In: Chemistry
Consider the reaction: CO(g) + H2O(g)<------> CO(g) + H2(g) and K = 0.118 at 4000 K. A reaction mixture initially contains a CO partial pressure of 1344 mbar and a H2O partial pressure of 1766 mbar at 4000K. Calculate the equilibrium partial pressures of each of the products.
Solution :-
Lets set up the ICE table for the reaction.
CO(g) + H2O(g) <------> CO2(g) + H2(g) and K = 0.118
1344 mbar 1766mbar 0 0
-x -x +x +x
1344-x 1766-x x x
equilibrium constant equation is as follows.
K= [CO2] [H2]/ [CO][H2O]
0.118 = [x][x]/[1344-x][1766-x]
0.118 * [1344-x][1766-x] = x^2
Solving for x we get
X= 393
So the now lets calculate the equilibrium partial pressures of each
[CO] =1344-x = 1344-393 = 951 mbar
[H2O] = 1766 –x = 1766 – 393 = 1373 mbar
[CO2] = x = 393 mbar
[H2] = x = 393 mbar