Question

Consider the reaction: CO(g)+2H2(g)⇌CH3OH(g) A reaction mixture in a 5.25 −L flask at 500 K contains 9.02 g CO and 0.57 g of H2. At equilibrium, the flask contains 2.34 g CH3OH. Calculate the equilibrium constant at this temperature.

Answer 1

moles of CO = 9.02 g/28.01 g/mol = 0.322 mols

moles of H2 = 0.57 g/2.016 g/mol = 0.283 mols

moles of CH3OH = 2.34 g/32.04 g/mol = 0.073 mols

Reaction :              CO( g)           +          2H2(g)               ---->        CH3OH(g)

at equilibrium,   (0.322 - 0.073)          (0.283 - 2 x 0.073)                    0.073

thus,

Keq = [CH3OH]/[CO].[H2]^2

        = (0.073/5.25)/((0.322 - 0.073)/5.25).((0.283 - 0.146)/5.25)^2)

        = 430.45