The equilibrium constant for decomposition of H2O(g) H2O(g) H2(g) + 12O2(g); G = 92.34 kJ mol-1 is 8.0 10-3 at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ( ) of water is 10-2 (nearest integer value). [Assume is negligible with respect to 1]
Answer
5
Solution
H2O(g) H2(g) + 12O2(g)
Moles at equilibrium (1 mol water taken): 1 - , , 2; total = 1 + 2 1
With P = 1 bar: pH2O 1, pH2 , pO2 2
Kp = pH2 (pO2)1/2pH2O = ( 2 )1/2 = 3/2 2
3 = 2Kp2 = 2 (8.0 10-3)2 = 128 10-6
= (128)1/3 10-2 = 5.04 10-2
5 10-2
(The given G agrees with K: -RT K = -8.314 2300 (8.0 10-3) J 92.3 kJ mol-1; it is not needed.)
Answer: 5