Post Assessment Equilibrium Part 2 #30

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Sabrina Dunbar 1I
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Post Assessment Equilibrium Part 2 #30

Postby Sabrina Dunbar 1I » Sun Nov 26, 2017 3:33 pm

30. 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. Calculate the equilibrium constant Kc of this reaction.

A. 5.0 x 10-3

B. 6.5 x 10-4

C. 0.16

D. 1.2
Can anyone help me on this? I converted everything to molarity values but I don't know what to do after that point. Thanks so much.

Kelly Kiremidjian 1C
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

Re: Module Part 2 #30

Postby Kelly Kiremidjian 1C » Sun Nov 26, 2017 3:47 pm

Hey Sabrina,

so when you convert everything to molarity values you get 0.167 M of Carbon and .05 M of H20. Those are the initial amount for of C and H20 in the ice table. Since Carbon is a solid it does not have an effect on the equilibrium values. We know the equilibrium concentration of H2 is 0.040 M and since it is a balanced equation with all 1:1 molar ratios, the change in equilibrium can be calculated by multiplying the equilibrium value of H2 by the molar ratio(x1). Since H20 is a reactant, the change in H20 would be -.040 and the change for CO and H2 would be +0.40
C(s) + H2O (g) ⇌ CO (g) + H2 (g)
I .167 M .05 M 0 0
C. --- -.04 +.04 +.04
E. ---- .01 .04 .04

Once you have completed the table you can solve for K:
which is C.

hope this helps!!

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

Re: Post Assessment Equilibrium Part 2 #30

Postby Adriana Rangel 1A » Sun Nov 26, 2017 3:52 pm

Ok So you first calculated everything to molarity values which is good. You should have gotten the concentration of H20 to be .05 M.
So you are given an initial concentration value of H20 and equilibrium concentration value of H2.
Kc= [H2][CO]/[H20] (Omit C in your setup as it is a pure solid and they are not included when finding K!)
Then, because you are given the initial concentration and equilibrium concentration, set up an ICE table:
H20 CO H2
I .05 0 0
C -x +x +x
E .05-x x .040

When you solve for x you should get x= .040. When you substitute x into your equations you should get the equilibrium concentration for H20=.01, CO= .04 and H2= .04. When you substitute these values into Kc={H2][CO]/[H20]= .16 = Kc

Hope this helps :)

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