## Module 2

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FrancoNancy_Sec1L
Posts: 26
Joined: Fri Sep 20, 2013 3:00 am

### Module 2

I am not sure how to approach this problem: A researcher fills a 1.00 L reaction vessel with 1.84 x 10-4 mol of BrCl gas and heats it to 500 K. At equilibrium, only 18.3 % of the BrCl gas remains. Calculate the equilibrium constant, assuming the following reaction is taking place.
2BrCl(g) ⇌ Br2 (g) + Cl2(g)

How do we calculate the amount of Br2 and Cl2 produced to find the equilibrium constant?
Can anyone please explain the step by step process? Thank you.

Krista 3L
Posts: 10
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Module 2

Krista Lum 3a

If 18.3% of the BrCl gas remains at equilibrium, and we have the initial amount, we can calculate the amount of BrCl left at equilibrium. So .183(1.84x10^-4)=3.36 x10^-5. This gives us the ice table below.

2BrCl(g) ⇌ Br2 (g) + Cl2(g)
I 1.84x10^-4 0 0
C -2x +x +x
E 3.36 x10^-5 x x
Now, we can solve for x. 1.84x10^-4 - 2x= 3.36 x10^-5 . The answer x also happens to be the amount of Br2 and Cl2 left. Use that to solve for the equilibrium constant.

melissa_dis4K
Posts: 106
Joined: Fri Sep 28, 2018 12:28 am

### Re: Module 2

I did the same thing as explained above and I get Kc=[BrCl]^2/[Br2][Cl2]=[3.37 x 10^-5]^2/[7.52 x 10^-5][7.52 x 10^-5].
My answer: 0.199637845, however, is not an answer choice can someone please explain what I did wrong?
Thank you!

Hadji Yono-Cruz 2L
Posts: 64
Joined: Fri Sep 28, 2018 12:26 am

### Re: Module 2

melissa_dis4K wrote:I did the same thing as explained above and I get Kc=[BrCl]^2/[Br2][Cl2]=[3.37 x 10^-5]^2/[7.52 x 10^-5][7.52 x 10^-5].
My answer: 0.199637845, however, is not an answer choice can someone please explain what I did wrong?
Thank you!

I made this mistake as well. The question should be answered in terms of partial pressure. Use P=nRT/V to find the partial pressures and use the ICE chart like how you would in your previous method, just in terms of pressure and not concentration.

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