, 60
degree, 60 degree triangle.

00:02:22.540 --> 00:02:24.790
And so all the sides are
going to have the same length,

00:02:24.790 --> 00:02:27.000
so it's going to be one, one and one.

00:02:27.000 --> 00:02:29.500
And if these two sides are congruent

00:02:29.500 --> 00:02:30.960
of the smaller triangles,

00:02:30.960 --> 00:02:32.600
of the smaller right triangles,

00:02:32.600 --> 00:02:35.310
well then this right over
here must be one half,

00:02:35.310 --> 00:02:39.770
and then this right over here
must be one half as well.

00:02:39.770 --> 00:02:41.310
Now, that's going to be useful

00:02:41.310 --> 00:02:44.900
for figuring out what this
length right over here

00:02:44.900 --> 00:02:45.960
is going to be.

00:02:45.960 --> 00:02:48.450
Because we have two right triangles,

00:02:48.450 --> 00:02:49.360
we could use either one,

00:02:49.360 --> 00:02:51.600
but if we just use this
bottom right triangle here

00:02:51.600 --> 00:02:53.530
the Pythagorean theorem tells us

00:02:53.530 --> 00:02:56.700
that one half squared, let's call this B,

00:02:56.700 --> 00:02:58.550
so plus B squared,

00:02:58.550 --> 00:02:59.990
I'm just pattern matching,

00:02:59.990 --> 00:03:02.440
A squared plus B squared
is equal to C squared

00:03:02.440 --> 00:03:04.660
where C is the length of the hypotenuse,

00:03:04.660 --> 00:03:06.650
is equal to one squared.

00:03:06.650 --> 00:03:11.270
And so we get that one
fourth plus B squared

00:03:11.270 --> 00:03:12.580
is equal to one

00:03:12.580 --> 00:03:14.970
or subtracting one fourth from both sides.

00:03:14.970 --> 00:03:17.540
B squared is equal to three-fourths,

00:03:17.540 --> 00:03:19.580
and then taking the
principle root of both sides,

00:03:19.580 --> 00:03:23.750
we get B is equal to the
square root of three over two.

00:03:23.750 --> 00:03:25.460
So just like that, we have figured out

00:03:25.460 --> 00:03:28.870
what all the lengths of this
30, 60, 90 triangle are.

00:03:28.870 --> 00:03:33.240
So B here is equal to square
root of three over two.

00:03:33.240 --> 00:03:35.200
Now, I said this would
be useful as we go into

00:03:35.200 --> 00:03:38.150
the unit circle definitions
of sine, cosine and tangent.

00:03:38.150 --> 00:03:39.910
And we're about to see why.

00:03:39.910 --> 00:03:42.790
So here I have two different unit circles,

00:03:42.790 --> 00:03:45.200
I'm going to use one for
each of these angles.

00:03:45.200 --> 00:03:48.310
So first, let's think about
pi over three radians.

00:03:48.310 --> 00:03:50.340
And so pi over three,

00:03:50.340 --> 00:03:53.513
would look something like this,

00:03:54.710 --> 00:03:58.940
this is pi over three radians.

00:03:58.940 --> 00:04:02.240
And the cosine and sine can be determined

00:04:02.240 --> 00:04:04.860
by the X and Y coordinates of this point

00:04:04.860 --> 00:04:08.200
where this radius intersects
the actual unit circle.

00:04:08.200 --> 00:04:10.887
The coordinates here
are going to be cosine

00:04:10.887 --> 00:04:13.006
of pi over three radians

00:04:13.006 --> 00:04:15.839
and sine of pi over three radians.

00:04:16.810 --> 00:04:18.400
Or another way to think about it is,

00:04:18.400 --> 00:04:20.780
I can set up a 30, 60, 90 triangle here,

00:04:20.780 --> 00:04:22.740
so I'm going to drop a perpendicular.

00:04:22.740 --> 00:04:26.370
This would be 90 degrees
or pi over two radians.

00:04:26.370 --> 00:04:27.740
And then this angle over here,

00:04:27.740 --> 00:04:30.260
this is 60, this is 90,
this is going to be 30,

00:04:30.260 --> 00:04:31.320
or another way of thinking about it,

00:04:31.320 --> 00:04:34.390
it's going to be pi over six radians.

00:04:34.390 --> 00:04:36.650
It's going to be just like
one of these triangles here.

00:04:36.650 --> 00:04:38.260
And so the X coordinate,

00:04:38.260 --> 00:04:41.460
which is going to be the
same thing as the cosine

00:04:41.460 --> 00:04:42.650
of pi over three,

00:04:42.650 --> 00:04:46.740
is going to be the length of
this side, right over here.

00:04:46.740 --> 00:04:48.320
Well, what's that going to be?

00:04:48.320 --> 00:04:50.470
Well, when your hypotenuse is one,

00:04:50.470 --> 00:04:52.270
we know that the shorter side,

00:04:52.270 --> 00:04:56.280
the side opposite the pi over
six radians, is one half.

00:04:56.280 --> 00:04:58.390
So just like that, we have
been able to establish

00:04:58.390 --> 00:05:03.310
that cosine of pi over three
radians is equal to one half.

00:05:03.310 --> 00:05:04.920
This right over here is one half,

00:05:04.920 --> 00:05:06.410
that is the X coordinate

00:05:06.410 --> 00:05:09.140
where this radius
intersects the units circle.

00:05:09.140 --> 00:05:11.540
Now, what about the Y coordinate?

00:05:11.540 --> 00:05:13.970
What is sine of pi over three going to be?

00:05:13.970 --> 00:05:15.710
Well, the Y coordinate is the same thing

00:05:15.710 --> 00:05:18.200
as the length of this side,

00:05:18.200 --> 00:05:21.100
and once again, it goes
back to being this triangle.

00:05:21.100 --> 00:05:22.800
If this is one, this is one half,

00:05:22.800 --> 00:05:24.390
this is one, this is one half,

00:05:24.390 --> 00:05:26.330
this other side is going to be square root

00:05:26.330 --> 00:05:27.540
of three over two.

00:05:27.540 --> 00:05:29.140
So sine of pi over three

00:05:29.140 --> 00:05:31.780
is going to be square
root of three over two,

00:05:31.780 --> 00:05:32.840
so let me write that down.

00:05:32.840 --> 00:05:35.610
Sine of pi over three

00:05:35.610 --> 00:05:37.840
is equal to square root of three over two.

00:05:37.840 --> 00:05:39.250
And these are good ones to know.

00:05:39.250 --> 00:05:41.400
I never say really memorize things,

00:05:41.400 --> 00:05:43.770
it's always good to know
how to derive things

00:05:43.770 --> 00:05:45.340
in case you forget.

00:05:45.340 --> 00:05:47.200
But if you have to memorize them

00:05:47.200 --> 00:05:49.320
I would highly recommend memorizing these,

00:05:49.320 --> 00:05:52.320
and then of course from these
we can figure out the tangent.

00:05:52.320 --> 00:05:54.470
The tangent is just going to
be the sine over the cosine,

00:05:54.470 --> 00:05:57.210
so let me write it down here.

00:05:57.210 --> 00:06:01.840
The tangent of pi over three

00:06:01.840 --> 00:06:03.350
is going to be the sine,

00:06:03.350 --> 00:06:05.950
which is square root of three over two,

00:06:05.950 --> 00:06:09.130
over the cosine which is one half,

00:06:09.130 --> 00:06:10.550
got a little squanchy down there,

00:06:10.550 --> 00:06:12.093
and so this is just going to be

00:06:12.093 --> 00:06:13.760
square root of three over two times two

00:06:13.760 --> 00:06:16.860
is just going to be square root of three.

00:06:16.860 --> 00:06:19.780
So now let's just use that
same logic for pi over six.

00:06:19.780 --> 00:06:21.410
And in fact, I encourage
you to pause this video

00:06:21.410 --> 00:06:23.353
and see if you can do that on your own.

00:06:24.430 --> 00:06:25.840
All right, now let's draw a radius

00:06:25.840 --> 00:06:30.190
that forms a pi over six radian angle

00:06:30.190 --> 00:06:32.340
with a positive X axis,
might look like that.

00:06:32.340 --> 00:06:35.670
So if that's going to
be pi over six radians,

00:06:35.670 --> 00:06:38.330
you might imagine it's interesting

00:06:38.330 --> 00:06:40.590
to drop a perpendicular here

00:06:40.590 --> 00:06:43.030
and see what type of
triangle we've constructed.

00:06:43.030 --> 00:06:44.380
So this has length one,

00:06:44.380 --> 00:06:46.240
this is pi over six radians,

00:06:46.240 --> 00:06:47.990
this is a right angle.

00:06:47.990 --> 00:06:50.960
So this again, is going to
follow the same pattern.

00:06:50.960 --> 00:06:54.610
This will be pi over three radians.

00:06:54.610 --> 00:06:57.150
And so the sides are
exactly the exact same

00:06:57.150 --> 00:06:59.910
as this top blue triangle here.

00:06:59.910 --> 00:07:02.280
So we know that this length over here

00:07:02.280 --> 00:07:04.610
is going to be one half.

00:07:04.610 --> 00:07:07.480
We know that this length over here

00:07:07.480 --> 00:07:11.670
is going to be square
root of three over two.

00:07:11.670 --> 00:07:13.910
And that's useful because that tells us

00:07:13.910 --> 00:07:15.110
the coordinates here.

00:07:15.110 --> 00:07:16.370
The coordinates here,

00:07:16.370 --> 00:07:18.480
the X coordinate of this
point where the radius

00:07:18.480 --> 00:07:20.060
intersects the unit circle

00:07:20.060 --> 00:07:22.840
is square root of three over two,

00:07:22.840 --> 00:07:26.450
and then the Y coordinate is one half.

00:07:26.450 --> 00:07:28.360
And that immediately tells us the cosine

00:07:28.360 --> 00:07:30.390
and the sine of pi over six,

00:07:30.390 --> 00:07:31.770
let's just write it down.

00:07:31.770 --> 00:07:35.350
So this tells us that
cosine of pi over six

00:07:35.350 --> 00:07:38.490
is equal to square root of three over two.

00:07:38.490 --> 00:07:41.930
And sine of pi over six

00:07:43.230 --> 00:07:45.540
is equal to one half.

00:07:45.540 --> 00:07:48.530
Notice, we actually just
swap these two things around

00:07:48.530 --> 00:07:50.380
because now the angle that we're taking

00:07:50.380 --> 00:07:51.530
the sine or cosine of,

00:07:51.530 --> 00:07:54.870
is a different angle on
a 30, 60, 90 triangle,

00:07:54.870 --> 00:07:57.900
but we're essentially utilizing
the same side measure,

00:07:57.900 --> 00:07:59.090
just one way to think about it.

00:07:59.090 --> 00:08:01.340
And then what's the tangent going to be?

00:08:01.340 --> 00:08:02.730
I'll write it down here.

00:08:02.730 --> 00:08:06.250
The tangent of pi over six

00:08:06.250 --> 00:08:09.520
is going to be the sine over the cosine

00:08:09.520 --> 00:08:11.660
square root of three over two,

00:08:11.660 --> 00:08:14.320
and so that's going to
be equal to one half

00:08:14.320 --> 00:08:17.580
times two over the square root of three,

00:08:17.580 --> 00:08:21.150
which is equal to one over
the square root of three.

00:08:21.150 --> 00:08:24.210
Now some people sometimes
don't like radicals

00:08:24.210 --> 00:08:25.043
in the denominator

00:08:25.043 --> 00:08:27.220
and so you can multiply the
numerator and the denominator

00:08:27.220 --> 00:08:29.210
by square root of three if you like

00:08:29.210 --> 00:08:30.270
to get something like this,

00:08:30.270 --> 00:08:31.580
you multiply the numerator and denominator

00:08:31.580 --> 00:08:32.413
by square root of three

00:08:32.413 --> 00:08:34.660
you get square root of three over three,

00:08:34.660 --> 00:08:37.560
which is another way of
writing tangent of pi over six.

00:08:37.560 --> 00:08:38.880
But either way, we're done,

00:08:38.880 --> 00:08:41.820
it's very useful to know
the cosine, sine and tangent

00:08:41.820 --> 00:08:44.570
of both pi over three and pi over six.

00:08:44.570 --> 00:08:46.823
And now you also know how to derive it.


Rule of 70 to approximate population doubling time

WEBVTT
Kind: captions
Language: en

00:00:01.230 --> 00:00:02.063
- [Instructor] When we're dealing

00:00:02.063 --> 00:00:03.280
with population growth rates

00:00:03.280 --> 00:00:05.050
an interesting question is,

00:00:05.050 --> 00:00:06.440
how long would it take

00:00:06.440 --> 00:00:10.650
for a given rate for the
population to double?

00:00:10.650 --> 00:00:13.120
So we're gonna think about doubling time.

00:00:13.120 --> 00:00:15.640
Now if you were to actually
calculate it precisely

00:00:15.640 --> 00:00:19.020
mathematically precisely,
it gets a little bit mathy.

00:00:19.020 --> 00:00:20.960
You need to use a little bit of logarithms

00:00:20.960 --> 00:00:23.220
and you'll probably need a calculator,

00:00:23.220 --> 00:00:25.070
but I did that here in this spreadsheet

00:00:25.070 --> 00:00:27.130
by calculating the exact doubling time.

00:00:27.130 --> 00:00:29.790
So this is saying that if
a population is growing

00:00:29.790 --> 00:00:31.800
at 1% a year,

00:00:31.800 --> 00:00:34.750
it's going to take almost 70 years

00:00:34.750 --> 00:00:36.700
for that population to double.

00:00:36.700 --> 00:00:41.250
But if that population
is growing at 5% per year

00:00:41.250 --> 00:00:44.130
then it's going to take
a little over 14 years

00:00:44.130 --> 00:00:46.240
for that population to double.

00:00:46.240 --> 00:00:48.790
If the population is growing at 10%,

00:00:48.790 --> 00:00:51.160
we know mathematically it's
going to take a little bit

00:00:51.160 --> 00:00:54.400
over seven years for that
population to double.

00:00:54.400 --> 00:00:56.840
Now, I was able to calculate
this as I just mentioned

00:00:56.840 --> 00:00:59.420
using a little bit of fancy math,

00:00:59.420 --> 00:01:01.000
but what we see in this next column

00:01:01.000 --> 00:01:03.120
is there's actually a pretty

00:01:03.120 --> 00:01:06.490
easy way to approximate doubling time.

00:01:06.490 --> 00:01:08.840
And this is known as the rule of 70.

00:01:08.840 --> 00:01:12.480
And the rule of 70 is used
in a lot of different areas,

00:01:12.480 --> 00:01:13.550
a lot of different subjects,

00:01:13.550 --> 00:01:15.780
people in finance would
use it because once again,

00:01:15.780 --> 00:01:17.710
you're thinking about things growing

00:01:17.710 --> 00:01:19.190
at a certain percent every year,

00:01:19.190 --> 00:01:20.530
but you can also use it

00:01:20.530 --> 00:01:23.540
for things like population growth rates.

00:01:23.540 --> 00:01:26.120
So what we see with the rule of 70,

00:01:26.120 --> 00:01:27.580
and let me just write that down,

00:01:27.580 --> 00:01:32.580
rule of 70 is that you can
approximate the doubling time

00:01:32.930 --> 00:01:34.960
by taking the number 70

00:01:34.960 --> 00:01:38.890
and dividing it by the, not
actually the percentage,

00:01:38.890 --> 00:01:41.110
but just the number of the percentage.

00:01:41.110 --> 00:01:42.480
So for example,

00:01:42.480 --> 00:01:47.480
this right over here is 70
divided by this one here,

00:01:48.920 --> 00:01:50.980
which is equal to 70.

00:01:50.980 --> 00:01:55.330
And notice this 70 is
pretty close to 69.7.

00:01:55.330 --> 00:01:56.760
If you wanted to figure out

00:01:56.760 --> 00:01:59.750
or you wanted to approximate
the doubling time,

00:01:59.750 --> 00:02:02.800
if the population is growing at 7% a year,

00:02:02.800 --> 00:02:04.330
well what you would say is, all right,

00:02:04.330 --> 00:02:08.090
what is 70 divided by seven?

00:02:08.090 --> 00:02:09.280
Well, that is equal to 10.

00:02:09.280 --> 00:02:10.950
So this would be your approximation.

00:02:10.950 --> 00:02:13.440
And if you were to do it in
a mathematically precise way,

00:02:13.440 --> 00:02:15.400
it would be 10.2.

00:02:15.400 --> 00:02:17.400
So if you're taking,

00:02:17.400 --> 00:02:20.000
say an AP environmental science course

00:02:20.000 --> 00:02:22.040
and they're asking you for the,

00:02:22.040 --> 00:02:24.400
how long it takes for something to double

00:02:24.400 --> 00:02:27.150
let's say a population
that's growing 7% a year,

00:02:27.150 --> 00:02:31.563
they're probably expecting
you to use the rule of 70.

00:02:32.830 --> 00:02:35.370
So let's say that we have a population

00:02:35.370 --> 00:02:40.370
that is growing at 14% per year,

00:02:41.390 --> 00:02:44.320
and that would actually be
a very huge growth rate.

00:02:44.320 --> 00:02:45.970
What I want you to do is pause this video

00:02:45.970 --> 00:02:48.210
and approximate how long would it take

00:02:48.210 --> 00:02:50.853
for that population to double?

00:02:52.460 --> 00:02:54.240
All right, now let's work
through this together.

00:02:54.240 --> 00:02:56.210
So as I mentioned, we're approximating,

00:02:56.210 --> 00:02:58.380
we don't have to do calculate
the exact doubling time.

00:02:58.380 --> 00:03:00.170
So if we're approximating,

00:03:00.170 --> 00:03:05.170
it's going to be 70 divided
by the rate of growth.

00:03:05.330 --> 00:03:09.603
So in this situation, this is
going to be 70 divided by 14,

00:03:11.240 --> 00:03:13.600
which is equal to five.

00:03:13.600 --> 00:03:15.550
So if a population is growing at 14%,

00:03:15.550 --> 00:03:18.863
it'll take it roughly
five years to double.


Worked example: Using bond enthalpies to calculate enthalpy of reaction

WEBVTT
Kind: captions
Language: en

00:00:00.510 --> 00:00:02.330
- [Educator] Bond enthalpies can be used

00:00:02.330 --> 00:00:05.100
to estimate the standard
change in enthalpy

00:00:05.100 --> 00:00:07.490
for a chemical reaction.

00:00:07.490 --> 00:00:08.900
Let's use bond enthalpies

00:00:08.900 --> 00:00:11.740
to estimate the enthalpy of combustion

00:00:11.740 --> 00:00:13.100
of ethanol.

00:00:13.100 --> 00:00:14.510
Looking at our balanced equation,

00:00:14.510 --> 00:00:16.840
we have one mole of ethanol reacting

00:00:16.840 --> 00:00:18.920
with three moles of oxygen gas

00:00:18.920 --> 00:00:21.600
to produce two moles of carbon dioxide

00:00:21.600 --> 00:00:25.990
and three moles of water
in the gaseous state.

00:00:25.990 --> 00:00:27.720
To find the standard change in enthalpy

00:00:27.720 --> 00:00:29.110
for this chemical reaction,

00:00:29.110 --> 00:00:31.980
we need to sum the bond enthalpies

00:00:31.980 --> 00:00:33.900
of the bonds that are broken.

00:00:33.900 --> 00:00:36.400
And from that, we subtract the sum

00:00:36.400 --> 00:00:38.670
of the bond enthalpies of the bonds

00:00:38.670 --> 00:00:41.550
that are formed in this chemical reaction.

00:00:41.550 --> 00:00:43.170
To figure out which bonds are broken

00:00:43.170 --> 00:00:44.450
and which bonds are formed,

00:00:44.450 --> 00:00:46.590
it's helpful to look at the dot structures

00:00:46.590 --> 00:00:48.130
for our molecules.

00:00:48.130 --> 00:00:51.180
So let's start with the ethanol molecule.

00:00:51.180 --> 00:00:52.770
We're gonna approach this problem first

00:00:52.770 --> 00:00:55.840
like we're breaking all of
the bonds in these molecules.

00:00:55.840 --> 00:00:58.570
And we're also not gonna worry
about units until the end,

00:00:58.570 --> 00:01:01.160
just to save some space on the screen.

00:01:01.160 --> 00:01:03.130
So looking at the ethanol molecule,

00:01:03.130 --> 00:01:06.820
we would need to break
a carbon-carbon bond.

00:01:06.820 --> 00:01:08.530
So let's go ahead and
write this down here.

00:01:08.530 --> 00:01:10.850
Right now, we're summing
the bond enthalpies

00:01:10.850 --> 00:01:12.860
of the bonds that are broken.

00:01:12.860 --> 00:01:15.490
So we have one carbon-carbon bond.

00:01:15.490 --> 00:01:16.323
So let's write in here,

00:01:16.323 --> 00:01:20.590
the bond enthalpy for
a carbon-carbon bond.

00:01:20.590 --> 00:01:24.630
Next, we have five carbon-hydrogen bonds

00:01:24.630 --> 00:01:26.390
that we need to break.

00:01:26.390 --> 00:01:27.640
So to this,

00:01:27.640 --> 00:01:30.140
we're going to write in here, a five,

00:01:30.140 --> 00:01:32.740
and then the bond enthalpy

00:01:32.740 --> 00:01:35.920
of a carbon-hydrogen bond.

00:01:35.920 --> 00:01:39.820
Next, we have to break a
carbon-oxygen single bond.

00:01:39.820 --> 00:01:41.840
So we write a one,

00:01:41.840 --> 00:01:43.930
and then the bond enthalpy

00:01:43.930 --> 00:01:47.310
for a carbon-oxygen single bond.

00:01:47.310 --> 00:01:49.260
And then for this ethanol molecule,

00:01:49.260 --> 00:01:52.650
we also have an
oxygen-hydrogen single bond.

00:01:52.650 --> 00:01:54.850
So we'll write in here, a one,

00:01:54.850 --> 00:01:56.880
and the bond enthalpy

00:01:56.880 --> 00:01:59.920
for an oxygen-hydrogen single bond.

00:01:59.920 --> 00:02:01.150
We saw in the balanced equation

00:02:01.150 --> 00:02:02.870
that one mole of ethanol reacts

00:02:02.870 --> 00:02:05.310
with three moles of oxygen gas.

00:02:05.310 --> 00:02:07.550
So to represent the three
moles of oxygen gas,

00:02:07.550 --> 00:02:08.500
I've drawn in here,

00:02:08.500 --> 00:02:11.370
three molecules of O2.

00:02:11.370 --> 00:02:13.970
And we can see in each molecule of O2,

00:02:13.970 --> 00:02:16.160
there's an oxygen-oxygen double bond.

00:02:16.160 --> 00:02:20.610
So we would need to break three
oxygen-oxygen double bonds.

00:02:20.610 --> 00:02:21.840
So to this,

00:02:21.840 --> 00:02:26.840
we're going to add a three
times the bond enthalpy

00:02:27.140 --> 00:02:30.450
of an oxygen-oxygen double bond.

00:02:30.450 --> 00:02:34.650
And this now gives us the
sum of the bond enthalpies

00:02:34.650 --> 00:02:37.360
for all the bonds that need to be broken.

00:02:37.360 --> 00:02:39.240
It takes energy to break a bond.

00:02:39.240 --> 00:02:41.340
So the summation of the bond enthalpies

00:02:41.340 --> 00:02:42.720
of the bonds that are broken

00:02:42.720 --> 00:02:45.230
is going to be a positive value.

00:02:45.230 --> 00:02:47.130
And since it takes energy to break bonds,

00:02:47.130 --> 00:02:49.580
energy is given off when bonds form.

00:02:49.580 --> 00:02:51.950
So next, we're gonna
sum the bond enthalpies

00:02:51.950 --> 00:02:53.390
of the bonds that are formed.

00:02:53.390 --> 00:02:55.900
And notice we have this
negative sign in here

00:02:55.900 --> 00:02:58.160
because this energy is given off.

00:02:58.160 --> 00:03:00.320
So we're gonna write a minus sign in here,

00:03:00.320 --> 00:03:02.910
and then we're gonna put some brackets

00:03:02.910 --> 00:03:05.620
because next we're going
to sum the bond enthalpies

00:03:05.620 --> 00:03:07.580
of the bonds that are formed.

00:03:07.580 --> 00:03:08.550
In our balanced equation,

00:03:08.550 --> 00:03:12.250
we formed two moles of carbon dioxide.

00:03:12.250 --> 00:03:14.530
So to represent those two moles,

00:03:14.530 --> 00:03:15.400
I've drawn in here,

00:03:15.400 --> 00:03:18.620
two molecules of CO2.

00:03:18.620 --> 00:03:22.110
And we can see that in
each molecule of CO2,

00:03:22.110 --> 00:03:27.110
we're going to form two
carbon-oxygen double bonds.

00:03:27.540 --> 00:03:31.350
So that's a total of four
carbon-oxygen double bonds.

00:03:31.350 --> 00:03:32.780
So down here,

00:03:32.780 --> 00:03:37.780
we're going to write a four
times the bond enthalpy

00:03:38.020 --> 00:03:41.890
of a carbon-oxygen double bond.

00:03:41.890 --> 00:03:44.910
We also formed three moles of H2O.

00:03:44.910 --> 00:03:47.580
And in each molecule of
water that's drawn here,

00:03:47.580 --> 00:03:51.950
we form two oxygen-hydrogen single bonds.

00:03:51.950 --> 00:03:54.290
And since we have three moles,

00:03:54.290 --> 00:03:58.340
we have a total of six
oxygen-hydrogen single bonds.

00:03:58.340 --> 00:03:59.173
So to this,

00:03:59.173 --> 00:04:03.900
we're going to add six
times the bond enthalpy

00:04:03.900 --> 00:04:07.623
of an oxygen-hydrogen single bond.

00:04:08.810 --> 00:04:11.120
The next step is to look
up the bond enthalpies

00:04:11.120 --> 00:04:13.090
of all of these different bonds.

00:04:13.090 --> 00:04:14.640
For example, the bond enthalpy

00:04:14.640 --> 00:04:19.480
for a carbon-carbon single
bond is about 348 kilojoules

00:04:19.480 --> 00:04:20.313
per mole.

00:04:20.313 --> 00:04:21.540
You might see a different value,

00:04:21.540 --> 00:04:23.620
if you look in a different textbook.

00:04:23.620 --> 00:04:26.640
However, we're gonna go
with 348 kilojoules per mole

00:04:26.640 --> 00:04:28.350
for our calculation.

00:04:28.350 --> 00:04:29.450
And we're gonna multiply this

00:04:29.450 --> 00:04:32.960
by one mole of carbon-carbon single bonds.

00:04:32.960 --> 00:04:34.470
Next, we look up the bond enthalpy

00:04:34.470 --> 00:04:36.640
for our carbon-hydrogen single bond.

00:04:36.640 --> 00:04:39.840
And that's about 413 kilojoules per mole

00:04:39.840 --> 00:04:41.390
of carbon-hydrogen bonds.

00:04:41.390 --> 00:04:43.060
And we're multiplying this by five.

00:04:43.060 --> 00:04:45.560
And we continue with everything else

00:04:45.560 --> 00:04:47.820
for the summation of
the the bond enthalpies

00:04:47.820 --> 00:04:49.560
of the bonds broken.

00:04:49.560 --> 00:04:51.000
When we do this,

00:04:51.000 --> 00:04:56.000
we get positive 4,719 kilojoules.

00:04:56.530 --> 00:04:57.670
Next, we do the same thing

00:04:57.670 --> 00:04:59.430
for the bond enthalpies

00:04:59.430 --> 00:05:00.590
of the bonds that are formed.

00:05:00.590 --> 00:05:01.450
So the bond enthalpy

00:05:01.450 --> 00:05:05.460
for our carbon-oxygen double
bond is 799 kilojoules

00:05:05.460 --> 00:05:06.293
per mole,

00:05:06.293 --> 00:05:08.020
and we multiply that by four.

00:05:08.020 --> 00:05:10.760
The bonds enthalpy for an
oxygen hydrogen single bond

00:05:10.760 --> 00:05:13.030
is 463 kilojoules per mole,

00:05:13.030 --> 00:05:14.940
and we multiply that by six.

00:05:14.940 --> 00:05:15.880
When we add these together,

00:05:15.880 --> 00:05:20.350
we get 5,974.

00:05:20.350 --> 00:05:23.210
So for the final standard
change in enthalpy

00:05:23.210 --> 00:05:25.090
for our chemical reaction,

00:05:25.090 --> 00:05:30.090
it's positive 4,719 minus 5,974,

00:05:31.540 --> 00:05:36.540
which gives us negative 1,255 kilojoules.

00:05:37.280 --> 00:05:39.460
Notice that we got a negative value

00:05:39.460 --> 00:05:41.420
for the change in enthalpy.

00:05:41.420 --> 00:05:43.860
And that means the combustion of ethanol

00:05:43.860 --> 00:05:46.160
is an exothermic reaction.

00:05:46.160 --> 00:05:50.840
And 1,255 kilojoules
of energy are given off

00:05:50.840 --> 00:05:52.880
for the combustion of one mole

00:05:52.880 --> 00:05:54.390
of ethanol.

00:05:54.390 --> 00:05:56.750
Also notice that the sum
of the bond enthalpies

00:05:56.750 --> 00:05:57.950
of the bonds formed,

00:05:57.950 --> 00:06:01.040
which is 5,974,

00:06:01.040 --> 00:06:03.790
is greater than the sum
of the bond enthalpies

00:06:03.790 --> 00:06:04.830
of the bonds broken,

00:06:04.830 --> 00:06:07.480
which is 4,719.

00:06:07.480 --> 00:06:10.270
And since we're
subtracting a larger number

00:06:10.270 --> 00:06:11.500
from a smaller number,

00:06:11.500 --> 00:06:12.980
we get that negative sign

00:06:12.980 --> 00:06:15.240
for the change in enthalpy.

00:06:15.240 --> 00:06:17.160
If the sum of the bond enthalpies

00:06:17.160 --> 00:06:18.480
of the bonds that are broken,

00:06:18.480 --> 00:06:21.180
if this number is larger than the sum

00:06:21.180 --> 00:06:22.030
of the bond enthalpies

00:06:22.030 --> 00:06:23.120
of the bonds that have formed,

00:06:23.120 --> 00:06:24.700
we would've gotten a positive value

00:06:24.700 --> 00:06:26.480
for the change in enthalpy.

00:06:26.480 --> 00:06:29.950
And that would be true for
an endothermic reaction.

00:06:29.950 --> 00:06:30.783
We did this problem,

00:06:30.783 --> 00:06:32.290
assuming that all of the bonds

00:06:32.290 --> 00:06:34.540
that we drew in our dots
structures were broken

00:06:34.540 --> 00:06:35.470
and all of the bonds

00:06:35.470 --> 00:06:37.890
that we drew in the dot
structures were formed.

00:06:37.890 --> 00:06:39.950
However, if we look
closely to dots structures

00:06:39.950 --> 00:06:41.810
or just look closely
to what we wrote here,

00:06:41.810 --> 00:06:46.810
we show breaking one oxygen-hydrogen
single bonds over here,

00:06:47.640 --> 00:06:48.780
and we show the formation

00:06:48.780 --> 00:06:53.270
of six oxygen-hydrogen
single bonds over here.

00:06:53.270 --> 00:06:55.320
So we could have just canceled out one

00:06:55.320 --> 00:06:57.840
of those oxygen-hydrogen single bonds.

00:06:57.840 --> 00:06:59.600
So we could have canceled this out.

00:06:59.600 --> 00:07:01.300
And instead of showing a six here,

00:07:01.300 --> 00:07:04.920
we could have written a
five times the bond enthalpy

00:07:04.920 --> 00:07:08.520
of an oxygen-hydrogen single bond.

00:07:08.520 --> 00:07:10.630
We still would have ended
up with the same answer

00:07:10.630 --> 00:07:14.250
of negative 1,255 kilojoules.

00:07:14.250 --> 00:07:15.910
So if you look at your dot structures,

00:07:15.910 --> 00:07:18.930
if you see a bond that's the
same on the reactant side

00:07:18.930 --> 00:07:21.116
and the same on the product side,

00:07:21.116 --> 00:07:23.640
you don't have to show the breaking

00:07:23.640 --> 00:07:25.360
and forming of that bond.

00:07:25.360 --> 00:07:27.410
You can make the problem
a little bit shorter,

00:07:27.410 --> 00:07:28.373
if you want to.

00:07:29.280 --> 00:07:31.400
Finally, let's show how we get our units.

00:07:31.400 --> 00:07:34.420
So to get kilojoules as your final answer,

00:07:34.420 --> 00:07:35.790
if we go back up to here,

00:07:35.790 --> 00:07:38.770
we wrote a one times 348.

00:07:38.770 --> 00:07:42.400
The one is referring to breaking one mole

00:07:42.400 --> 00:07:44.870
of carbon-carbon single bonds.

00:07:44.870 --> 00:07:46.690
And the 348, of course,

00:07:46.690 --> 00:07:48.160
is the bond enthalpy

00:07:48.160 --> 00:07:50.030
for a carbon-carbon single bond.

00:07:50.030 --> 00:07:54.340
So this was 348 kilojoules per one mole

00:07:54.340 --> 00:07:57.220
of carbon-carbon single bonds.

00:07:57.220 --> 00:07:59.140
When you multiply these two together,

00:07:59.140 --> 00:08:02.530
the moles of carbon-carbon
single bonds cancels

00:08:02.530 --> 00:08:06.317
and this gives you 348 kilojoules.

00:08:06.317 --> 00:08:07.630
And so, that's how to end up

00:08:07.630 --> 00:08:11.270
with kilojoules as your final answer.

00:08:11.270 --> 00:08:14.880
You also might see kilojoules
per mole of reaction

00:08:14.880 --> 00:08:16.880
as the units for this.

00:08:16.880 --> 00:08:19.580
And, kilojoules per mole reaction means

00:08:19.580 --> 00:08:21.510
how the reaction is written.

00:08:21.510 --> 00:08:22.670
So for the combustion

00:08:22.670 --> 00:08:24.780
of one mole of ethanol,

00:08:24.780 --> 00:08:29.720
1,255 kilojoules of energy are released.

00:08:29.720 --> 00:08:33.980
To get kilojoules per mole
of reaction as our units,

00:08:33.980 --> 00:08:36.620
the balanced equation had
a one as the coefficient

00:08:36.620 --> 00:08:38.630
in front of ethanol.

00:08:38.630 --> 00:08:41.070
Therefore, you're breaking one mole

00:08:41.070 --> 00:08:45.080
of carbon-carbon single bonds per one mole

00:08:45.080 --> 00:08:46.160
of reaction.

00:08:46.160 --> 00:08:48.950
So we can use this conversion factor.

00:08:48.950 --> 00:08:51.410
Now, when we multiply through the moles

00:08:51.410 --> 00:08:52.900
of carbon-carbon single bonds,

00:08:52.900 --> 00:08:57.900
cancel and this gives us
348 kilojoules per mole

00:08:59.000 --> 00:09:00.533
of reaction.


Bond enthalpies

WEBVTT
Kind: captions
Language: en

00:00:00.960 --> 00:00:03.300
- [Instructor] Bond enthalpy
is the change in enthalpy

00:00:03.300 --> 00:00:06.200
or delta H for breaking a particular bond

00:00:06.200 --> 00:00:08.710
in one mole of a gaseous substance.

00:00:08.710 --> 00:00:11.040
If we think about the
diatomic chlorine molecule,

00:00:11.040 --> 00:00:14.280
so Cl2, down here is a
little picture of Cl2,

00:00:14.280 --> 00:00:16.540
each of the green spheres
is a chlorine atom

00:00:16.540 --> 00:00:19.510
and they're bonded together
by a single covalent bond.

00:00:19.510 --> 00:00:21.990
It would take energy to break this bond

00:00:21.990 --> 00:00:23.840
in diatomic chlorine gas

00:00:23.840 --> 00:00:26.570
and turn diatomic chlorine gas, Cl2

00:00:26.570 --> 00:00:29.210
in to two individual chlorine atoms.

00:00:29.210 --> 00:00:33.580
So we're going from Cl2 in
the gaseous state to 2Cl.

00:00:33.580 --> 00:00:37.280
Bond enthalpy can be
symbolized by the letters BE.

00:00:37.280 --> 00:00:41.100
So the bond enthalpy of the
chlorine-chlorine single bond

00:00:41.100 --> 00:00:45.900
is equal to +242 kilojoules per mole.

00:00:45.900 --> 00:00:46.810
And what this means is,

00:00:46.810 --> 00:00:49.160
if we have one mole of
chlorine-chlorine bonds,

00:00:49.160 --> 00:00:54.160
it takes +242 kilojoules of
energy to break those bonds.

00:00:55.008 --> 00:00:57.150
Bond enthalpies are always positive

00:00:57.150 --> 00:01:00.370
because it takes energy to break bonds.

00:01:00.370 --> 00:01:04.270
Another name for bond enthalpy
is bond dissociation energy.

00:01:04.270 --> 00:01:07.600
So you might see this symbolized as BDE

00:01:07.600 --> 00:01:10.800
or just simply the letter D.

00:01:10.800 --> 00:01:12.680
Bond enthalpies are often found

00:01:12.680 --> 00:01:15.965
in the appendices of chemistry textbooks.

00:01:15.965 --> 00:01:19.610
For example, we just saw the
chlorine-chlorine single bond,

00:01:19.610 --> 00:01:23.290
the bond enthalpy is
242 kilojoules per mole.

00:01:23.290 --> 00:01:26.160
Whereas to break a
carbon-carbon single bond

00:01:26.160 --> 00:01:29.300
takes 348 kilojoules of energy per mole

00:01:29.300 --> 00:01:31.620
of carbon-carbon single bonds.

00:01:31.620 --> 00:01:34.100
A carbon-carbon double
bond has a bond enthalpy

00:01:34.100 --> 00:01:37.310
of 614 kilojoules per mole.

00:01:37.310 --> 00:01:40.320
Since the carbon-carbon
double bond is stronger

00:01:40.320 --> 00:01:42.000
than a carbon-carbon single bond,

00:01:42.000 --> 00:01:44.910
it takes more energy to
break the double bond.

00:01:44.910 --> 00:01:47.010
And that's why the
carbon-carbon double bond

00:01:47.010 --> 00:01:50.320
has a high higher bond enthalpy.

00:01:50.320 --> 00:01:52.460
So the higher the value
for the bond enthalpy,

00:01:52.460 --> 00:01:54.410
the stronger the bond.

00:01:54.410 --> 00:01:56.810
Notice that these are
average bond enthalpies.

00:01:56.810 --> 00:01:58.400
So the average bond enthalpy

00:01:58.400 --> 00:02:00.480
for a carbon-carbon single bond

00:02:00.480 --> 00:02:03.530
is around 348 kilojoules per mole.

00:02:03.530 --> 00:02:05.920
You might see slightly
different values for this,

00:02:05.920 --> 00:02:08.100
depending on which
textbook you're looking in,

00:02:08.100 --> 00:02:10.700
but they're all pretty
close to the same value.

00:02:10.700 --> 00:02:12.905
The reason why these are
average bond enthalpies

00:02:12.905 --> 00:02:16.640
is because if we look at two
different molecules down here,

00:02:16.640 --> 00:02:20.060
this is ethane on the left
and propane on the right,

00:02:20.060 --> 00:02:23.250
if we break a carbon-carbon
single bond in ethane,

00:02:23.250 --> 00:02:25.460
the bond enthalpy is slightly different

00:02:25.460 --> 00:02:28.580
from breaking a carbon-carbon
single bond in propane.

00:02:28.580 --> 00:02:32.830
And that's why we use
average bond enthalpies.

00:02:32.830 --> 00:02:35.740
We've already seen that it
takes energy to break bonds.

00:02:35.740 --> 00:02:39.100
So to break the
chlorine-chlorine single bond

00:02:39.100 --> 00:02:44.100
in diatomic chlorine gas takes
+242 kilojoules per mole.

00:02:44.500 --> 00:02:46.320
If it takes energy to break bonds,

00:02:46.320 --> 00:02:49.890
that means energy is
given off when bonds form.

00:02:49.890 --> 00:02:53.510
So when two individual chlorine
gas atoms come together

00:02:53.510 --> 00:02:56.740
to form a chlorine-chlorine bond,

00:02:56.740 --> 00:02:58.270
so let's go into highlight that in here.

00:02:58.270 --> 00:03:02.020
So this bond is forming,
energy is given off.

00:03:02.020 --> 00:03:06.310
The magnitude of energy is
still 242 kilojoules per mole,

00:03:06.310 --> 00:03:08.700
however, now we have this
negative sign in here

00:03:08.700 --> 00:03:12.660
to indicate the energy is
given off when bonds form.

00:03:12.660 --> 00:03:14.660
Bond enthalpies can be used to estimate

00:03:14.660 --> 00:03:16.240
enthalpies of reactions.

00:03:16.240 --> 00:03:17.970
So to find the change in the enthalpy

00:03:17.970 --> 00:03:20.577
for a chemical reaction, you take the sum

00:03:20.577 --> 00:03:23.870
of the bond enthalpies
of the bonds broken.

00:03:23.870 --> 00:03:26.120
And from that you subtract the sum

00:03:26.120 --> 00:03:29.400
of the bond enthalpies
of the bonds formed.

00:03:29.400 --> 00:03:31.240
The minus sign is in there because energy

00:03:31.240 --> 00:03:33.920
is given off when bonds form.

00:03:33.920 --> 00:03:35.940
A good way to remember this equation

00:03:35.940 --> 00:03:40.020
is to remember that B comes
before F in the alphabet.

00:03:40.020 --> 00:03:42.700
So B before F therefore it's bonds broken

00:03:42.700 --> 00:03:44.860
minus bonds formed.

00:03:44.860 --> 00:03:47.100
Let's use bond enthalpies to estimate

00:03:47.100 --> 00:03:50.400
the enthalpy of reaction for
the following reaction here,

00:03:50.400 --> 00:03:54.380
methane with chlorine gas
to form chloro methane

00:03:54.380 --> 00:03:56.950
and hydrogen chloride gas.

00:03:56.950 --> 00:03:59.040
It's often helpful to draw dot structures

00:03:59.040 --> 00:04:00.690
for these kinds of problems.

00:04:00.690 --> 00:04:03.330
If we look at the methane dot structure,

00:04:03.330 --> 00:04:07.670
we would need to break one
carbon-hydrogen single bond

00:04:07.670 --> 00:04:09.350
in order to get to our products.

00:04:09.350 --> 00:04:13.220
We would also need to break a
chlorine-chlorine single bond.

00:04:13.220 --> 00:04:18.130
Next, one of the chlorine
goes over to the CH3

00:04:18.130 --> 00:04:19.200
to form CH3Cl.

00:04:19.200 --> 00:04:21.920
So therefore we are forming

00:04:21.920 --> 00:04:24.830
one carbon-chlorine single bonds,

00:04:24.830 --> 00:04:27.480
and the other Cl goes with the hydrogen.

00:04:27.480 --> 00:04:32.090
So we also need to form one
hydrogen-chlorine single bond.

00:04:32.090 --> 00:04:35.150
The next step is to
sum the bond enthalpies

00:04:35.150 --> 00:04:37.200
of the bonds broken.

00:04:37.200 --> 00:04:38.330
So let's think about this.

00:04:38.330 --> 00:04:40.430
For our reactants we're breaking bonds.

00:04:40.430 --> 00:04:42.170
So we have one mole of methane

00:04:42.170 --> 00:04:45.170
reacting with one mole of chlorine.

00:04:45.170 --> 00:04:48.840
And since we're breaking one
carbon-hydrogen single bond

00:04:48.840 --> 00:04:51.000
for every one molecule of methane,

00:04:51.000 --> 00:04:53.670
since we have one mole
of methane molecules,

00:04:53.670 --> 00:04:58.330
we're breaking one mole of
carbon-hydrogen single bonds.

00:04:58.330 --> 00:05:00.070
Therefore we can write down here,

00:05:00.070 --> 00:05:03.660
one mole of carbon-hydrogen bonds,

00:05:03.660 --> 00:05:07.720
and the bond enthalpy for a
carbon-hydrogen single bond

00:05:07.720 --> 00:05:12.323
is 413 kilojoules per mole.

00:05:14.510 --> 00:05:17.520
Since there's one
chlorine-chlorine single bond

00:05:17.520 --> 00:05:19.540
for every Cl2 molecule,

00:05:19.540 --> 00:05:22.630
and we have one mole
of chlorine molecules,

00:05:22.630 --> 00:05:26.120
we're breaking one mole of
chlorine-chlorine single bonds.

00:05:26.120 --> 00:05:28.830
So to this we're gonna add one mole

00:05:28.830 --> 00:05:31.170
of chlorine-chlorine single bonds.

00:05:31.170 --> 00:05:34.860
And the bond enthalpy for a
chlorine-chlorine single bond

00:05:34.860 --> 00:05:39.363
is 242 kilojoules per mole.

00:05:40.550 --> 00:05:43.700
Moles cancel out and we get that the sum

00:05:43.700 --> 00:05:46.050
of the bond enthalpies of the bonds broken

00:05:46.050 --> 00:05:51.050
is equal to 655 kilojoules.

00:05:51.240 --> 00:05:53.081
Next, we need to sum the bond enthalpies

00:05:53.081 --> 00:05:54.950
of the bonds formed.

00:05:54.950 --> 00:05:58.150
So we're forming one
mole of chloro methane

00:05:58.150 --> 00:06:01.970
and one mole of hydrogen chloride gas.

00:06:01.970 --> 00:06:05.840
And since we're forming one
carbon-chlorine single bond

00:06:05.840 --> 00:06:08.620
for every molecule of chloro methane,

00:06:08.620 --> 00:06:10.900
since we're forming one
mole of chloral methane,

00:06:10.900 --> 00:06:15.070
we're reforming one mole of
carbon-chlorine single bonds.

00:06:15.070 --> 00:06:17.640
So let's write down here,
we're forming one mole

00:06:17.640 --> 00:06:21.014
of carbon-chlorine bonds
and the bond enthalpy

00:06:21.014 --> 00:06:23.550
for a carbon-chlorine single bond

00:06:23.550 --> 00:06:28.550
is equal to 328 kilojoules per mole.

00:06:30.980 --> 00:06:34.112
And since we form one
hydrogen-chlorine single bond

00:06:34.112 --> 00:06:36.760
for every molecule of hydrogen chloride,

00:06:36.760 --> 00:06:39.370
since we're making one
mole of hydrogen chloride,

00:06:39.370 --> 00:06:43.310
we're forming one mole of
hydrogen-chlorine single bonds.

00:06:43.310 --> 00:06:46.140
So to this we add one mole

00:06:46.140 --> 00:06:50.120
of hydrogen-chlorine single
bonds and the bond enthalpy

00:06:50.120 --> 00:06:52.080
for a hydrogen-chlorine single bond

00:06:52.080 --> 00:06:57.063
is 431 kilojoules per mole.

00:06:58.660 --> 00:07:02.530
Moles cancel and we get that
the sum of the bond enthalpies

00:07:02.530 --> 00:07:07.530
of the bonds formed is
equal to 759 kilojoules.

00:07:09.690 --> 00:07:11.966
Next, we're ready to find
the change in enthalpy

00:07:11.966 --> 00:07:14.720
for our chemical reaction.

00:07:14.720 --> 00:07:18.020
The sum of the enthalpy
of the bonds broken,

00:07:18.020 --> 00:07:21.670
we found that was equal to 655 kilojoules.

00:07:21.670 --> 00:07:24.049
And from that we subtract the
sum of the bond enthalpies

00:07:24.049 --> 00:07:29.049
of the bonds formed, which
we found was 759 kilojoules.

00:07:29.420 --> 00:07:34.420
So 655 minus 759 gives -104 kilojoules.

00:07:37.430 --> 00:07:40.710
Sometimes we see kilojoules
or kilojoules per mole

00:07:40.710 --> 00:07:43.330
or kilojoules per mole of reaction.

00:07:43.330 --> 00:07:45.460
Kilojoules per mole of reaction just means

00:07:45.460 --> 00:07:48.180
how the balanced equation is written.

00:07:48.180 --> 00:07:50.170
And let's see how we can look at the units

00:07:50.170 --> 00:07:52.570
to get kilojoules per mole of reaction

00:07:52.570 --> 00:07:54.680
when we do the calculations.

00:07:54.680 --> 00:07:56.190
If you go back to breaking

00:07:56.190 --> 00:08:00.200
the carbon hydrogen bond over here,

00:08:00.200 --> 00:08:03.780
we've seen there's one mole
of carbon-hydrogen bonds

00:08:03.780 --> 00:08:06.950
that we need to break for
how the equation is written.

00:08:06.950 --> 00:08:09.030
Therefore, we can write
a conversion factor

00:08:09.030 --> 00:08:11.320
of one mole of carbon-hydrogen bonds

00:08:11.320 --> 00:08:14.160
per one mole of reaction as it's written.

00:08:14.160 --> 00:08:16.900
And then we multiply
that by the bond enthalpy

00:08:16.900 --> 00:08:21.190
as 413 kilojoules per mole
for a carbon-hydrogen bond,

00:08:21.190 --> 00:08:24.120
this cancels out moles
of carbon-hydrogen bonds

00:08:24.120 --> 00:08:27.010
and this gives us kilojoules
per mole of reaction

00:08:27.010 --> 00:08:28.480
as our units.

00:08:28.480 --> 00:08:30.530
So it's more time consuming
to write it this way

00:08:30.530 --> 00:08:31.860
but we could do that for all

00:08:31.860 --> 00:08:33.540
of our different bond enthalpies

00:08:33.540 --> 00:08:35.450
to get kilojoules per mole of reaction

00:08:35.450 --> 00:08:37.600
for the units for our final answer.

00:08:37.600 --> 00:08:39.300
When everything is under
standard conditions,

00:08:39.300 --> 00:08:41.400
we need to add a superscript of note.

00:08:41.400 --> 00:08:44.770
So this would be the
standard change in enthalpy

00:08:44.770 --> 00:08:46.930
for a chemical reaction.

00:08:46.930 --> 00:08:48.740
So for the value we just calculated,

00:08:48.740 --> 00:08:52.520
- 140 kilojoules per mole of reaction,

00:08:52.520 --> 00:08:53.870
this is under standard of conditions.

00:08:53.870 --> 00:08:56.330
So this is actually the
standard change in enthalpy

00:08:56.330 --> 00:08:58.850
for this chemical reaction.

00:08:58.850 --> 00:09:01.950
Remember that bond
enthalpies are only averages.

00:09:01.950 --> 00:09:05.340
And so this value that we
calculated is only an estimate

00:09:05.340 --> 00:09:07.050
for the standard change in enthalpy

00:09:07.050 --> 00:09:08.950
for this chemical reaction.

00:09:08.950 --> 00:09:11.350
A more accurate way of
finding this standard change

00:09:11.350 --> 00:09:13.580
in enthalpy for a chemical reaction

00:09:13.580 --> 00:09:16.476
is to use standard
enthalpies of formation.

00:09:16.476 --> 00:09:19.580
And when you use standard
enthalpies of formation

00:09:19.580 --> 00:09:21.460
to find the standard change in enthalpy

00:09:21.460 --> 00:09:23.330
for this particular chemical reaction,

00:09:23.330 --> 00:09:28.330
you get -99.8 kilojoules
per mole of reaction.

00:09:28.560 --> 00:09:32.413
So -104 is pretty close to -99.8.