A mixture of 1 mol CO, and 1 mol of water vapour is undergoing the water gas shift reaction at a temperature of 1100K and a pressure of 1 bar. CO (g) + H2O (g) -> CO2 + H2 (g) The equilibrium constant for the reaction is K=1. Assume that the gas mixture behaves as ideal gas. Calculate (a) the fractional dissociation of steam.

Question-Answer › Category: Chemical Thermodynamics › A mixture of 1 mol CO, and 1 mol of water vapour is undergoing the water gas shift reaction at a temperature of 1100K and a pressure of 1 bar. CO (g) + H2O (g) –> CO2 + H2 (g) The equilibrium constant for the reaction is K=1. Assume that the gas mixture behaves as ideal gas. Calculate (a) the fractional dissociation of steam.

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GCT asked 1 year ago

A mixture of 1 mol CO, and 1 mol of water vapour is undergoing the water gas shift reaction at a temperature of 1100K and a pressure of 1 bar.
CO (g) + H2O (g) –> CO2 + H2 (g)
The equilibrium constant for the reaction is K=1. Assume that the gas mixture behaves as ideal gas. Calculate
(a) the fractional dissociation of steam.

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MG answered 1 year ago

$The mole fractions of the species at equilibrium to the equilibrium constant which is given by related Ky=k pv (9.65) kф wher$ $the mixture behaves as an to the n= ti-1-1=0 As gas ideal gas kg = 1. Equation (965) gives Ky = kaloky is related mole fracti$
From this we can conclude that 50% of steam has converted in this reaction.