1
00:00:00,410 --> 00:00:01,670
有个反应是

2
00:00:01,690 --> 00:00:03,420
1mol的甲烷

3
00:00:03,430 --> 00:00:05,590
和2mol的氧气反应

4
00:00:05,610 --> 00:00:08,220
会生成1mol的二氧化碳

5
00:00:08,240 --> 00:00:09,430
和2mol的水

6
00:00:09,440 --> 00:00:11,630
这集里 我们想判断

7
00:00:11,650 --> 00:00:14,940
这个反应是不是自发的

8
00:00:14,960 --> 00:00:16,200
上次我们已经学过

9
00:00:16,210 --> 00:00:17,750
怎么判断自发性啦

10
00:00:17,760 --> 00:00:19,700
这时候就要利用吉布斯自由能

11
00:00:19,710 --> 00:00:21,330
或吉布斯自由能变啦

12
00:00:21,340 --> 00:00:23,520
而吉布斯自由能变ΔG

13
00:00:23,530 --> 00:00:28,140
等于反应的焓变ΔH

14
00:00:28,160 --> 00:00:30,700
减去反应的温度T

15
00:00:30,710 --> 00:00:32,820
乘以熵变ΔS

16
00:00:32,840 --> 00:00:35,010
如果ΔG<0

17
00:00:35,030 --> 00:00:39,400
那反应就是自发的了

18
00:00:39,410 --> 00:00:41,860
我先给大伙开个好头

19
00:00:41,880 --> 00:00:44,040
我刚刚已经把

20
00:00:44,060 --> 00:00:46,080
反应的焓变算出来了

21
00:00:46,090 --> 00:00:47,460
就在这里呢

22
00:00:47,480 --> 00:00:48,440
大家都知道怎么求ΔH了吧

23
00:00:48,450 --> 00:00:50,170
几集前我们讲过的

24
00:00:50,180 --> 00:00:52,360
先查出来每个产物的

25
00:00:52,380 --> 00:00:54,480
生成热

26
00:00:54,490 --> 00:00:56,650
例如水 你要把生成热乘以2

27
00:00:56,670 --> 00:00:58,170
因为反应生成了2mol的水

28
00:00:58,180 --> 00:01:00,710
这样就有了产物的生成热之和

29
00:01:00,730 --> 00:01:02,070
然后再减去

30
00:01:02,090 --> 00:01:03,940
反应物的生成热之和

31
00:01:03,960 --> 00:01:07,100
当然啦 O2的生成热是0

32
00:01:07,110 --> 00:01:08,510
所以式子里面没有这项

33
00:01:08,520 --> 00:01:11,910
算出来就是 -890.3kJ

34
00:01:11,930 --> 00:01:12,060
好啦

35
00:01:12,070 --> 00:01:14,950
这就说明 反应是放热的

36
00:01:14,970 --> 00:01:18,380
方程式这边的能量小于…

37
00:01:18,400 --> 00:01:19,630
你也可以这样想的…

38
00:01:19,650 --> 00:01:20,360
比那边的能量小

39
00:01:20,370 --> 00:01:22,650
所以必须释放能量才行

40
00:01:22,670 --> 00:01:25,430
可以在这里写 +e e代表能量

41
00:01:25,440 --> 00:01:25,920
我写上

42
00:01:25,930 --> 00:01:28,080
加上释放出来的能量

43
00:01:28,100 --> 00:01:29,720
这就是反应放热的原因啦

44
00:01:29,740 --> 00:01:31,960
但是问题是 反应是不是自发的呢？

45
00:01:31,970 --> 00:01:33,550
想要判断反应的自发性

46
00:01:33,570 --> 00:01:39,240
首先要算出ΔS

47
00:01:39,260 --> 00:01:41,380
为了计算ΔS的值呢

48
00:01:41,400 --> 00:01:43,190
我提前就查好了

49
00:01:43,210 --> 00:01:48,140
这里每种分子的标准摩尔熵

50
00:01:48,160 --> 00:01:49,610
比如说 标准…

51
00:01:49,630 --> 00:01:51,410
我换个颜色表示

52
00:01:51,420 --> 00:01:53,360
标准

53
00:01:53,380 --> 00:01:56,790
小小讲点拓展 这里没有Δ

54
00:01:56,800 --> 00:02:00,510
我擦了吧 还能补救

55
00:02:00,530 --> 00:02:02,850
标准

56
00:02:02,870 --> 00:02:05,240
这里画个圈里面带个横表示

57
00:02:05,260 --> 00:02:07,400
标准摩尔熵Sm

58
00:02:07,410 --> 00:02:10,850
“标准”指的是在298°K下

59
00:02:10,860 --> 00:02:12,660
实际不应该说“度开尔文”

60
00:02:12,680 --> 00:02:14,590
就是298K

61
00:02:14,600 --> 00:02:16,420
用开尔文K的时候

62
00:02:16,440 --> 00:02:17,450
不用说度°

63
00:02:17,460 --> 00:02:19,240
所以反应温度是289K

64
00:02:19,260 --> 00:02:20,650
也就是25°C

65
00:02:20,660 --> 00:02:22,100
相当于室温

66
00:02:22,110 --> 00:02:24,300
所以用289K作标准状态

67
00:02:24,320 --> 00:02:29,240
所以室温下 甲烷的标准摩尔熵

68
00:02:29,260 --> 00:02:31,030
就等于这个数

69
00:02:37,810 --> 00:02:40,380
所以如果有1mol的甲烷

70
00:02:40,400 --> 00:02:43,890
就有186J/K的熵

71
00:02:43,900 --> 00:02:46,130
如果有2mol的甲烷 就乘以2

72
00:02:46,140 --> 00:02:48,420
如果有3mol 就乘以3

73
00:02:48,430 --> 00:02:53,460
所以这个反应的总熵变

74
00:02:53,470 --> 00:02:58,020
就是产物标准熵之和

75
00:02:58,040 --> 00:03:00,690
减去反应物标准熵之和

76
00:03:00,700 --> 00:03:02,550
就跟算ΔHr差不多

77
00:03:02,560 --> 00:03:09,760
所以熵变就等于213.6 加上…

78
00:03:09,770 --> 00:03:12,390
产物里有2mol的水

79
00:03:12,400 --> 00:03:15,920
所以就是加上2乘以…

80
00:03:15,940 --> 00:03:17,850
就取70好了

81
00:03:17,860 --> 00:03:19,800
69.9 约等于70

82
00:03:19,820 --> 00:03:21,910
加上2×70

83
00:03:21,920 --> 00:03:23,790
然后再减去

84
00:03:23,800 --> 00:03:26,110
反应物的熵之和

85
00:03:26,120 --> 00:03:28,910
也就是方程式这边的这些

86
00:03:28,920 --> 00:03:31,770
1molCH4的熵

87
00:03:31,780 --> 00:03:42,860
等于186 加上2×205

88
00:03:42,880 --> 00:03:44,250
大概心算一下

89
00:03:44,270 --> 00:03:45,670
这个数非常接近这个数

90
00:03:45,690 --> 00:03:48,020
但是这个数比这个数大得多

91
00:03:48,040 --> 00:03:50,330
液态水的熵…

92
00:03:50,350 --> 00:03:51,990
这是液态水的熵

93
00:03:52,000 --> 00:03:54,650
它的熵远远小于氧气的熵

94
00:03:54,670 --> 00:03:55,760
这很合理呀

95
00:03:55,770 --> 00:03:58,570
因为液态水的微观状态数比氧气少得多

96
00:03:58,580 --> 00:04:02,370
液态水都沉在容器底了

97
00:04:02,390 --> 00:04:03,040
气体就不同

98
00:04:03,050 --> 00:04:04,670
气体能膨胀 随空间变换形状

99
00:04:04,690 --> 00:04:06,030
所以理所当然 气体的熵

100
00:04:06,050 --> 00:04:08,150
比液体的熵大的多

101
00:04:08,170 --> 00:04:09,260
简单心算

102
00:04:09,270 --> 00:04:12,280
就已经能看出来产物的熵

103
00:04:12,300 --> 00:04:14,020
比反应物的熵小

104
00:04:14,030 --> 00:04:15,460
所以熵变应该是负的

105
00:04:15,480 --> 00:04:19,430
不过我们还是确认一下

106
00:04:19,440 --> 00:04:28,540
这个是213.6加上…

107
00:04:28,560 --> 00:04:30,750
是加上140 对嘛？

108
00:04:30,760 --> 00:04:31,380
是2×70

109
00:04:31,390 --> 00:04:35,550
加上140 就等于353.6

110
00:04:35,560 --> 00:04:39,900
这部分等于353.6

111
00:04:39,910 --> 00:04:43,580
然后从这里减去…

112
00:04:43,590 --> 00:04:52,660
所以186 加上2×205

113
00:04:52,670 --> 00:04:54,430
就等于596

114
00:04:54,440 --> 00:04:57,170
所以就是减去596

115
00:04:57,190 --> 00:04:58,600
最后等于什么？

116
00:04:58,620 --> 00:05:06,430
-596 加上353.6

117
00:05:06,440 --> 00:05:10,520
等于-242.4

118
00:05:10,530 --> 00:05:17,610
所以它就等于-242.4J/K

119
00:05:17,630 --> 00:05:21,200
这就是ΔS 负的

120
00:05:21,220 --> 00:05:24,060
所以系统的熵减少了这么多

121
00:05:24,080 --> 00:05:25,980
你可能对熵的单位大小没有概念

122
00:05:25,990 --> 00:05:28,950
不过只要知道是某个大小就可以

123
00:05:28,960 --> 00:05:29,620
但是你可以说 喏

124
00:05:29,630 --> 00:05:30,830
反应之后系统更有序啦

125
00:05:30,840 --> 00:05:32,760
这很合理 因为开始是一大堆气体

126
00:05:32,770 --> 00:05:35,350
开始是3个单独的分子

127
00:05:35,360 --> 00:05:38,300
有1个甲烷 还有2个氧气

128
00:05:38,310 --> 00:05:40,080
后来还是3个分子

129
00:05:40,100 --> 00:05:42,390
但是这个水是液态的

130
00:05:42,400 --> 00:05:45,520
所以 反应后熵减小是有道理的

131
00:05:45,530 --> 00:05:48,570
尤其液态物质 它的微观状态数很少

132
00:05:48,590 --> 00:05:49,430
我们来判断一下

133
00:05:49,450 --> 00:05:51,210
这个反应是不是自发的

134
00:05:51,220 --> 00:05:57,510
ΔG等于ΔH…

135
00:05:57,530 --> 00:06:00,880
反应放热 所以就是-890

136
00:06:00,900 --> 00:06:02,540
我把小数省略掉了

137
00:06:02,550 --> 00:06:03,990
我们不用那么精确

138
00:06:04,010 --> 00:06:05,930
减去温度

139
00:06:05,950 --> 00:06:08,280
假设反应是在室温下进行的

140
00:06:08,290 --> 00:06:10,210
所以温度是298°K

141
00:06:10,230 --> 00:06:13,390
就是… 我应该说“298K”

142
00:06:13,400 --> 00:06:14,420
我要改掉坏习惯

143
00:06:14,430 --> 00:06:15,910
在用K表示温度的时候 不说“°”

144
00:06:15,930 --> 00:06:18,710
298K 也就是25°C

145
00:06:18,730 --> 00:06:22,100
再乘以熵变

146
00:06:22,120 --> 00:06:24,920
这项是负的

147
00:06:24,940 --> 00:06:27,460
你可能会说 好的 是-242

148
00:06:27,470 --> 00:06:28,540
直接把这个数放进去

149
00:06:28,550 --> 00:06:30,360
但是你要非常非常非常的小心

150
00:06:30,370 --> 00:06:33,040
它的单位是千焦kJ

151
00:06:33,050 --> 00:06:34,940
可是它的单位是焦耳J

152
00:06:34,960 --> 00:06:37,630
所以如果都以千焦做单位的话

153
00:06:37,640 --> 00:06:38,870
因为前面写了kJ

154
00:06:38,890 --> 00:06:40,480
我们把这个也换算成千焦吧

155
00:06:40,500 --> 00:06:46,990
所以它就是0.242kJ/K

156
00:06:47,000 --> 00:06:48,100
前面放个小数点

157
00:06:48,110 --> 00:06:50,110
这里的0.45擦掉

158
00:06:50,120 --> 00:06:51,890
单位是kJ/k

159
00:06:51,910 --> 00:06:55,510
所以吉布斯自由能变

160
00:06:55,530 --> 00:07:00,250
就是-890kJ 减去298…

161
00:07:00,260 --> 00:07:02,660
负负得正

162
00:07:02,690 --> 00:07:03,880
完全合理

163
00:07:03,890 --> 00:07:05,650
因为熵的这项

164
00:07:05,660 --> 00:07:08,390
会使吉布斯自由能变得更正

165
00:07:08,410 --> 00:07:09,410
因为

166
00:07:09,430 --> 00:07:12,080
我们想让ΔG<0

167
00:07:12,090 --> 00:07:14,070
但是ΔS>0会降低自发性

168
00:07:14,090 --> 00:07:18,840
现在我们来看这项能不能抵消ΔH

169
00:07:18,850 --> 00:07:20,590
也就是放热的影响

170
00:07:20,610 --> 00:07:21,670
目测好像是不行

171
00:07:21,670 --> 00:07:23,710
因为一个小数乘以它

172
00:07:23,730 --> 00:07:25,130
得到的数肯定更小

173
00:07:25,150 --> 00:07:26,720
我们算算看

174
00:07:26,730 --> 00:07:31,040
所以除以 1,2,3 3个0

175
00:07:31,050 --> 00:07:34,320
系统的熵变

176
00:07:34,330 --> 00:07:37,810
乘以298 这是系统的温度

177
00:07:37,820 --> 00:07:40,160
等于-72

178
00:07:40,170 --> 00:07:42,610
所以这项就等于…

179
00:07:42,630 --> 00:07:43,790
因为前面还有个减号…

180
00:07:43,810 --> 00:07:46,860
所以就是加上72.2

181
00:07:46,870 --> 00:07:50,010
所以这就是标准温度下的熵的项

182
00:07:50,030 --> 00:07:51,360
最后就等于它咯

183
00:07:51,370 --> 00:07:53,040
而这是焓项

184
00:07:53,050 --> 00:07:54,160
这样我们就能看出来

185
00:07:54,180 --> 00:07:57,020
焓变的绝对值

186
00:07:57,040 --> 00:07:59,040
比T×ΔS的绝对值

187
00:07:59,050 --> 00:08:00,370
大得多

188
00:08:00,380 --> 00:08:04,530
所以这项压倒性胜利了

189
00:08:04,550 --> 00:08:06,910
虽然反应是个熵减的反应

190
00:08:06,920 --> 00:08:09,110
但是反应放出的热量太多了

191
00:08:09,130 --> 00:08:10,930
所以反应仍然是自发

192
00:08:10,950 --> 00:08:12,890
这个数显然小于0

193
00:08:12,910 --> 00:08:17,340
所以这是个自发反应

194
00:08:17,350 --> 00:08:19,400
如你所见 这些吉布斯自由能的问题

195
00:08:19,420 --> 00:08:20,710
其实没那么难

196
00:08:20,730 --> 00:08:23,570
只要知道这几项的值就行啦

197
00:08:23,580 --> 00:08:27,130
这几项的值要么直接给出

198
00:08:27,150 --> 00:08:27,970
比如ΔH

199
00:08:27,990 --> 00:08:29,830
不过我们也知道怎么求出来

200
00:08:29,850 --> 00:08:31,240
只要查到产物的

201
00:08:31,250 --> 00:08:32,540
生成热

202
00:08:32,550 --> 00:08:34,640
再减去反应物的生成热

203
00:08:34,650 --> 00:08:37,770
当然还要各自乘以相应的化学计量数

204
00:08:37,780 --> 00:08:40,170
然后 用同样的方法

205
00:08:40,180 --> 00:08:41,010
算出熵变

206
00:08:41,030 --> 00:08:43,650
查到每种产物的标准摩尔熵

207
00:08:43,670 --> 00:08:46,000
分别乘以相应的化学计量数

208
00:08:46,020 --> 00:08:47,810
再减去反应物的总熵

209
00:08:47,820 --> 00:08:49,970
然后把数代入这个式子中

210
00:08:49,990 --> 00:08:51,980
最后就得到了吉布斯自由能变

211
00:08:51,990 --> 00:08:54,560
这个例子里 ΔG是负的

212
00:08:54,570 --> 00:08:56,170
现在 大家可以想象一下

213
00:08:56,190 --> 00:08:57,780
温度极高的情况

214
00:08:57,800 --> 00:09:00,200
比如太阳表面之类的

215
00:09:00,220 --> 00:09:04,000
温度就不是298K啦

216
00:09:04,010 --> 00:09:07,980
温度一下子变成了2000K或者4000K

217
00:09:08,000 --> 00:09:09,920
这时候就有意思啦

218
00:09:09,940 --> 00:09:11,430
比如说

219
00:09:11,450 --> 00:09:15,660
反应温度是40000K

220
00:09:15,670 --> 00:09:17,520
那么熵这一项

221
00:09:17,540 --> 00:09:19,870
也就是熵减 影响就可大啦

222
00:09:19,890 --> 00:09:22,300
所以正的这一项

223
00:09:22,310 --> 00:09:23,190
就抵消这一项

224
00:09:23,210 --> 00:09:25,650
所以在超高温下

225
00:09:25,670 --> 00:09:28,010
反应可能就无法自发进行啦

226
00:09:28,020 --> 00:09:29,110
换个角度

227
00:09:29,120 --> 00:09:34,230
一个反应放出热量…

228
00:09:34,240 --> 00:09:36,300
环境温度已经非常高

229
00:09:36,310 --> 00:09:38,180
分子的动能已经很大了的时候

230
00:09:38,190 --> 00:09:40,020
放出的热量就没什么影响了

231
00:09:40,040 --> 00:09:41,460
如果环境温度足够高

232
00:09:41,480 --> 00:09:43,890
这个反应就不是自发的了

233
00:09:43,910 --> 00:09:47,060
因为熵项会把焓抵消掉

234
00:09:47,080 --> 00:09:47,500
好啦

235
00:09:47,520 --> 00:09:49,020
我只是想带大家算一次

236
00:09:49,040 --> 00:09:51,350
就是想让大家知道 这没那么难

237
00:09:51,370 --> 00:09:53,010
这些数据都可以从网上查到

238
00:09:53,020 --> 00:09:54,000
然后就能判断出

239
00:09:54,010 --> 00:09:56,320
反应是否可以自发进行了