WEBVTT

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有个反应是

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1mol的甲烷

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和2mol的氧气反应

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会生成1mol的二氧化碳

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和2mol的水

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这集里 我们想判断

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这个反应是不是自发的

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上次我们已经学过

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怎么判断自发性啦

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这时候就要利用吉布斯自由能

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或吉布斯自由能变啦

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而吉布斯自由能变ΔG

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等于反应的焓变ΔH

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减去反应的温度T

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乘以熵变ΔS

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如果ΔG<0

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那反应就是自发的了

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我先给大伙开个好头

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我刚刚已经把

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反应的焓变算出来了

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就在这里呢

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大家都知道怎么求ΔH了吧

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几集前我们讲过的

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先查出来每个产物的

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生成热

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例如水 你要把生成热乘以2

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因为反应生成了2mol的水

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这样就有了产物的生成热之和

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然后再减去

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反应物的生成热之和

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当然啦 O2的生成热是0

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所以式子里面没有这项

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算出来就是 -890.3kJ

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好啦

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这就说明 反应是放热的

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方程式这边的能量小于…

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你也可以这样想的…

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比那边的能量小

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所以必须释放能量才行

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可以在这里写 +e e代表能量

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我写上

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加上释放出来的能量

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这就是反应放热的原因啦

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但是问题是 反应是不是自发的呢？

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想要判断反应的自发性

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首先要算出ΔS

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为了计算ΔS的值呢

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我提前就查好了

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这里每种分子的标准摩尔熵

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比如说 标准…

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我换个颜色表示

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标准

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小小讲点拓展 这里没有Δ

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我擦了吧 还能补救

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标准

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这里画个圈里面带个横表示

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标准摩尔熵Sm

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“标准”指的是在298°K下

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实际不应该说“度开尔文”

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就是298K

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用开尔文K的时候

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不用说度°

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所以反应温度是289K

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也就是25°C

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相当于室温

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所以用289K作标准状态

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所以室温下 甲烷的标准摩尔熵

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就等于这个数

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所以如果有1mol的甲烷

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就有186J/K的熵

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如果有2mol的甲烷 就乘以2

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如果有3mol 就乘以3

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所以这个反应的总熵变

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就是产物标准熵之和

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减去反应物标准熵之和

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就跟算ΔHr差不多

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所以熵变就等于213.6 加上…

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产物里有2mol的水

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所以就是加上2乘以…

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就取70好了

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69.9 约等于70

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加上2×70

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然后再减去

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反应物的熵之和

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也就是方程式这边的这些

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1molCH4的熵

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等于186 加上2×205

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大概心算一下

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这个数非常接近这个数

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但是这个数比这个数大得多

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液态水的熵…

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这是液态水的熵

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它的熵远远小于氧气的熵

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这很合理呀

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因为液态水的微观状态数比氧气少得多

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液态水都沉在容器底了

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气体就不同

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气体能膨胀 随空间变换形状

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所以理所当然 气体的熵

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比液体的熵大的多

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简单心算

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就已经能看出来产物的熵

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比反应物的熵小

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所以熵变应该是负的

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不过我们还是确认一下

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这个是213.6加上…

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是加上140 对嘛？

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是2×70

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加上140 就等于353.6

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这部分等于353.6

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然后从这里减去…

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所以186 加上2×205

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就等于596

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所以就是减去596

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最后等于什么？

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-596 加上353.6

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等于-242.4

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所以它就等于-242.4J/K

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这就是ΔS 负的

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所以系统的熵减少了这么多

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你可能对熵的单位大小没有概念

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不过只要知道是某个大小就可以

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但是你可以说 喏

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反应之后系统更有序啦

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这很合理 因为开始是一大堆气体

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开始是3个单独的分子

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有1个甲烷 还有2个氧气

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后来还是3个分子

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但是这个水是液态的

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所以 反应后熵减小是有道理的

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尤其液态物质 它的微观状态数很少

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我们来判断一下

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这个反应是不是自发的

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ΔG等于ΔH…

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反应放热 所以就是-890

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我把小数省略掉了

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我们不用那么精确

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减去温度

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假设反应是在室温下进行的

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所以温度是298°K

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就是… 我应该说“298K”

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我要改掉坏习惯

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在用K表示温度的时候 不说“°”

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298K 也就是25°C

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再乘以熵变

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这项是负的

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你可能会说 好的 是-242

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直接把这个数放进去

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但是你要非常非常非常的小心

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它的单位是千焦kJ

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可是它的单位是焦耳J

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所以如果都以千焦做单位的话

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因为前面写了kJ

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我们把这个也换算成千焦吧

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所以它就是0.242kJ/K

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前面放个小数点

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这里的0.45擦掉

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单位是kJ/k

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所以吉布斯自由能变

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就是-890kJ 减去298…

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负负得正

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完全合理

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因为熵的这项

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会使吉布斯自由能变得更正

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因为

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我们想让ΔG<0

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但是ΔS>0会降低自发性

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现在我们来看这项能不能抵消ΔH

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也就是放热的影响

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目测好像是不行

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因为一个小数乘以它

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得到的数肯定更小

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我们算算看

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所以除以 1,2,3 3个0

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系统的熵变

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乘以298 这是系统的温度

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等于-72

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所以这项就等于…

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因为前面还有个减号…

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所以就是加上72.2

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所以这就是标准温度下的熵的项

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最后就等于它咯

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而这是焓项

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这样我们就能看出来

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焓变的绝对值

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比T×ΔS的绝对值

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大得多

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所以这项压倒性胜利了

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虽然反应是个熵减的反应

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但是反应放出的热量太多了

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所以反应仍然是自发

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这个数显然小于0

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所以这是个自发反应

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如你所见 这些吉布斯自由能的问题

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其实没那么难

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只要知道这几项的值就行啦

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这几项的值要么直接给出

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比如ΔH

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不过我们也知道怎么求出来

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只要查到产物的

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生成热

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再减去反应物的生成热

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当然还要各自乘以相应的化学计量数

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然后 用同样的方法

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算出熵变

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查到每种产物的标准摩尔熵

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分别乘以相应的化学计量数

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再减去反应物的总熵

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然后把数代入这个式子中

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最后就得到了吉布斯自由能变

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这个例子里 ΔG是负的

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现在 大家可以想象一下

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温度极高的情况

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比如太阳表面之类的

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温度就不是298K啦

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温度一下子变成了2000K或者4000K

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这时候就有意思啦

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比如说

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反应温度是40000K

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那么熵这一项

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也就是熵减 影响就可大啦

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所以正的这一项

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就抵消这一项

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所以在超高温下

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反应可能就无法自发进行啦

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换个角度

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一个反应放出热量…

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环境温度已经非常高

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分子的动能已经很大了的时候

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放出的热量就没什么影响了

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如果环境温度足够高

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这个反应就不是自发的了

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因为熵项会把焓抵消掉

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好啦

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我只是想带大家算一次

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就是想让大家知道 这没那么难

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这些数据都可以从网上查到

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然后就能判断出

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反应是否可以自发进行了