## Chemical Equilibrium Part 2 Module Question #27

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Nancy 1B
Posts: 24
Joined: Wed Sep 21, 2016 2:58 pm

### Chemical Equilibrium Part 2 Module Question #27

A mixture of 2.5 moles H2O and 100 g of C are placed in a 50 L container and allowed to come to equilibrium subject to the following reaction:
C(s) + H2O(g) = CO(g) + H2(g)
The equilibrium concentration of hydrogen is found to be [H2]=0.040 M. What is the equilibrium concentration of water, [H2O]?

Can someone please help work through this problem, I'm not sure how to go about solving it. Thanks!

Seth_Evasco1L
Posts: 54
Joined: Thu Jul 27, 2017 3:00 am

### Re: Chemical Equilibrium Part 2 Module Question #27

1. Create an ICE table for the reaction.
2. Find the molarity for H2O since it gives you the value in moles. (No need to find the molarity for C because it's a solid and is not involved in the equilibrium constant)
3. The initial concentration for CO and H2 is going to be 0.
4. You know the equilibrium concentration for H2 is 0.040 M which means the change is 0.040 M.
5. You subtract that change from the initial concentration to get the equilibrium concentration of H2O.

Christy Zhao 1H
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

### Re: Chemical Equilibrium Part 2 Module Question #27

Adding on, you should get 0.05 M H2O for the initial concentration, and you know x=0.040 from H2. The equilibrium concentration for H2O is 0.05-x, and plugging in x you would get 0.01 M H2O.

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