Worksheet Practice for Test 1

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Emily Vainberg 1D
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Joined: Sat Jul 20, 2019 12:15 am

Worksheet Practice for Test 1

Postby Emily Vainberg 1D » Tue Jan 21, 2020 2:30 pm

For the test 1 practice worksheet, I don't understand how to solve problem 6.

1 M H2 and 2 M Cl2 are mixed with the reaction H2(g) + Cl2(g) ⇔ 2HCl(g). Kc=50.
What are the equilibrium concentrations?

Brian Tangsombatvisit 1C
Posts: 119
Joined: Sat Aug 17, 2019 12:15 am

Re: Worksheet Practice for Test 1

Postby Brian Tangsombatvisit 1C » Tue Jan 21, 2020 3:20 pm

You would set up an ice table. Initial concentrations: 1 for H2, 2 for Cl2, and 0 for HCl. We don't know how much the change is, so we represent this with -x, -x, and +2x, respectively. Therefore, the expressions for the equilibrium are: H2 = 1-x, Cl2 = 2-x, and HCl = 4x^2. Write the expression for K in terms of x to obtain (4x^2)/(1-x)(2-x), then set this equal to 50 and solve for x using the quadratic formula. Plug x back into the ICE table to find the equilibrium concentrations for the reactants and product.

Lizette Noriega 1H
Posts: 105
Joined: Wed Sep 18, 2019 12:15 am

Re: Worksheet Practice for Test 1

Postby Lizette Noriega 1H » Tue Jan 21, 2020 8:19 pm

Brian Tangsombatvisit 1C wrote:You would set up an ice table. Initial concentrations: 1 for H2, 2 for Cl2, and 0 for HCl. We don't know how much the change is, so we represent this with -x, -x, and +2x, respectively. Therefore, the expressions for the equilibrium are: H2 = 1-x, Cl2 = 2-x, and HCl = 4x^2. Write the expression for K in terms of x to obtain (4x^2)/(1-x)(2-x), then set this equal to 50 and solve for x using the quadratic formula. Plug x back into the ICE table to find the equilibrium concentrations for the reactants and product.


How did you get 4x^2 for the equilibrium expression for HCl?

Brooke Yasuda 2J
Posts: 102
Joined: Sat Jul 20, 2019 12:17 am

Re: Worksheet Practice for Test 1

Postby Brooke Yasuda 2J » Tue Jan 21, 2020 9:25 pm

It's because of the ICE table that is set up. The equation is H2 + Cl2 --> 2HCl. So, when you do an ICE table, you represent the change as some variable like x. So the H2 and Cl2 concentrations will decrease both by x, and the HCl concentration will increase by 2x. It is 2x because there is a 1:2 ratio between the reactants and products. Therefore, when you set up the equilibrium constant representation, 2x is in the numerator and it is squared because you know that in K representations, the stoichiometric coefficient becomes the exponent. Therefore, you have 4x^2


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