Question

PS8.2. The equilibrium constant, KP, for the reaction CO2(g) + H2(g) H2O(g) + CO(g) is 0.138. Calculate the partial pressure of all species at equilibrium for each of the following original mixtures: a) 1.36 atm of CO2 and 1.36 atm of H2. b) 0.87 atm of CO2, 0.87 atm of H2 and 0.87 atm of H2O(g). c) 0.64 atm of H2O and 0.64 atm of CO.

Answer 1

For the given reaction CO2+ H2 -<--->H2O+CO

Kp = [PH2] [PCO]/ [PCO2] [PH2] =0.138, P indicates partial pressure

Let P= drop in partial pressure of CO2(or H2) to reach equilibrium

At equilibrium PH2= 1.36-P, PCO=1.36-P, PH2O= PCO= P

P²/1.36-P)²= 0.138, P/(1.36-P)= 0.37, P= 1.36*0.37-0.37P, 1.37P= 1.36*0.37, P= 0.37atm

At equilibrium PH2=PCO= 1.36-0.37= 0.99 atm and PH2O= PCO=0.37 atm

) 0.87 atm of CO2, 0.87 atm of H2 and 0.87 atm of H2O

CO2+ H2 ----àH2O+CO

At equilibrium PCO2= PH2= 0.87-x, PH2O= 0.87+x and PCO= x

Hence x*(0.87+x)/ (0.87-x)²= 0.138, when solved using excel, x= 0.088

At equilibrium PCO2= PH2=0.87-0.088=0.782, PH2O=0.87+0.088=0.958, PCO= 0.088 ( all in atm)

) 0.64 atm of H2O and 0.64 atm of CO.

1/Kp = [CO2] [H2]/[H2O][CO] = 1/0.138= 7.25 ( since only products are available)

PCO2=PH2= P , PH2O= PCO= 0.64-P

At equilibrium P²/(0.64-P)²= 7.25, P/(0.64-P)= 2.7, P= 0.64*2.7-2.7P, P= 0.64*2.7/3.7= 0.47

At equilibrijm PH2O= PCO=0.64-0.47= 0.17, PCO2= PH2=0.47