Question

Problem 15.31

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 3.82×10⁻² mol CH3OH, 0.160 mol CO, and 0.302 mol H2 at 500 K.

Part A

Calculate Kc at this temperature.

Kc =

Answer 1

CO(g)+2H2(g)←−→CH3OH(g).

Kc expression for the given reaction is

Kc = [CH3OH] / [CO] [H2]²

but concentration he has given in moles but we need in molarities

chabge th emoles in to molarity using the following formula

molarity = no of moles/ volume in liters

M of CH3OH = 3.82×10⁻² / 2 = 0.0191 M

M of CO = 0.16 / 2 = 0.08 M

M of H2 = 0.302 / 2 = 0.151 M

put all these values in above expression

Kc = [0.0191] / [0.08] 0.151]²

Kc = [0.0191] / 0.001824

Kc = 10.471