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Consider the following equilibrium process at 686 degree C: CO2 (g) + H2O (g) >=> CO (g) + H2O (g) The equilibrium concentrations of the reacting species are [CO] = 0.050 M, [H2] = 0.045 M, [CO2] = 0.086 M, and [H2O] = 0.040 M. (a) Calculate Kc for the reaction at 686°C. (b) If we add CO2 to increase its concentration to 0.50 mol/L, what will the concentrations of all the gases be when equilibrium is reestablished?

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Answered 5 months ago
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Knowns\textbf{Knowns}

We have a reversible reaction with equilibrium concentrations of

[CO]\ce{[CO]} = 0.050 MM

[HX2]\ce{[H2]} = 0.045 MM

[COX2]\ce{[CO2]} = 0.086 MM

[HX2O]\ce{[H2O]} = 0.040 MM.

COX2(g)+HX2(g)CO(g)+HX2O(g)\ce{ CO2$(g)$ + H2$(g)$ <=> CO$(g)$ + H2O$(g)$ }

We need to find the following

(a) KcK_c at this temperature.

(b) equilibrium composition if we we make [COX2\ce{CO2}] = 0.50 MM

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