If we add CO2 to increase its concentration to 0.30 mol/L, what will the concentrations of all the gases be when equilibrium is reestablished?

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?

The answers to the question are

(a) 0.52

(b) The concentrations of the gases when equilibrium is reestablished will be

[CO₂] = 0.28 M

[H₂]  = 0.0295 M

[CO] = 0.068 M

[H₂O]  = 0.058 M

Explanation:

(a) To find we have to write out the expression for equilibrium as follows

(b) When the concentration of CO₂ is increase we have go over the previous step to check the reaction quotient as follows

Since Q < K the reaction is positive that is to go forward

The change in concentration of the reactants and products at equilibrium is given by x [CO₂] = [0.3 -x], [H₂] = 0.045 -x, [CO] = 0.05 + x and [H₂O] = 0.040+x

to solve for x we have

expanding and collecting like terms, we have

0.48x² +0.2694x-0.00502 = 0

factorising gives (x+0.579)(x-1.805)×0.48=0

x = -0.579 or x = 1.805×10⁻²

[CO₂] = (0.3 -1.805×10⁻²) = 0.28 M

[H₂] = 0.045 -x = 0.0295 M

[CO] = 0.05 + x = 0.068 M

[H₂O] = 0.040+x = 0.058 M

am i too late?

explanation: