#29 Chem Equilibrium Pt. 2 Module

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Charlene D 3H
Posts: 103
Joined: Wed Sep 30, 2020 9:54 pm

#29 Chem Equilibrium Pt. 2 Module

Postby Charlene D 3H » Wed Jan 06, 2021 5:33 pm

the problem states: A researcher fills a 1.00L reaction vessel with 1.84 x 10^-4 mol of BrCl gas and heats it to 500K. At equilibrium, only 18.3% of BrCl gas remains. Calculate the equilibrium constant, assuming the following reaction is taking place 2BrCl (g) -> Br2(g) + Cl2(g).

I'm wondering if I made a computational or procedural error for this problem.
I used the 18.3% to find the remaining moles of BrCl at equilibrium and then solved for the change in moles for BrCl. To find the amount of moles of each product created, I divided the calculated # of moles of BrCl by 2 bc of the molar ratio. I then put all the values over 1.00L and plugged the concentrations in to find the equilibrium constant.

If it's easier to just explain how you approached it, that works, too!
Thank you in advance! (:

Bella Wachter 1A
Posts: 88
Joined: Wed Nov 18, 2020 12:17 am
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Re: #29 Chem Equilibrium Pt. 2 Module

Postby Bella Wachter 1A » Wed Jan 06, 2021 6:33 pm

I'll explain the process I went through. Hopefully this helps!

For this problem, I began by finding the initial concentration of BrCl (dividing the initial moles by 1.00L) to get 1.84 x 10^-4 M.

At equilibrium, only 18.3% of this initial amount remains, so we have 0.183 (1.84 x 10^-4) or 3.37 x 10^-5 M of BrCl at equilibrium.

We know the "change" row for BrCl is -2x, so we solve (1.84x10^-4) - 2x = 3.37x10^-5 to find x, the change in concentration.

Using that x, we can find the equilibrium concentrations of Br2 and Cl2, which are both equal to x based on their stoichiometric coefficients of 1.

Then we can plug in these values into the Kc expression and solve.

Madison Muggeo 3H
Posts: 101
Joined: Wed Sep 30, 2020 9:35 pm

Re: #29 Chem Equilibrium Pt. 2 Module

Postby Madison Muggeo 3H » Thu Jan 07, 2021 1:21 pm

I had a similar issue with this question thank you for replying! :)


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