Consider the equilibrium: CO(g) + H2O(g) <---> CO2(g) + H2(g) At T = 800K, the c
ID: 799229 • Letter: C
Question
Consider the equilibrium: CO(g) + H2O(g) <---> CO2(g) + H2(g)
At T = 800K, the composition of the reaction mixture is:
Substace, Mole Fraction
CO2(g), 0.291
H2(g), 0.291
CO(g), 0.209
H2O(g), 0.209
Please explain! Thank you! :)
Explanation / Answer
Kp=0.0611 at 2000 K
This was provided in the question.
CO(g) + H2O(g) ? CO2(g ) + H2(g)
Kp = (P(products))/(P(reactants)) = ((P(CO2))*(P(H2)))/((P(CO))*(P(H2O)))
Set up an ICE table:
Cmpd...Initial...Change...Equil...
CO.......1326......-x........(1326-x) ...
H2O.......1772....-x........(1772-x).....
CO2.......0..........+x...........x
H2..........0...........+x..........x
Substitute these Equil values into the Kp expression.
Kp = ((x)*(x))/((1326-x)*(1772-x)) = 0.0611
Multiply this out to obtain a quadratic equation
((x)*(x))/((1326-x)*(1772-x)) = 0.0611
x^2 = 0.0611*((1326-x)*(1772-x)) = 0.0611*(2349672 - 3098x + x^2)
x^2 = 143564.96 - 189.2878x + 0.0611x^2
0.9389x^2 + 189.2878x - 143564.96 = 0
Using the quadratic formula, we solve this for x: The negative root is a negative number which is not realistic and is discarded. The positive root gives
x = 303 Torr (to three sig figs)
From the ICE table, we see that this is the partial pressure for CO2 at equilibrium:
P(CO2) = x = 303 Torr