Haber process and solving for Kc
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The Haber process for ammonia synthesis is exothermic:
N2(g) + 3H2(g) <--(double arrow for equilibrium) --> 2NH3(g)
Change in H (degrees) = -92KJ
If the equilibrium is constant Kc for this process at 500 Degrees Celsius is 6.0 X 10^-2, what is its value at 300 Degrees Celsius.
I am not sure which equation I should be using to solve for Kc.
My book has the example equation of:
Kc= [C]^c *[D]^d / [A]^a * [B]^b
Where product is divided by reactants, but I don't think that is the correct equation when temperature is changing also.
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Solution Summary
It shows how to solve for the equilibrium contant Kc for the Harber process for ammonia synthesis.
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If we determine the equilibrium constant for a reaction at one temperature (Kt1), and also its delta H (dH), we can then estimate the equilibrium constant at a second temperature (Kt2) using what is called the "van't Hoff equation."
ln (Kt2/Kt1) = dH(T2 - T1)/(R*T2*T1)
First, ...
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