WEBVTT

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有個反應是

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1mol的甲烷

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和2mol的氧氣反應

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會生成1mol的二氧化碳

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和2mol的水

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這集裏 我們想判斷

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這個反應是不是自發的

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上次我們已經學過

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怎麽判斷自發性啦

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這時候就要利用吉布斯自由能

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或吉布斯自由能變啦

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而吉布斯自由能變ΔG

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等於反應的焓變ΔH

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減去反應的溫度T

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乘以熵變ΔS

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如果ΔG<0

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那反應就是自發的了

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我先給大夥開個好頭

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我剛剛已經把

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反應的焓變算出來了

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就在這裡呢

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大家都知道怎麽求ΔH了吧

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幾集前我們講過的

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先查出來每個產物的

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生成熱

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例如水 你要把生成熱乘以2

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因爲反應生成了2mol的水

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這樣就有了產物的生成熱之和

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然後再減去

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反應物的生成熱之和

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當然啦 O2的生成熱是0

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所以式子裏面沒有這項

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算出來就是 -890.3kJ

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好啦

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這就說明 反應是放熱的

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方程式這邊的能量少於…

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你也可以這樣想的…

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比那邊的能量小

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所以必須釋放能量才行

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可以在這裡寫 +e e代表能量

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我寫上

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加上釋放出來的能量

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這就是反應放熱的原因啦

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但是問題是 反應是不是自發的呢？

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想要判斷反應的自發性

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首先要算出ΔS

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爲了計算ΔS的值呢

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我提前就查好了

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這裡每種分子的標準莫耳熵

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比如說 標準…

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我換個顏色表示

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標準

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小小講點拓展 這裡沒有Δ

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我擦了吧 還能補救

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標準

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這裡畫個圈裏面帶個橫表示

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標準莫耳熵Sm

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“標準”指的是在298°K下

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實際不應該說“度克耳文”

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就是298K

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用克耳文K的時候

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不用說度°

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所以反應溫度是289K

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也就是25°C

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相當於室溫

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所以用289K作標準狀態

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所以室溫下 甲烷的標準莫耳熵

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就等於這個數

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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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所以這個反應的總熵變

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就是產物標準熵之和

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減去反應物標準熵之和

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就跟算ΔHr差不多

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所以熵變就等於213.6 加上…

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產物裏有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 約等於70

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加上2×70

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然後再減去

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反應物的熵之和

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也就是方程式這邊的這些

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1molCH4的熵

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等於186 加上2×205

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大概心算一下

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這個數非常接近這個數

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但是這個數比這個數大得多

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液態水的熵…

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這是液態水的熵

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它的熵遠遠少於氧氣的熵

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這很合理呀

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因爲液態水的微觀狀態數比氧氣少得多

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液態水都沈在容器底了

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氣體就不同

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氣體能膨脹 隨空間變換形狀

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所以理所當然 氣體的熵

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比液體的熵大的多

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簡單心算

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就已經能看出來產物的熵

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比反應物的熵小

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所以熵變應該是負的

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不過我們還是確認一下

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這個是213.6加上…

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是加上140 對嘛？

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是2×70

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加上140 就等於353.6

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這部分等於353.6

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然後從這裡減去…

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所以186 加上2×205

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就等於596

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所以就是減去596

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最後等於什麽？

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-596 加上353.6

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等於-242.4

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所以它就等於-242.4J/K

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這就是ΔS 負的

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所以係統的熵減少了這麽多

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你可能對熵的單位大小沒有概念

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不過只要知道是某個大小就可以

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但是你可以說 喏

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反應之後係統更有序啦

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這很合理 因爲開始是一大堆氣體

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開始是3個單獨的分子

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有1個甲烷 還有2個氧氣

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後來還是3個分子

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但是這個水是液態的

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所以 反應後熵減小是有道理的

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尤其液態物質 它的微觀狀態數很少

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我們來判斷一下

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這個反應是不是自發的

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ΔG等於ΔH…

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反應放熱 所以就是-890

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我把小數省略掉了

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我們不用那麽精確

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減去溫度

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假設反應是在室溫下進行的

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所以溫度是298°K

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就是… 我應該說“298K”

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我要改掉壞習慣

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在用K表示溫度的時候 不說“°”

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298K 也就是25°C

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再乘以熵變

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這項是負的

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你可能會說 好的 是-242

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直接把這個數放進去

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但是你要非常非常非常的小心

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它的單位是千焦kJ

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可是它的單位是焦耳J

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所以如果都以千焦做單位的話

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因爲前面寫了kJ

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我們把這個也換算成千焦吧

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所以它就是0.242kJ/K

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前面放個小數點

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這裡的0.45擦掉

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單位是kJ/k

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所以吉布斯自由能變

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就是-890kJ 減去298…

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負負得正

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完全合理

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因爲熵的這項

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會使吉布斯自由能變得更正

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因爲

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我們想讓ΔG<0

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但是ΔS>0會降低自發性

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現在我們來看這項能不能抵消ΔH

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也就是放熱的影響

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目測好像是不行

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因爲一個小數乘以它

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得到的數肯定更小

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我們算算看

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所以除以 1,2,3 3個0

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係統的熵變

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乘以298 這是係統的溫度

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等於-72

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所以這項就等於…

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因爲前面還有個減號…

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所以就是加上72.2

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所以這就是標準溫度下的熵的項

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最後就等於它咯

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而這是焓項

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這樣我們就能看出來

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焓變的絕對值

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比T×ΔS的絕對值

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大得多

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所以這項壓倒性勝利了

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雖然反應是個熵減的反應

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但是反應放出的熱量太多了

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所以反應仍然是自發

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這個數顯然少於0

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所以這是個自發反應

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如你所見 這些吉布斯自由能的問題

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其實沒那麽難

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只要知道這幾項的值就行啦

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這幾項的值要麽直接給出

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比如ΔH

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不過我們也知道怎麽求出來

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只要查到產物的

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生成熱

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再減去反應物的生成熱

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當然還要各自乘以相應的化學計量數

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然後 用同樣的方法

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算出熵變

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查到每種產物的標準莫耳熵

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分別乘以相應的化學計量數

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再減去反應物的總熵

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然後把數代入這個式子中

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最後就得到了吉布斯自由能變

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這個例子裏 ΔG是負的

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現在 大家可以想象一下

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溫度極高的情況

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比如太陽表面之類的

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溫度就不是298K啦

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溫度一下子變成了2000K或者4000K

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這時候就有意思啦

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比如說

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反應溫度是40000K

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那麽熵這一項

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也就是熵減 影響就可大啦

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所以正的這一項

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就抵消這一項

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所以在超高溫下

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反應可能就無法自發進行啦

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換個角度

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一個反應放出熱量…

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周圍溫度已經非常高

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分子的動能已經很大了的時候

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放出的熱量就沒什麽影響了

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如果周圍溫度足夠高

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這個反應就不是自發的了

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因爲熵項會把焓抵消掉

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好啦

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我只是想帶大家算一次

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就是想讓大家知道 這沒那麽難

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這些數據都可以從網上查到

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然後就能判斷出

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反應是否可以自發進行了