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。

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

喔喔喔我看懂了 感謝認真的回答啊!!!!謝謝!

推c大A_A