Respuesta :
Answer:
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) [tex]K_{c} =[/tex] 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
[tex]K_{c} = \frac{[CO][H_{2}O] }{[CO_{2}][H_{2} ] } = \frac{(0.050)(0.040)}{(0.086)(0.045)} = 0.52[/tex]
(b) When the concentration of CO₂ is increase we have go over the previous step to check the reaction quotient as follows
[tex]K_{c} = \frac{[CO][H_{2}O] }{[CO_{2}][H_{2} ] } = \frac{(0.050)(0.040)}{(0.30)(0.045)} = 0.14815[/tex]
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
[tex]\frac{(0.05+x)(0.04+x)}{(0.3-x)(0.045-x)} = 0.52[/tex]
expanding and collecting like terms, we have
[tex]\frac{x^{2} +0.9x+0.002}{x^{2}-0.345x+0.135 } = 0.52[/tex]
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
