For the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g),
Kp = 0.0611 = PCO2 . PH2 /PCO .
PH2O
We see the changes in pressure in of the species, as th reaction
proceeds
                  Â
   CO(g)   +  Â
H2O(g)   ⇌  CO2(g)  Â
+Â Â Â Â H2(g)
Initial partial pressure (torr)
         Â
1360Â Â Â Â Â Â Â Â Â Â Â
1768Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
Change                                         Â
-x               Â
-x               Â
x                    Â
x
At
equilibrium                           Â
(1360-x)Â Â Â Â Â Â Â Â Â
(1768-x)Â Â Â Â Â Â Â Â Â Â Â
x                    Â
x
Put equilibrium partial pressures in expression of Kp
Kp = 0.0611 = x.x / (1360-x). (1768-x)
x2 = 0.0611 (1360-x). (1768-x)
x2 = 0.0611 (2404480 -1360x -1768x +
x2)
x2 = (146913.728 -83.096x -108.025x
+0.06112x2)
0.9389x2 + 191.121x - 146913.728 = 0
Solve the quadratic equation, find the value of x.
x = 306 torr
PCO2 = PH2 = x = 306
torr