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.
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
Consider the reaction: CO(g)+2H2(g)⇌CH3OH(g) A reaction mixture in a 5.25 −L flask at 500 K contains...
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