Consider this equilibrium process at 686°C.
CO2(g) + H2(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 the concentration of CO2 to raised to 0.40 mol/L by the addition of CO2, what would be the concentrations of all the gases when equilibrium is reestablished?
3 answers
Can you do the Kc; i.e., the first part?
Yes, the answer Kc=0.517
........CO2(g) + H2(g)==> CO(g) + H2O(g)
E......0.086....0.045.....0.050...0.040
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I.......0.400....0.045...0.050...0.040
C.........-x......-x.......+x.....+x
E.....0.400-x..0.045-x..0.050-x..0.040+x
Substitute the E line into Keq expression and solve. Post your work if get stuck.
E......0.086....0.045.....0.050...0.040
---------------------------------------
I.......0.400....0.045...0.050...0.040
C.........-x......-x.......+x.....+x
E.....0.400-x..0.045-x..0.050-x..0.040+x
Substitute the E line into Keq expression and solve. Post your work if get stuck.