F.5 Chemical equilibrium

2011-02-20 3:33 am
Can you help me to solve the following questions?
1)

圖片參考:http://imgcld.yimg.com/8/n/HA00328041/o/701102190111113873382160.jpg


圖片參考:http://imgcld.yimg.com/8/n/HA00328041/o/701102190111113873382171.jpg

I wanna to ask why my ans is wrong?
2)

圖片參考:http://imgcld.yimg.com/8/n/HA00328041/o/701102190111113873382172.jpg

回答 (1)

2011-02-20 6:28 am
✔ 最佳答案
1.
2CO2(g) ⇌ 2CO(g) + O2(g)
(The initial concentration of CO2 is 0.001 mol dm^-3, but you have multiplied it by 2. Consequently, you have got a wrong answer.)

At eqm:
[CO] = 1 x 10^-4 = 0.0001 mol dm^-3
[CO2] = 0.001 - 0.0001 = 0.0009 mol dm^-3
[O2] = 0.0001/2 = 0.00005 mol dm^-3

Kc = [CO]^2 [O2] / [CO2]^2
= (0.0001)^2 (0.00005) / (0.0009)^2
= 6.17 x 10^-7 mol dm^-3


2.
CO(g) + Cl2(g) ⇌ COCl2(g)
[COCl2]o = 0.2/8 = 0.025 mol dm^-3

At eqm:Let [CO] = y mol dm^-3
Then [Cl2] = y mol dm^-3
and [COCl2] = (0.025 - y) mol dm^-3 ≈ 0.025 mol dm^-3
(assuming that 0.025 >> y)

Kc = [COCl2] / [CO] [Cl2]
0.025 / y^2 = 4.7 x 10^9
y = 2.31 x 10^-6

At eqm:
[CO] = 2.31 x 10^-6 mol dm^-3
[Cl­2] = 2.31 x 10^-6 mol dm^-3
[COCl2] = 0.025 - (2.31 x 10^-6) = 0.025 mol dm^-3
參考: micatkie


收錄日期: 2021-04-13 17:50:09
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110219000051KK01111

檢視 Wayback Machine 備份