A mixture of 0.2063 mol of Cl2, 0.2093 mol of H2O, 0.1887 mol of HCl, and...

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.

Solutions

Expert Solution

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

queen_honey_blossom answered 1 year ago

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