Chemical equilibrium

[匿名]

2013-11-06 1:07 am

1)For the reaction
When 40.64g of I2 and 2.00g of H2 are heated at a certain temperature and keep constant, the equilibrium mixture contains 2.54g of I2.Calculate Kc.
H2(g) + I2(g) to form 2HI(g) it is a reversible reaction

2)For the reaction
When 159.8g of Br2 and 0.2g of H2 are heated at a certain temperature and keep constant , the equilibrium mixture contains 0.05g of H2. Calculate Kc.
H2(g) + Br2(g) to form 2HBr(g) it is a reversible reaction

更新1:

Let me try 2)Initial no. of mole of H2: Br2 : 2HBr 0.1 : 1 : 0 no. of mole at eqm H2 : Br2 : 2HBr 0.1-x : 1-x : 2x that contain 0,05 g of H2 in eqm mixture 0.1-x=0.05/2 x=0.075

更新2:

so that , no. of mole at eqm H2 : Br2 : 2HBr 0.025:0.925:0.15 Kc=(0.15/v)^2/(0.925/v)(0.025/v) Kc=0.97

更新3:

thank you

回答 (1)

用戶4366

2013-11-06 1:42 am

✔ 最佳答案

1.H₂(g) + I₂(g) <===> 2HI(g)
init.2.0040.64
eqm.2.54

Initial no. of mole of I₂ = 40.64/(126.9*2) = 0.160
no. of mole of I₂ at eqm. = 2.54/(126.9*2) = 0.01
no. of mole of I₂ reacted to form HI = 0.160 - 0.01 = 0.15
no. of mole of HI producted = 2*0.15 = 0.3
Initial no. of mole of H₂ = 2.00/(1.008*2) = 0.992
no. of mole of H₂ at eqm. = 0.992 - 0.15 = 0.842
At eqm.

[HI] = 0.3/V
[H₂] = 0.842/V
[I₂] = 0.01/V

Kc = [HI]² / [H₂][I₂]

Kc = (0.3/V)²/(0.842/V)(0.01/V) = (0.3)²/(0.842)(0.01) = 10.7

Question 2 and 1 are very similar. Try it yourself first. If you still cannot do it. say something in 補充，then, I'll do it for you.

2013-11-05 17:44:27 補充：
1. ......... H₂(g) + I₂(g) <===> 2HI(g)
init. ...... 2.00 ......... 40.64
eqm. ....................... 2.54

2013-11-05 17:45:32 補充：
V is the volume of the container.

2013-11-05 19:05:12 補充：
That is very good. I get the same result as you for question 2.

So. now you know how to handle such kind of question.

congratulation !!!

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