In: Chemistry
CO(g)+H2O(g)⇌CO2(g)+H2(g)
Kp=0.0611 at 2000 K
A reaction mixture initially contains a CO partial pressure of 1360
torr and a H2O partial pressure of 1768 torr at 2000
K.
Part A
Calculate the equilibrium partial pressure of CO2 and H2
? torr
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
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
Initial partial pressure (torr) 1360 1768 0 0
Change -x -x x x
At equilibrium (1360-x) (1768-x) x 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