Question

the value of Kp for the reaction,CO2(g)+C(s)->2CO(g) is 3.0 at 1000K. if initially pCO2=0.48bar and pCO=0bar and pure graphite is present, calulate the equilibrium partial pressures of CO and CO2.

Answer 1

The value of Kp for the reaction,CO2(g)+C(s)->2CO(g) is 3.0 at 1000K. if initially pCO2=0.48bar and pCO=0bar and pure graphite is present, calulate the equilibrium partial pressures of CO and CO2.

                    CO2(g)+C(s)------->2CO(g)

Initial                 0.48                         0

at equillibrium   0.48 - x                   2x

So the expression of kp is (CO)²/(CO2)

             (2x)²/(0.48-x) = 3

but in the expression (0.48 - x), x is very negligible hence ignored

(2x)²/0.48 = 3

So x = 0.6

then at equilibrium, P CO2 is 0.12 bar

                                 P CO is 1.2