Asked by sibabalwe
A 1.0 L reaction vessel contained 0.750 mol of CO(g) and 0.275 mol of H2O(g). After 1.0 h, equilibrium was reached. Analysis showed that 0.25 mol of CO2(g) was present. The equation for the reaction is:
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
Find Keq for the reaction.
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
Find Keq for the reaction.
Answers
Answered by
DrBob222
(CO) = mols/L = 0.750/1.0 = 0.750 M
(H2O) = 0.275 M
(CO2) = 0.25 M
.........................CO(g) + H2O(g) ↔ CO2(g) + H2(g)
I........................0.750.......0.275...........0
C..........................-x...........-x.................+x
E....................0.750-x.....0.275-x...........0.250
Keq = (CO2)/(CO)(H2O)
Plug the E line into Keq expression and solve for x. Then evaluate the concentrations of CO, H2O and CO2 and solve for Keq. Post your work if you get stuck.
(H2O) = 0.275 M
(CO2) = 0.25 M
.........................CO(g) + H2O(g) ↔ CO2(g) + H2(g)
I........................0.750.......0.275...........0
C..........................-x...........-x.................+x
E....................0.750-x.....0.275-x...........0.250
Keq = (CO2)/(CO)(H2O)
Plug the E line into Keq expression and solve for x. Then evaluate the concentrations of CO, H2O and CO2 and solve for Keq. Post your work if you get stuck.