Chemical Equilibrium module Part 2 #27?

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Adriana Rangel 1A
Posts: 96
Joined: Fri Sep 29, 2017 7:04 am

Chemical Equilibrium module Part 2 #27?

Postby Adriana Rangel 1A » Sat Nov 25, 2017 7:25 pm

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]?

I figured out that the concentration of H2O is .05 (2.5mol/50L) and that Kc=[H2][CO]/[H2O] But I don't know what the question is asking or how to solve this?

Ammar Amjad 1L
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Joined: Thu Jul 27, 2017 3:00 am
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Re: Chemical Equilibrium module Part 2 #27?

Postby Ammar Amjad 1L » Sat Nov 25, 2017 11:02 pm

Hey,

The question is asking for the equilibrium concentration of H20. Since you know the initial concentration of H20 and the equilibrium concentration of H2 you can setup an ICE table and find the value of X. The products start off with initial concentration of zero and end up with an equilibrium concentration; therefore, since you're given the equilibrium value of H2 you can calculate the value of X in the ICE Table. The value of X would be .04, and to find the equilibrium value of H20 you would subtract .04 from .05 and you would end up with .01 M as the equilibrium value of H20.

Hope this helps.

Adriana Rangel 1A
Posts: 96
Joined: Fri Sep 29, 2017 7:04 am

Re: Chemical Equilibrium module Part 2 #27?

Postby Adriana Rangel 1A » Sun Nov 26, 2017 11:25 am

That was helpful thank you!


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