6th edition 11.53

Moderators: Chem_Mod, Chem_Admin

Jeremiah Hutauruk
Posts: 67
Joined: Fri Sep 28, 2018 12:28 am

6th edition 11.53

Postby Jeremiah Hutauruk » Tue Jan 15, 2019 11:03 pm

11.53 A reaction mixture that consisted of 0.400 mol H2 and 1.60 mol I2 was introduced into a 3.00-L ask and heated. At equilibrium, 60.0% of the hydrogen gas had reacted. What is the equilibrium constant K for the reaction H2(g) I2(g) ∆ 2 HI(g) at this temperature?

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

Re: 6th edition 11.53

Postby Andrea Zheng 1H » Wed Jan 16, 2019 12:27 am

First, you would find the initial concentration of each substance. For H2, it is 0.4 mol/3L = 0.133M. For I2, it is 1.6mol/3L=0.533M. For HI, it is 0M. Using the balanced chemical reaction, change in H2 is -x, change in I2 is -x, and change in HI is +2x. This means the eq composition of each are: H2=0.133-x, I2=0.533-x, and HI=2x.

It says that 60% of H2 reacted, meaning that there is still 40% of the gas left at equilibrium. This means you can set 40% of the initial concentration of H2 equal to the eq composition H2. Therefore, 0.4*0.133 = 0.133 -x. Solving for x, you get x=0.080. You can then plug this x into the eq composition of each of the substances and get [H2]=0.053 M, [I2]=0.453 M, [HI]=0.16 M.

With these equilibrium concentrations, you can find K, because K is the [products at eq]/[reactants at eq]. Therefore, K= (0.16)^2/(0.453)(0.053) = 1.07

Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 2 guests