8 An industrial chemist introduces 20 atm of H2 and 20 atm o

8. An industrial chemist introduces 2.0 atm of H2 and 2.0 atm of CO2 into 1.00 L container at 25°C. Then the temperature of the container is raised to 700°C, at which Kc = 0.534: H2 (g) +CO2 (g) H20(g)+CO(g) How many grams of H2 are present at equilibrium in the 700°C container?

Solution

          H2   + CO2 <----> H2O + CO

initial 2        2            0   0

change   -x      -x           +x   +x

equil    2-x    2-x            x    x

Kp = KC*(RT)^Dn

Dn = 2-2 = 0

Kp = Kc

Kp = pH2O*pCO/pH2*pCO2

0.534 = x^2/(2-x)^2

x = 0.844

at equilibrium, pH2 = 2-0.844 = 1.156 atm

No of mol of H2 = pv/RT

                = 1.156*1/(0.0821*973.15)

                = 0.0145 mol

mass of H2 = 0.0145*2 = 0.029 g