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### Post Assessment Chemical Equilibrium Part 2 #29

Posted: Fri Nov 24, 2017 10:00 pm
Can anyone explain #29 from the Chemical Equilibrium Part 2 Post Assessment? Here it is

29. 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)

A. 19.5

B. 8.39 x 10-5

C. 4.98

D. 1.68 x 10-4

thanks!

### Re: Post Assessment Chemical Equilibrium Part 2 #29

Posted: Sat Nov 25, 2017 10:51 pm
Hey,
So in this problem you're given the initial concentration of BrCl, 1.84x10-4 mol/L. The question also tells you that at equilibrium the molar concentration is 18.3% of the initial concentration, which would be 1.84x10^-4 mol/L times .183 which gives you 3.36x10^-5. Now that you have the initial and equilibrium concentrations of BrCl, you would set up the ICE table and solve for X to find the equilibrium constant.

2BrCl(g) ⇌ Br2 (g) + Cl2(g)
1.84x10^-4 0 0
-2x +x +x
3.36x10^-5 x x

Then you can find the value of x to find the equilibrium constant. X = 7.52x10^-5.

K= [Br2][Cl2]/[BrCl]^2

From there you can plug the values in, and you end up with the final answer as C.

### Re: Post Assessment Chemical Equilibrium Part 2 #29

Posted: Sun Nov 26, 2017 3:27 pm
how did you solve for x without knowing the value of k?

### Re: Post Assessment Chemical Equilibrium Part 2 #29

Posted: Sun Nov 26, 2017 7:58 pm
1.84x10^-4 - 2x = 3.36x10^-5

Using this equation you can find X, since you know the initial and equilibrium concentrations.

### Re: Post Assessment Chemical Equilibrium Part 2 #29

Posted: Sun Nov 26, 2017 11:48 pm
oh I see. thank you!

### Re: Post Assessment Chemical Equilibrium Part 2 #29

Posted: Mon Nov 27, 2017 10:16 pm
Why don't you just take the difference of 1.84x10^-4 - 3.36x10^-5 ? Why is there a 2x involved?

### Re: Post Assessment Chemical Equilibrium Part 2 #29  [ENDORSED]

Posted: Wed Nov 29, 2017 9:39 pm
You need the 2x, because that's the ratio of BrCl. You can use x instead of 2x for BrCl, but then you would need to use 1/2x for Br2 and Cl2. The 2x is involved in the equation, because that is the change from initial to equilibrium.

Hope this helps.