Question

A mixture of 0.2063 mol of Cl2, 0.2093 mol of H2O, 0.1887 mol of HCl, and 0.09871 mol of O2 is placed in a 1.0-L steel pressure vessel at 575 K. The following equilibrium is established: 2 Cl2(g) + 2 H2O(g) 4 HCl(g) + 1 O2(g) At equilibrium 0.05821 mol of O2 is found in the reaction mixture. (a) Calculate the equilibrium partial pressures of Cl2, H2O, HCl, and O2. (b) Calculate KP for this reaction.

Answer 1

               2 Cl2(g) + 2 H2O(g) ----> 4 HCl(g) + 1 O2(g)

initial        0.2063 mol   0.2093 mol    0.1887 mol 0.09871 mol

change        2*0.0405 mol 2*0.0405 mol 4*0.0405 mol 0.0405 mol
     
equilibrium    0.2873 mol    0.2873 mol   0.0267 mol 0.05821 mol

at equilibrium

ntotal = 0.2873 +0.2873 +0.0267 +0.05821 = 0.66 mol

Ptotal = ntotal*RT/V

       = 0.66*0.0821*575/1

       = 31.16 atm

partial pressure of Cl2(pCl2) = (ncl2/ntotal)*ptotal

                               = 0.2873/0.66*31.16
          
               = 13.56 atm
    pH2O = 13.56 atm

   pHCl = (0.0267/0.66)*31.16 = 1.26 atm

   pO2 = (0.05821/0.66)*31.16 = 2.75 atm

   Kp = pHCl^4*pO2 / pCl2^2*pH2O^2

       = (1.26^4*2.75/(13.56^2*13.56^2))

Kp = 2.05*10^-4