A mixture of 2.90e-02 mol of Cl2, 1.20e-01 mol of H2O, 2.53e-02 mol of HCl, and

Question

A mixture of 2.90e-02 mol of Cl2, 1.20e-01 mol of H2O, 2.53e-02 mol of HCl, and 4.87e-02 mol of O2 is placed in a 1.0-L steel pressure vessel at 593 K. The following equilibrium is established:
2 Cl2(g) + 2 H2O(g) 4 HCl(g) + O2(g)

At equilibrium 8.19e-03 mol of HCl is found in the reaction mixture. Calculate the equilibrium pressures of all gases in the reaction vessel and the value of KP for the reaction. Pick the correct statement from the multiple choices. Use the value R = 0.0821 L-atm/mol-K for the gas constant.

a) The equilibrium pressure of Cl2 is PCl2 = 0.994 atm.
b) The equilibrium constant is KP = 2.4e+03 .
c) The equilibrium pressure of H2O is PH2O = 5.41 atm.
d) The equilibrium constant is KP = 8.4e-05 .
e) The equilibrium pressure of O2 is PO2 = 2.16 atm.

Explanation / Answer

2Cl2 + 2H2O ----------> 4HCl + O2
2.9e-2 1.2e-1 2.53e-2 4.87e-2
2.9e-2 -2x 1.2e-1 -2x 2.53e-2 + 4x 4.87e-2+x

so 2.53e-2 +4x= 8.19e-3
x= -4.2775 e^-3 moles

eqlbrm conc.
cl2 = 3.755 e-2
h2o = 1.28555e-1
hcl = 8.19e-3
o2 =4.44e-2

corresponding pressures
cl2 = 1.828 atm
h20 = 6.258 atm
hcl = 0.3987 atm
o2=2.16 atm

e) is the correct option

kp = 4.174 x 10^-4

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