1
00:00:00,000 --> 00:00:00,410


2
00:00:00,410 --> 00:00:02,730
I have this reaction here
where if I had a mole of

3
00:00:02,730 --> 00:00:05,970
methane, and I react that with
two moles of oxygen, I'll

4
00:00:05,970 --> 00:00:08,045
produce a mole of carbon
dioxide and

5
00:00:08,045 --> 00:00:09,680
two moles of water.

6
00:00:09,680 --> 00:00:12,380
And what we want to answer in
this video is whether this

7
00:00:12,380 --> 00:00:14,760
reaction is spontaneous.

8
00:00:14,760 --> 00:00:17,290
And we learned in the last
video that to answer that

9
00:00:17,290 --> 00:00:19,810
question, we have to turn to
Gibbs free energy, or the

10
00:00:19,810 --> 00:00:21,620
change in Gibbs free energy.

11
00:00:21,620 --> 00:00:25,320
And the change in Gibbs free
energy is equal to the

12
00:00:25,320 --> 00:00:29,300
enthalpy change for the reaction
minus the temperature

13
00:00:29,300 --> 00:00:32,820
at which it is occurring, times
the change in entropy.

14
00:00:32,820 --> 00:00:35,840
And if this is less than
zero, then it's

15
00:00:35,840 --> 00:00:39,240
a spontaneous reaction.

16
00:00:39,240 --> 00:00:41,860
So I gave us a little
bit of a head start.

17
00:00:41,860 --> 00:00:46,070
I just calculated the change in
enthalpy for this reaction,

18
00:00:46,070 --> 00:00:47,530
and that's right here.

19
00:00:47,530 --> 00:00:48,460
And we know how to do that.

20
00:00:48,460 --> 00:00:50,230
We've done that several
videos ago.

21
00:00:50,230 --> 00:00:52,900
You could just look up the heats
of formation of each of

22
00:00:52,900 --> 00:00:54,880
these products.

23
00:00:54,880 --> 00:00:56,950
For water you'll multiply
it by 2, since you

24
00:00:56,950 --> 00:00:58,180
have 2 moles of it.

25
00:00:58,180 --> 00:01:00,510
And so you have the heats of
formation of all the products,

26
00:01:00,510 --> 00:01:02,500
and then you subtract
out the heats of

27
00:01:02,500 --> 00:01:03,780
formation of all the reactants.

28
00:01:03,780 --> 00:01:07,270
And of course the heat of
formation of O2 is O, so this

29
00:01:07,270 --> 00:01:11,040
won't even show up in it, and
you'll get minus 890.3

30
00:01:11,040 --> 00:01:11,590
kilojoules.

31
00:01:11,590 --> 00:01:15,090
Well, this tells us that this
is an exothermic reaction.

32
00:01:15,090 --> 00:01:18,490
That this side of the equation
has less energy in it-- you

33
00:01:18,490 --> 00:01:20,450
could kind of think of it
that way-- is that side.

34
00:01:20,450 --> 00:01:22,450
So some energy must have
been released.

35
00:01:22,450 --> 00:01:25,400
We could even put here, you
know, plus e for energy.

36
00:01:25,400 --> 00:01:27,980
Let me write, plus some energy
is going to be released.

37
00:01:27,980 --> 00:01:29,750
So that's why it's exothermic.

38
00:01:29,750 --> 00:01:32,010
But our question is, is
this spontaneous?

39
00:01:32,010 --> 00:01:34,100
So to figure out if it's
spontaneous, we also have to

40
00:01:34,100 --> 00:01:36,140
figure out our delta s.

41
00:01:36,140 --> 00:01:39,450


42
00:01:39,450 --> 00:01:42,690
And to help figure out the
delta s I, ahead of time,

43
00:01:42,690 --> 00:01:46,090
looked up the standard
molar entropies for

44
00:01:46,090 --> 00:01:48,130
each of these molecules.

45
00:01:48,130 --> 00:01:50,460
So for example, the standard--
I'll write it here in a

46
00:01:50,460 --> 00:01:51,710
different color.

47
00:01:51,710 --> 00:02:00,510


48
00:02:00,510 --> 00:02:04,980
The standard-- you put a little
naught symbol there--

49
00:02:04,980 --> 00:02:08,389
the standard molar entropy-- so
when we say standard, it's

50
00:02:08,389 --> 00:02:11,039
at 298 degrees Kelvin.

51
00:02:11,039 --> 00:02:12,730
Actually, I shouldn't
say degrees Kelvin.

52
00:02:12,730 --> 00:02:15,680
It's at 298 Kelvin You don't
use the word degrees,

53
00:02:15,680 --> 00:02:17,440
necessarily, when you
talk about Kelvin.

54
00:02:17,440 --> 00:02:21,020
So it's at 298 Kelvin, which is
25 degrees Celsius, so it's

55
00:02:21,020 --> 00:02:21,880
at room temperature.

56
00:02:21,880 --> 00:02:24,690
So that's why it's considered
standard temperature.

57
00:02:24,690 --> 00:02:29,700
So the standard entropy of
methane at room temperature is

58
00:02:29,700 --> 00:02:30,980
equal to this number
right here.

59
00:02:30,980 --> 00:02:37,880
186 joules per Kelvin mole.

60
00:02:37,880 --> 00:02:42,320
So if I have 1 mole of methane,
I have 186 joules per

61
00:02:42,320 --> 00:02:43,930
Kelvin of entropy.

62
00:02:43,930 --> 00:02:46,050
If I have 2 moles, I
multiply that by 2.

63
00:02:46,050 --> 00:02:48,780
If I have 3 moles, I
multiply that by 3.

64
00:02:48,780 --> 00:02:54,870
So the total change in entropy
of this reaction is the total

65
00:02:54,870 --> 00:02:58,950
standard entropies of the
products minus the total

66
00:02:58,950 --> 00:03:00,610
standard entropies
of the reactants.

67
00:03:00,610 --> 00:03:02,450
Just like what we did
with enthalpy.

68
00:03:02,450 --> 00:03:11,650
So that's going to be equal to
213.6 plus-- I have 2 moles of

69
00:03:11,650 --> 00:03:12,190
water here.

70
00:03:12,190 --> 00:03:17,810
So it's plus 2 times-- let's
just write 70 there.

71
00:03:17,810 --> 00:03:20,090
69.9, almost 70.

72
00:03:20,090 --> 00:03:24,720
Plus 2 times 70, and then I
want to subtract out the

73
00:03:24,720 --> 00:03:28,920
entropy of the reactants, or
this side of the reaction.

74
00:03:28,920 --> 00:03:42,900
So the entropy of 1 mole of CH4
is 186 plus 2 times 205.

75
00:03:42,900 --> 00:03:45,430
So just eyeballing it already,
this number is close to this

76
00:03:45,430 --> 00:03:48,220
number, but this number is much
larger than this number.

77
00:03:48,220 --> 00:03:50,620
Liquid water has a much
lower-- this is

78
00:03:50,620 --> 00:03:51,650
liquid water's entropy.

79
00:03:51,650 --> 00:03:54,970
It has a much lower entropy
than oxygen gas.

80
00:03:54,970 --> 00:03:55,730
And that makes sense.

81
00:03:55,730 --> 00:04:00,150
Because liquid formed, there's
a lot fewer states.

82
00:04:00,150 --> 00:04:02,630
It all falls to the bottom of
the container, as opposed to

83
00:04:02,630 --> 00:04:04,680
kind of taking the shape of
the room and expanding.

84
00:04:04,680 --> 00:04:06,600
So a gas is naturally going
to have much higher

85
00:04:06,600 --> 00:04:08,230
entropy than a liquid.

86
00:04:08,230 --> 00:04:11,830
So just eyeballing it, we can
already see that our products

87
00:04:11,830 --> 00:04:13,930
are going to have a lower
entropy than our reactants.

88
00:04:13,930 --> 00:04:15,400
So this is probably going
to be a negative number.

89
00:04:15,400 --> 00:04:19,430
But let's confirm that.

90
00:04:19,430 --> 00:04:30,610
So I have 200, 213.6 plus--
well, plus 140, right?

91
00:04:30,610 --> 00:04:31,270
2 times 70.

92
00:04:31,270 --> 00:04:35,540
Plus 140 is equal to 353.6.

93
00:04:35,540 --> 00:04:39,930
So this is 353.6.

94
00:04:39,930 --> 00:04:47,580
And then from that, I'm going to
subtract out-- so 186 plus

95
00:04:47,580 --> 00:04:54,790
2 times 205 is equal to 596.

96
00:04:54,790 --> 00:04:58,900
So minus 596, and what
is that equal to?

97
00:04:58,900 --> 00:05:06,780
So we put the minus 596, and
then plus the 353.6, and we

98
00:05:06,780 --> 00:05:10,910
have minus 242.4.

99
00:05:10,910 --> 00:05:18,220
So this is equal to minus 242.4
joules per Kelvin is our

100
00:05:18,220 --> 00:05:21,820
delta s minus.

101
00:05:21,820 --> 00:05:23,800
So we lose that much entropy.

102
00:05:23,800 --> 00:05:26,020
And those units might not make
sense to you right now, and

103
00:05:26,020 --> 00:05:28,780
actually you know these are
but of arbitrary units.

104
00:05:28,780 --> 00:05:30,750
But you can just say, hey, this
is getting more ordered.

105
00:05:30,750 --> 00:05:33,710
And it makes sense, because
we have a ton of gas.

106
00:05:33,710 --> 00:05:36,080
We have 3 separate molecules,
1 here and 2

107
00:05:36,080 --> 00:05:38,230
molecules of oxygen.

108
00:05:38,230 --> 00:05:40,840
And then we go to 3 molecules
again, but the

109
00:05:40,840 --> 00:05:42,310
water is now liquid.

110
00:05:42,310 --> 00:05:45,770
So it makes sense to me
that we lose entropy.

111
00:05:45,770 --> 00:05:47,330
There's fewer states
that the liquid,

112
00:05:47,330 --> 00:05:48,790
especially, can take on.

113
00:05:48,790 --> 00:05:51,650
But let's figure out whether
this reaction is spontaneous.

114
00:05:51,650 --> 00:05:57,440
So our delta g is equal
to our delta h.

115
00:05:57,440 --> 00:06:01,140
We're releasing energy,
so it's minus 890.

116
00:06:01,140 --> 00:06:02,510
I'll just get rid
of the decimals.

117
00:06:02,510 --> 00:06:04,490
We don't have to be
that precise.

118
00:06:04,490 --> 00:06:06,350
Minus our temperature.

119
00:06:06,350 --> 00:06:09,220
We're assuming that we're at
room temperature, or 298

120
00:06:09,220 --> 00:06:10,340
degrees Kelvin.

121
00:06:10,340 --> 00:06:13,400
That's 28-- I should just
say, 298 Kelvin.

122
00:06:13,400 --> 00:06:14,590
I should get in the habit
of not saying

123
00:06:14,590 --> 00:06:15,960
degrees when I say Kelvin.

124
00:06:15,960 --> 00:06:22,750
Which is 25 degrees Celsius,
times our change in entropy.

125
00:06:22,750 --> 00:06:25,050
Now, this is going
to be a minus.

126
00:06:25,050 --> 00:06:27,760
Now you might say, OK, minus
242, you might want to put

127
00:06:27,760 --> 00:06:28,530
that there.

128
00:06:28,530 --> 00:06:30,280
But you have to be very,
very, very careful.

129
00:06:30,280 --> 00:06:33,160
This right here is
in kilojoules.

130
00:06:33,160 --> 00:06:35,190
This right here is in joules.

131
00:06:35,190 --> 00:06:37,540
So if we want to write
everything in kilojoules,

132
00:06:37,540 --> 00:06:39,660
since we already wrote that
down, let's write this in

133
00:06:39,660 --> 00:06:40,370
kilojoules.

134
00:06:40,370 --> 00:06:46,620
So it's 0.242 kilojoules
per Kelvin.

135
00:06:46,620 --> 00:06:52,370


136
00:06:52,370 --> 00:06:55,790
And so now our Gibbs free energy
right here is going to

137
00:06:55,790 --> 00:07:00,880
be minus 890 kilojoules minus
290-- so the minus and the

138
00:07:00,880 --> 00:07:02,760
minus, you get a plus.

139
00:07:02,760 --> 00:07:05,770
And that makes sense, that the
entropy term is going to make

140
00:07:05,770 --> 00:07:08,480
our Gibbs free energy
more positive.

141
00:07:08,480 --> 00:07:11,480
Which, as we know, since we want
to get this thing below

142
00:07:11,480 --> 00:07:13,980
0, this is going to fight
the spontaneity.

143
00:07:13,980 --> 00:07:18,840
But let's see if it can
overwhelm the actual enthalpy,

144
00:07:18,840 --> 00:07:20,470
the exothermic nature of it.

145
00:07:20,470 --> 00:07:22,480
And it seems like it will,
because you multiply a

146
00:07:22,480 --> 00:07:24,010
fraction times this,
it's going to be a

147
00:07:24,010 --> 00:07:25,090
smaller number than that.

148
00:07:25,090 --> 00:07:27,560
But let's just figure it out.

149
00:07:27,560 --> 00:07:31,830
So divided by 1, 2, 3.

150
00:07:31,830 --> 00:07:37,170
That's our change in entropy
times 298, that's our

151
00:07:37,170 --> 00:07:40,410
temperature, is minus 72.

152
00:07:40,410 --> 00:07:43,710
So this term becomes-- and then
we put a minus there-- so

153
00:07:43,710 --> 00:07:47,360
it's plus 72.2.

154
00:07:47,360 --> 00:07:50,300
So this is the entropy term
at standard temperature.

155
00:07:50,300 --> 00:07:51,370
It turns into that.

156
00:07:51,370 --> 00:07:52,970
And this is our enthalpy term.

157
00:07:52,970 --> 00:07:55,900
So we can already see that the
enthalpy is a much more

158
00:07:55,900 --> 00:07:58,320
negative number than our
positive term from our

159
00:07:58,320 --> 00:08:00,480
temperature times our
change in entropy.

160
00:08:00,480 --> 00:08:04,530
So this term is going
to win out.

161
00:08:04,530 --> 00:08:08,330
Even though we lose entropy in
this reaction, it releases so

162
00:08:08,330 --> 00:08:10,840
much energy that's going
to be spontaneous.

163
00:08:10,840 --> 00:08:14,340
This is definitely less than
0, so this is going to be a

164
00:08:14,340 --> 00:08:17,360
spontaneous reaction.

165
00:08:17,360 --> 00:08:19,520
As you can see, these Gibbs free
energy problems, they're

166
00:08:19,520 --> 00:08:20,630
really not too difficult.

167
00:08:20,630 --> 00:08:23,900
You just really need to
find these values.

168
00:08:23,900 --> 00:08:27,230
And to find these values, it'll
either be given, the

169
00:08:27,230 --> 00:08:29,670
delta h, but we know how to
solve for the delta h.

170
00:08:29,670 --> 00:08:31,720
You just look up the heats
of formations of all the

171
00:08:31,720 --> 00:08:35,210
products, subtract out the
reactants, and of course you

172
00:08:35,210 --> 00:08:38,110
wait by the coefficients.

173
00:08:38,110 --> 00:08:40,490
And then, to figure out the
change in entropy, you do the

174
00:08:40,490 --> 00:08:40,909
same thing.

175
00:08:40,909 --> 00:08:44,020
You have to look up the standard
molar entropies of

176
00:08:44,020 --> 00:08:46,840
the products' weight by the
coefficients, subtract out the

177
00:08:46,840 --> 00:08:50,110
reactants, and then just
substitute in here, and then

178
00:08:50,110 --> 00:08:51,850
you essentially have the
Gibbs free energy.

179
00:08:51,850 --> 00:08:54,640
And in this case,
it was negative.

180
00:08:54,640 --> 00:08:56,570
Now, you could imagine a
situation where we're at a

181
00:08:56,570 --> 00:08:57,720
much higher temperature.

182
00:08:57,720 --> 00:09:00,890
Like the surface of the sun or
something, where all of a

183
00:09:00,890 --> 00:09:06,380
sudden, instead of a 298 here,
if you had like a 2,000 or a

184
00:09:06,380 --> 00:09:08,140
4,000 there.

185
00:09:08,140 --> 00:09:10,700
Then all of a sudden, things
become interesting.

186
00:09:10,700 --> 00:09:14,710
If you could imagine, if
you had a 40,000 Kelvin

187
00:09:14,710 --> 00:09:17,530
temperature here, then all of a
sudden the entropy term, the

188
00:09:17,530 --> 00:09:20,150
loss of entropy, is going
to matter a lot more.

189
00:09:20,150 --> 00:09:22,480
And so this term, this positive
term, is going to

190
00:09:22,480 --> 00:09:25,870
outweigh this, and maybe it
wouldn't be spontaneous at a

191
00:09:25,870 --> 00:09:27,960
very, very, very, very
high temperature.

192
00:09:27,960 --> 00:09:29,160
Another way to think about it.

193
00:09:29,160 --> 00:09:34,560
A reaction that generates heat
that lets out heat-- the heat

194
00:09:34,560 --> 00:09:37,090
being released doesn't matter so
much when there's already a

195
00:09:37,090 --> 00:09:39,980
lot of heat or kinetic energy
in the environment.

196
00:09:39,980 --> 00:09:42,920
If the temperature was high
enough, this reaction would

197
00:09:42,920 --> 00:09:46,050
not be spontaneous, because
maybe then the entropy term

198
00:09:46,050 --> 00:09:46,980
would win out.

199
00:09:46,980 --> 00:09:49,130
But anyway, I just wanted to do
this calculation for you to

200
00:09:49,130 --> 00:09:51,360
show you that there's nothing
too abstract here.

201
00:09:51,360 --> 00:09:53,950
You can look up everything on
the web, and then figure out

202
00:09:53,950 --> 00:09:56,320
if something is going
to be spontaneous.

203
00:09:56,320 --> 00:09:56,597