[化學]化學平衡

[blablawawa (bla)]

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 colagrange:初濃度[CO2]=[H2]=0.24M   [H2O]=[CO]=0.48M
→ colagrange:欲使[CO]=0.6M可知[CO]增加0.6-0.48=0.12M
→ colagrange:增加反應物CO2和H2由勒沙特列知平衡向右，達新平衡
→ colagrange:假設增加x mole,則等於濃度增加(x/2) M
→ colagrange:反應物濃度變成0.24+（x/2) M
→ colagrange:生成物[CO]增加量=反應物消耗量=0.12M
→ colagrange:故平衡後濃度[CO2]=[H2]=0.24+(x/2)-0.12
→ colagrange:[H2O]=[CO]=0.6，由Kc=4=[H2O][CO]/[CO2][H2]代入數據
→ colagrange:得4=[0.6][0.6]/[0.24+(x/2)-0.12][0.24+(x/2)-0.12]
→ colagrange:解得x=0.36mole-------這是第一題
→ colagrange:第二題一樣平衡後[CO]要=0.6就必須生成0.12M
→ colagrange:移走[H2O]由勒沙特列知平衡向右達新平衡
→ colagrange:設移走x mole,即移走(x/2)M又[CO]增加量=[H2O]增加量
→ colagrange:得[H2O]新平衡濃度=0.48-(x/2)+0.12，[CO]=0.6
→ colagrange:[CO2]和[H2]各消耗0.12平衡後[CO2]=[H2]=0.12
→ colagrange:代入Kc=4=[0.48-(x/2)+0.12][0.6]/[0.12][0.12]，解x
→ colagrange:得x=1.008
→ colagrange:不好意思手機不好發文只好一條條步驟慢慢推，希望你看
→ colagrange:得懂，祝你考試順利，加油！
→ blablawawa:喔喔喔我看懂了 感謝認真的回答啊!!!!謝謝!
推 mimipig:推c大A_A