Chemical Equilibrium Part 2 Question 19

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Adams Dis 1A
Posts: 23
Joined: Wed Nov 14, 2018 12:23 am

Chemical Equilibrium Part 2 Question 19

Postby Adams Dis 1A » Wed Jan 09, 2019 9:10 pm

On this particular question which asks
"Calculate the reaction quotient Qc, from the following equilibrium data collected in a 3 L sealed reaction vessel for the reaction
AsH3(g) = As(s) + H2(g)
AsH3= 5.55x10^-4 mol, As= 3.31 x 10^-3 mol, H2= 1.23 x10^-3 mole

- No matter how I divide it after I balanced it (2AsH3 = 2As + 3H2), I cannot get the correct answer of 2.01 x10^-3, I keep getting 4.91, does anybody know what I am doing wrong.

Andrea Zheng 1H
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am

Re: Chemical Equilibrium Part 2 Question 19

Postby Andrea Zheng 1H » Wed Jan 09, 2019 9:37 pm

First, it says it is in a 3L vessel, so you would divide all the given values by 3 to get the moles per 1 liter, as that is the molarity used in the calculation. Looking at the equation, the As is solid, so you would not need to account for it in the Qc. Solving the problem, you would divide the two remaining substances by 3 and get 4.1*10^-4 mol/L for H2 and 1.85*10^-4. Using the coefficients from your balanced equation, you could then set it up like [H2]^3/[AsH3]^2. You should get 2.01*10^-3.

Hope this helped!

Arianna Ko 2E
Posts: 33
Joined: Fri Sep 28, 2018 12:20 am

Re: Chemical Equilibrium Part 2 Question 19

Postby Arianna Ko 2E » Thu Jan 10, 2019 1:07 pm

Don't forget that you don't take solids into consideration when you are trying to find the value of the reaction quotient Q. Also, because they give you that you account for the gases in a 3.00L vessel, you should divide all of the moles by this 3.00L value.
You should end up with Q= ((1.23x10^-3)/3)^3/((5.55x10^-4)/3)^2 and then you should be able to get the answer 2.01x10^-3 after calculating! Hopefully this makes sense.

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