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posted by  ss83 on 7/13/2008 10:45:52 PM  |  status: Live  

Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant

Course Textbook Chapter Problem
General Chemistry N/A 14 N/A
Question Details:
Consider the following reaction:
{\rm{CO}}\left( g \right)\; + \;{\rm{H}}_2 {\rm{O}}\left( g \right)\; \rightleftharpoons \;{\rm{CO}}_2 \left( g \right)\; + \;{\rm{H}}_2 \left( g \right)
K_{\rm{c}} = 102 at 500 \rm K
A reaction mixture initially contains 0.130 M {\rm CO} and 0.130 M {\rm{H}}_2 {\rm{O}}.
What will be the equilibrium concentration of [{\rm CO}_2]?
  [{\rm CO}_2]  =
 ? \rm M

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Sage
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posted by Jesley on 7/13/2008 11:46:54 PM  |  status: Live
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Response Details:
the reaction is
{\rm{CO}}\left( g \right)\; + \;{\rm{H}}_2 {\rm{O}}\left( g \right)\; \rightleftharpoons \;{\rm{CO}}_2 \left( g \right)\; + \;{\rm{H}}_2 \left( g \right) 
let 'x' be the change in concentration
the I.C.E table for this reaction is
                          [CO]                  [H2O]                    [CO2]                      [H2]
initial:               0.130M                0.130M                     0                             0
change:                -x                         -x                          +x                          +x
equilibrium:      (0.130-x)              (0.130-x)                    x                             x
the equilibrium constant is related to the concentrations as
Kc = [CO2][H2]/[CO][H2O] = 102
(x*x)/((0.130-x)*(0.130-x)) = 102
on solving we get x= 0.118
therefore , the equilibrium concentration is
[CO2]=x=0.118M
this is the procedure, please check the initial concentrations and Kc values
Mentor
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posted by AJ <3 on 7/14/2008 12:41:12 AM  |  status: Live
Asker's Rating: Helpful   
ss83's comment:
"thanks so much!! "
Response Details:

Hey ss83, I helped you with this exact problem and gave you all the final concentrations at equilibrium. Don't you remember? You seem to be asking a lot of questions like this. I'd be happy to help further if there is something you're not getting about the concepts. Anyways, here's the same exact solution:

                       {\rm{CO}}\left( g \right)\; + \;{\rm{H}}_2 {\rm{O}}\left( g \right)\; \rightleftharpoons \;{\rm{CO}}_2 \left( g \right)\; + \;{\rm{H}}_2 \left( g \right) 
initial         :   0.130M        0.130M                 0                   0

change      :     - X                -X                   + X                + X
equilibrium:   0.130 - X      0.130 - X            X                     X

Use the formula from the equilibrium constant, Kc:

Plug everything in from the information obtained above:
Ok, Here's where the math tricks come in to make it easier to solve this problem. We can rewrite our problem like this:
Now we can just take the square root of both sides and look how our problem turns out:
Just continue on with the math and solve for X:
And here's our X, no need for the quadratic formula or estimations.
So lets solve for our concentrations:

[CO2] = X = .118 M
[H2] = X = .118 M
[CO] = .130 - X = .0120 M
[H2O] = .130 - X = .0120 M
Please ask for further assistance. AJ
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