Question

The value of Kp for the reaction below is 8.00 x 10^-6.

CH3OH(g) <----> 2 H2(g) + CO(g)

Calculate the partial pressure of the gases if initially CH3OH(g) is 0.205 atm and CO(g) is 0.114 atm, then the reaction is permitted to come to equilibrium in a closed vessel.

Answer 1

CH3OH(g)    <-------------------> 2 H2(g)    +   CO(g)

0.205                                           0                0.114

0.205 - x                                      2x             0.114+x

Kp = P²_H2 x P_CO / P _CH3OH

8 x 10^-6 = (2x)^2 (0.114+x) / (0.205 - x)

4x^3 + 0.456 x^2 + 8 x 10^-6 x -1.64 x 10^-6 = 0

x = 0.00187 atm

partial pressures at equailibrium :

paritail pressure of CH3OH= 0.205 - x   = 0.203 atm            

paritail pressure of H2 =   2x = 0.00374 atm

paritail pressure of CO = 0.114+x = 0.1159 atm