1
00:00:00,410 --> 00:00:01,670
有個反應是

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

3
00:00:03,430 --> 00:00:05,590
和2mol的氧氣反應

4
00:00:05,610 --> 00:00:08,220
會生成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
這集裏 我們想判斷

7
00:00:11,650 --> 00:00:14,940
這個反應是不是自發的

8
00:00:14,960 --> 00:00:16,200
上次我們已經學過

9
00:00:16,210 --> 00:00:17,750
怎麽判斷自發性啦

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

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

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

13
00:00:23,530 --> 00:00:28,140
等於反應的焓變ΔH

14
00:00:28,160 --> 00:00:30,700
減去反應的溫度T

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

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

17
00:00:35,030 --> 00:00:39,400
那反應就是自發的了

18
00:00:39,410 --> 00:00:41,860
我先給大夥開個好頭

19
00:00:41,880 --> 00:00:44,040
我剛剛已經把

20
00:00:44,060 --> 00:00:46,080
反應的焓變算出來了

21
00:00:46,090 --> 00:00:47,460
就在這裡呢

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

23
00:00:48,450 --> 00:00:50,170
幾集前我們講過的

24
00:00:50,180 --> 00:00:52,360
先查出來每個產物的

25
00:00:52,380 --> 00:00:54,480
生成熱

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

27
00:00:56,670 --> 00:00:58,170
因爲反應生成了2mol的水

28
00:00:58,180 --> 00:01:00,710
這樣就有了產物的生成熱之和

29
00:01:00,730 --> 00:01:02,070
然後再減去

30
00:01:02,090 --> 00:01:03,940
反應物的生成熱之和

31
00:01:03,960 --> 00:01:07,100
當然啦 O2的生成熱是0

32
00:01:07,110 --> 00:01:08,510
所以式子裏面沒有這項

33
00:01:08,520 --> 00:01:11,910
算出來就是 -890.3kJ

34
00:01:11,930 --> 00:01:12,060
好啦

35
00:01:12,070 --> 00:01:14,950
這就說明 反應是放熱的

36
00:01:14,970 --> 00:01:18,380
方程式這邊的能量少於…

37
00:01:18,400 --> 00:01:19,630
你也可以這樣想的…

38
00:01:19,650 --> 00:01:20,360
比那邊的能量小

39
00:01:20,370 --> 00:01:22,650
所以必須釋放能量才行

40
00:01:22,670 --> 00:01:25,430
可以在這裡寫 +e e代表能量

41
00:01:25,440 --> 00:01:25,920
我寫上

42
00:01:25,930 --> 00:01:28,080
加上釋放出來的能量

43
00:01:28,100 --> 00:01:29,720
這就是反應放熱的原因啦

44
00:01:29,740 --> 00:01:31,960
但是問題是 反應是不是自發的呢？

45
00:01:31,970 --> 00:01:33,550
想要判斷反應的自發性

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

47
00:01:39,260 --> 00:01:41,380
爲了計算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
這裡每種分子的標準莫耳熵

50
00:01:48,160 --> 00:01:49,610
比如說 標準…

51
00:01:49,630 --> 00:01:51,410
我換個顏色表示

52
00:01:51,420 --> 00:01:53,360
標準

53
00:01:53,380 --> 00:01:56,790
小小講點拓展 這裡沒有Δ

54
00:01:56,800 --> 00:02:00,510
我擦了吧 還能補救

55
00:02:00,530 --> 00:02:02,850
標準

56
00:02:02,870 --> 00:02:05,240
這裡畫個圈裏面帶個橫表示

57
00:02:05,260 --> 00:02:07,400
標準莫耳熵Sm

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

59
00:02:10,860 --> 00:02:12,660
實際不應該說“度克耳文”

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

61
00:02:14,600 --> 00:02:16,420
用克耳文K的時候

62
00:02:16,440 --> 00:02:17,450
不用說度°

63
00:02:17,460 --> 00:02:19,240
所以反應溫度是289K

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

65
00:02:20,660 --> 00:02:22,100
相當於室溫

66
00:02:22,110 --> 00:02:24,300
所以用289K作標準狀態

67
00:02:24,320 --> 00:02:29,240
所以室溫下 甲烷的標準莫耳熵

68
00:02:29,260 --> 00:02:31,030
就等於這個數

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
所以這個反應的總熵變

74
00:02:53,470 --> 00:02:58,020
就是產物標準熵之和

75
00:02:58,040 --> 00:03:00,690
減去反應物標準熵之和

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
所以熵變就等於213.6 加上…

78
00:03:09,770 --> 00:03:12,390
產物裏有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 約等於70

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

83
00:03:21,920 --> 00:03:23,790
然後再減去

84
00:03:23,800 --> 00:03:26,110
反應物的熵之和

85
00:03:26,120 --> 00:03:28,910
也就是方程式這邊的這些

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

87
00:03:31,780 --> 00:03:42,860
等於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
這個數非常接近這個數

90
00:03:45,690 --> 00:03:48,020
但是這個數比這個數大得多

91
00:03:48,040 --> 00:03:50,330
液態水的熵…

92
00:03:50,350 --> 00:03:51,990
這是液態水的熵

93
00:03:52,000 --> 00:03:54,650
它的熵遠遠少於氧氣的熵

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

95
00:03:55,770 --> 00:03:58,570
因爲液態水的微觀狀態數比氧氣少得多

96
00:03:58,580 --> 00:04:02,370
液態水都沈在容器底了

97
00:04:02,390 --> 00:04:03,040
氣體就不同

98
00:04:03,050 --> 00:04:04,670
氣體能膨脹 隨空間變換形狀

99
00:04:04,690 --> 00:04:06,030
所以理所當然 氣體的熵

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

101
00:04:08,170 --> 00:04:09,260
簡單心算

102
00:04:09,270 --> 00:04:12,280
就已經能看出來產物的熵

103
00:04:12,300 --> 00:04:14,020
比反應物的熵小

104
00:04:14,030 --> 00:04:15,460
所以熵變應該是負的

105
00:04:15,480 --> 00:04:19,430
不過我們還是確認一下

106
00:04:19,440 --> 00:04:28,540
這個是213.6加上…

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

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

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

110
00:04:35,560 --> 00:04:39,900
這部分等於353.6

111
00:04:39,910 --> 00:04:43,580
然後從這裡減去…

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
就等於596

114
00:04:54,440 --> 00:04:57,170
所以就是減去596

115
00:04:57,190 --> 00:04:58,600
最後等於什麽？

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

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

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

119
00:05:17,630 --> 00:05:21,200
這就是ΔS 負的

120
00:05:21,220 --> 00:05:24,060
所以係統的熵減少了這麽多

121
00:05:24,080 --> 00:05:25,980
你可能對熵的單位大小沒有概念

122
00:05:25,990 --> 00:05:28,950
不過只要知道是某個大小就可以

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

124
00:05:29,630 --> 00:05:30,830
反應之後係統更有序啦

125
00:05:30,840 --> 00:05:32,760
這很合理 因爲開始是一大堆氣體

126
00:05:32,770 --> 00:05:35,350
開始是3個單獨的分子

127
00:05:35,360 --> 00:05:38,300
有1個甲烷 還有2個氧氣

128
00:05:38,310 --> 00:05:40,080
後來還是3個分子

129
00:05:40,100 --> 00:05:42,390
但是這個水是液態的

130
00:05:42,400 --> 00:05:45,520
所以 反應後熵減小是有道理的

131
00:05:45,530 --> 00:05:48,570
尤其液態物質 它的微觀狀態數很少

132
00:05:48,590 --> 00:05:49,430
我們來判斷一下

133
00:05:49,450 --> 00:05:51,210
這個反應是不是自發的

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

135
00:05:57,530 --> 00:06:00,880
反應放熱 所以就是-890

136
00:06:00,900 --> 00:06:02,540
我把小數省略掉了

137
00:06:02,550 --> 00:06:03,990
我們不用那麽精確

138
00:06:04,010 --> 00:06:05,930
減去溫度

139
00:06:05,950 --> 00:06:08,280
假設反應是在室溫下進行的

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

141
00:06:10,230 --> 00:06:13,390
就是… 我應該說“298K”

142
00:06:13,400 --> 00:06:14,420
我要改掉壞習慣

143
00:06:14,430 --> 00:06:15,910
在用K表示溫度的時候 不說“°”

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

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

146
00:06:22,120 --> 00:06:24,920
這項是負的

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

148
00:06:27,470 --> 00:06:28,540
直接把這個數放進去

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
它的單位是千焦kJ

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

152
00:06:34,960 --> 00:06:37,630
所以如果都以千焦做單位的話

153
00:06:37,640 --> 00:06:38,870
因爲前面寫了kJ

154
00:06:38,890 --> 00:06:40,480
我們把這個也換算成千焦吧

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
前面放個小數點

157
00:06:48,110 --> 00:06:50,110
這裡的0.45擦掉

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

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

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

161
00:07:00,260 --> 00:07:02,660
負負得正

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

163
00:07:03,890 --> 00:07:05,650
因爲熵的這項

164
00:07:05,660 --> 00:07:08,390
會使吉布斯自由能變得更正

165
00:07:08,410 --> 00:07:09,410
因爲

166
00:07:09,430 --> 00:07:12,080
我們想讓ΔG<0

167
00:07:12,090 --> 00:07:14,070
但是ΔS>0會降低自發性

168
00:07:14,090 --> 00:07:18,840
現在我們來看這項能不能抵消ΔH

169
00:07:18,850 --> 00:07:20,590
也就是放熱的影響

170
00:07:20,610 --> 00:07:21,670
目測好像是不行

171
00:07:21,670 --> 00:07:23,710
因爲一個小數乘以它

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

173
00:07:25,150 --> 00:07:26,720
我們算算看

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

175
00:07:31,050 --> 00:07:34,320
係統的熵變

176
00:07:34,330 --> 00:07:37,810
乘以298 這是係統的溫度

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

178
00:07:40,170 --> 00:07:42,610
所以這項就等於…

179
00:07:42,630 --> 00:07:43,790
因爲前面還有個減號…

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
所以這就是標準溫度下的熵的項

182
00:07:50,030 --> 00:07:51,360
最後就等於它咯

183
00:07:51,370 --> 00:07:53,040
而這是焓項

184
00:07:53,050 --> 00:07:54,160
這樣我們就能看出來

185
00:07:54,180 --> 00:07:57,020
焓變的絕對值

186
00:07:57,040 --> 00:07:59,040
比T×ΔS的絕對值

187
00:07:59,050 --> 00:08:00,370
大得多

188
00:08:00,380 --> 00:08:04,530
所以這項壓倒性勝利了

189
00:08:04,550 --> 00:08:06,910
雖然反應是個熵減的反應

190
00:08:06,920 --> 00:08:09,110
但是反應放出的熱量太多了

191
00:08:09,130 --> 00:08:10,930
所以反應仍然是自發

192
00:08:10,950 --> 00:08:12,890
這個數顯然少於0

193
00:08:12,910 --> 00:08:17,340
所以這是個自發反應

194
00:08:17,350 --> 00:08:19,400
如你所見 這些吉布斯自由能的問題

195
00:08:19,420 --> 00:08:20,710
其實沒那麽難

196
00:08:20,730 --> 00:08:23,570
只要知道這幾項的值就行啦

197
00:08:23,580 --> 00:08:27,130
這幾項的值要麽直接給出

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

199
00:08:27,990 --> 00:08:29,830
不過我們也知道怎麽求出來

200
00:08:29,850 --> 00:08:31,240
只要查到產物的

201
00:08:31,250 --> 00:08:32,540
生成熱

202
00:08:32,550 --> 00:08:34,640
再減去反應物的生成熱

203
00:08:34,650 --> 00:08:37,770
當然還要各自乘以相應的化學計量數

204
00:08:37,780 --> 00:08:40,170
然後 用同樣的方法

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

206
00:08:41,030 --> 00:08:43,650
查到每種產物的標準莫耳熵

207
00:08:43,670 --> 00:08:46,000
分別乘以相應的化學計量數

208
00:08:46,020 --> 00:08:47,810
再減去反應物的總熵

209
00:08:47,820 --> 00:08:49,970
然後把數代入這個式子中

210
00:08:49,990 --> 00:08:51,980
最後就得到了吉布斯自由能變

211
00:08:51,990 --> 00:08:54,560
這個例子裏 ΔG是負的

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

213
00:08:56,190 --> 00:08:57,780
溫度極高的情況

214
00:08:57,800 --> 00:09:00,200
比如太陽表面之類的

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

216
00:09:04,010 --> 00:09:07,980
溫度一下子變成了2000K或者4000K

217
00:09:08,000 --> 00:09:09,920
這時候就有意思啦

218
00:09:09,940 --> 00:09:11,430
比如說

219
00:09:11,450 --> 00:09:15,660
反應溫度是40000K

220
00:09:15,670 --> 00:09:17,520
那麽熵這一項

221
00:09:17,540 --> 00:09:19,870
也就是熵減 影響就可大啦

222
00:09:19,890 --> 00:09:22,300
所以正的這一項

223
00:09:22,310 --> 00:09:23,190
就抵消這一項

224
00:09:23,210 --> 00:09:25,650
所以在超高溫下

225
00:09:25,670 --> 00:09:28,010
反應可能就無法自發進行啦

226
00:09:28,020 --> 00:09:29,110
換個角度

227
00:09:29,120 --> 00:09:34,230
一個反應放出熱量…

228
00:09:34,240 --> 00:09:36,300
周圍溫度已經非常高

229
00:09:36,310 --> 00:09:38,180
分子的動能已經很大了的時候

230
00:09:38,190 --> 00:09:40,020
放出的熱量就沒什麽影響了

231
00:09:40,040 --> 00:09:41,460
如果周圍溫度足夠高

232
00:09:41,480 --> 00:09:43,890
這個反應就不是自發的了

233
00:09:43,910 --> 00:09:47,060
因爲熵項會把焓抵消掉

234
00:09:47,080 --> 00:09:47,500
好啦

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

236
00:09:49,040 --> 00:09:51,350
就是想讓大家知道 這沒那麽難

237
00:09:51,370 --> 00:09:53,010
這些數據都可以從網上查到

238
00:09:53,020 --> 00:09:54,000
然後就能判斷出

239
00:09:54,010 --> 00:09:56,320
反應是否可以自發進行了