Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: cloud5 on July 05, 2010, 09:53:11 AM
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This is another question that I need help. Please show formula if can. :) Thanks for your help.
At temperature T K, the 3 equilibrium constants for each equilibrium are:
CO2(g) + H2(g) <--> CO(g)+ H2O(g)
Kc = 0.2
C(s) + CO2(g) <--> 2CO(g)
Kc=6.2
C(s) + 1/2 O2(g) <-->CO(g)
Kc=3.3 x 104
What is the numerical value of Kc for the following equilibrium at temperature T K?
3 CO2(g) + H2(g) <--> 3CO(g) + H2O(g) + O2(g)
The answer given is 7.1 x 10-9
Is this related to the Born-Haber cycle? Do I need that to do this?
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It is not related to the Born-Haber cycle.
In order to do this, we need to figure out the value of
([CO]^3 * [H2O] * [O2])/([CO2]^3 * [H2])
since that is defined to be Kc for that reaction.
Now we want to combine the three reactions in such a way so that the net reaction will be the one you want.
CO2(g) + H2(g) <--> CO(g) + H2O(g) [1] Kc = 0.2
C(s) + CO2(g) <--> 2CO(g) [2] Kc = 6.2
C(s) + 1/2 O2(g) <--> CO(g) [3] Kc = 3.3 x 10^4
It follows that adding the first equation and two times the second and two times the reverse of the third gives us the final reaction.
So now we have the three equations:
CO2(g) + H2(g) <--> CO(g) + H2O(g) [1'] Kc = 0.2
2C(s) + 2CO2(g) <--> 4CO(g) [2'] Kc = (6.2)^2 = 38.4
2CO(g) <--> 2C(s) + O2(g) [3'] Kc = 1/(3.3 x 10^4)^2 = 9.2 x 10^-10
We now multiply the Kc values together to get the final Kc value for the final reaction:
0.2 * 38.4 * 9.2 x 10^-10 = 7.1 x 10^-9
Hope this helped!
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When you tackle these kind of problems, you should be aware what happens to the value of the equilibrium constants when you add and subtract equations or multiply and divide the coefficients.
And eleven6eleven you should not just give away the whole process and the answer. Follow the forum rules please.