A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of Kp is?

Answer 1

CO2(g) + C(s) <===> 2CO(g)

0.5atm...0atm............0atm.....Initial

-x...........0..................+2x.......Change

0.5-x......0.................2x..........Equilibrium

At constant temperature and constant volume, a change in pressure will be proportional to a change in moles. Since total pressure = 0.8 atm, the partial pressures of CO2 + CO must equal 0.8 atm. Thus,..

0.8 atm = (0.5 -x)atm + 2x atm

0.8 atm = 0.5 atm + x atm

x = 0.3 atm, so partial pressure of the two gasses at equilibrium will be:

P_CO2 = 0.5 - 0.3 = 0.2 atm

P_CO = 2*0.3 = 0.6 atm

K_p = (PCO)²/(PCO₂)

Substitute partial pressures and solve for K_p. Hopefully, you'll find K_p = 1.8 atm