[化學] 化學平衡

看板TransBioChem作者 (bla)時間11年前 (2013/07/05 21:38), 編輯推噓2(2推 0噓 19→)
留言21則, 3人參與, 最新討論串1/1
1. at equilibrium CO2 + H2 → H2O + CO ← 2L vessel contains 0.48 mol each of CO2 and H2 and 0.96 mol each of H2O and CO Kc=4 (1)how many moles of H2 and CO2 must be added to bring the concentration of CO to 0.6 M ? (0.36 mol) (2)how many moles of H2O must be removed to breing the concentration of CO to 0.6 M ? (1.008 mol) 不知道怎麼算?括號是答案。 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.122.167.59
推
07/06 09:05, , 1F
初濃度[CO2]=[H2]=0.24M [H2O]=[CO]=0.48M
07/06 09:05, 1F
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07/06 09:07, , 2F
欲使[CO]=0.6M可知[CO]增加0.6-0.48=0.12M
07/06 09:07, 2F
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07/06 09:09, , 3F
增加反應物CO2和H2由勒沙特列知平衡向右，達新平衡
07/06 09:09, 3F
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07/06 09:11, , 4F
假設增加x mole,則等於濃度增加(x/2) M
07/06 09:11, 4F
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07/06 09:13, , 5F
反應物濃度變成0.24+（x/2) M
07/06 09:13, 5F
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07/06 09:14, , 6F
生成物[CO]增加量=反應物消耗量=0.12M
07/06 09:14, 6F
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07/06 09:16, , 7F
故平衡後濃度[CO2]=[H2]=0.24+(x/2)-0.12
07/06 09:16, 7F
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07/06 09:19, , 8F
[H2O]=[CO]=0.6，由Kc=4=[H2O][CO]/[CO2][H2]代入數據
07/06 09:19, 8F
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07/06 09:20, , 9F
得4=[0.6][0.6]/[0.24+(x/2)-0.12][0.24+(x/2)-0.12]
07/06 09:20, 9F
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07/06 09:21, , 10F
解得x=0.36mole-------這是第一題
07/06 09:21, 10F
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07/06 09:23, , 11F
第二題一樣平衡後[CO]要=0.6就必須生成0.12M
07/06 09:23, 11F
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07/06 09:25, , 12F
移走[H2O]由勒沙特列知平衡向右達新平衡
07/06 09:25, 12F
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07/06 09:32, , 13F
設移走x mole,即移走(x/2)M又[CO]增加量=[H2O]增加量
07/06 09:32, 13F
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07/06 09:35, , 14F
得[H2O]新平衡濃度=0.48-(x/2)+0.12，[CO]=0.6
07/06 09:35, 14F
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07/06 09:37, , 15F
[CO2]和[H2]各消耗0.12平衡後[CO2]=[H2]=0.12
07/06 09:37, 15F
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07/06 09:39, , 16F
代入Kc=4=[0.48-(x/2)+0.12][0.6]/[0.12][0.12]，解x
07/06 09:39, 16F
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07/06 09:40, , 17F
得x=1.008
07/06 09:40, 17F
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07/06 09:41, , 18F
不好意思手機不好發文只好一條條步驟慢慢推，希望你看
07/06 09:41, 18F
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07/06 09:42, , 19F
得懂，祝你考試順利，加油！
07/06 09:42, 19F
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07/06 10:58, , 20F
喔喔喔我看懂了 感謝認真的回答啊!!!!謝謝!
07/06 10:58, 20F
推
07/08 15:37, , 21F
推c大A_A
07/08 15:37, 21F
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