Question

Consider the reaction: CO(g)+2H2(g)⇌CH3OH(g)

A reaction mixture in a 5.21 −L flask at 500 K contains 9.04 g CO and 0.58 g of H2. At equilibrium, the flask contains 2.34 g CH3OH.

Calculate the equilibrium constant at this temperature.

Answer 1

moles of CO = 9.04 g/28.01 g/mol = 0.322 mols

moles of H2 = 0.58 g/2.016 g/mol = 0.288 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)          (0.288-2 *

0.073) 0.073

thus,

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

        = (0.073/5.21)/((0.322 - 0.073)/5.21).((0.288 - 0.146)/5.21)^2)

        = 394.66 ANSWER

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