Problem 11.37  [ENDORSED]


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HanSitoy
Posts: 18
Joined: Fri Sep 29, 2017 7:06 am

Problem 11.37

Postby HanSitoy » Sun Nov 26, 2017 9:22 pm

For the reaction N2(g) + 3H2 (g) === 2 NH3(g) at 400.K, K = 41. Find the value of K for each of the following reactions at the same temperature:
B) 1/2 N2(g) + 3/2H2(g) ==== NH3 (g)

Aijun Zhang 1D
Posts: 53
Joined: Tue Oct 10, 2017 7:13 am

Re: Problem 11.37

Postby Aijun Zhang 1D » Sun Nov 26, 2017 9:32 pm

The K expression for original reaction is [NH3]^2/([H2]^3*[n2]), and the K expression for b is [NH3]/([N2]^1/2*[H2]^2/3).
Say the K expression for (b) is K2.
K is the square of K2. So if you want to get the value of K2, you need to take the square root, which is square root of 41. The answer is 6.4.

isauramora3K
Posts: 20
Joined: Fri Sep 29, 2017 7:07 am

Re: Problem 11.37  [ENDORSED]

Postby isauramora3K » Mon Nov 27, 2017 10:57 am

The information for this question is on pages 439-440. Basically for (a) it is the reverse reaction so it is just the inverse of the forward reaction. For (b) and (c) you raise the equilibrium constant (K=41) to the number by which the reaction was multiplied. So in example since in part (b) it is multiplied by 1/2 the answer would be (41)^(1/2)

Catherine Yang 3G
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Re: Problem 11.37

Postby Catherine Yang 3G » Mon Nov 27, 2017 3:12 pm

I thought the constant always stayed the same unless you reversed the reaction or changed the temperature?


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