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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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$\textbf{Knowns}$

We have a reversible reaction with equilibrium concentrations of

$\ce{[CO]}$ = 0.050 $M$

$\ce{[H2]}$ = 0.045 $M$

$\ce{[CO2]}$ = 0.086 $M$

$\ce{[H2O]}$ = 0.040 $M$.

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

We need to find the following

(a) $K_c$ at this temperature.

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

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