Question #65136
Methanol (CH3OH) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen:
CO(g)+2H2(g)←−→CH3OH(g).
An equilibrium mixture in a 2.00 −L vessel is found to contain 4.33×10−2 mol CH3OH, 0.180 mol CO, and 0.303 mol H2 at 500 K.
Calculate Kc at this temperature.
CO(g)+2H2(g)←−→CH3OH(g).
An equilibrium mixture in a 2.00 −L vessel is found to contain 4.33×10−2 mol CH3OH, 0.180 mol CO, and 0.303 mol H2 at 500 K.
Calculate Kc at this temperature.
Expert's answer
Kc = [CH3OH]/[CO][H2]2
[CH3OH] = 7.64·10-2 mol / 1.50 L = 5.09·10-4 M
[CO] = 0.180 mol / 1.50 L = 0.120 M
[H2] = 0.302 mol / 1.50 L = 0.201 M
Kc = (5.09·10-4) / (0.120)(0.201)2 = 0.105
[CH3OH] = 7.64·10-2 mol / 1.50 L = 5.09·10-4 M
[CO] = 0.180 mol / 1.50 L = 0.120 M
[H2] = 0.302 mol / 1.50 L = 0.201 M
Kc = (5.09·10-4) / (0.120)(0.201)2 = 0.105
