Question

19) When lead (II) oxide, PbO, is allowed to react with oxygen, O2, at a temperature of 823 K and standard pressure through the reaction PbO(s) + 420 (g) = PbO (5) the partial pressure of oxygen at equilibrium is 11.6 Pa. Determine the standard Gibbs energy of reaction at this temperature. a) -31.0 kJ mol b) 62.0 kJ mol-' 9-20.8 kJ mol'. d) -41.4 kJ mol-'

Answer 1

ΔG°=−RTlnKp

Partial pressure of oxygen ;11.6 Pa = 0.0001144831 ATM

First write the equilibrium constant in form of pressure = product / reactnats

Kp = 1/ (O2)^1/2

FOR SOLID PRESSURE IS EQUAL = 1

Kp = 1/ 0.0001144831)^1/2

= 1/0.0107

= 93.46

Here T = 823 K AND R = 8.314 J / (mol. · K)

ΔG°=−RTlnKp

= - 8.314 J / (mol. · K) *823 K ln 93.46

=--31049.5865 J/ mole *1 kj/1000 J

= - 31.0KJ/ MOLE

THUS THE CORRECT ANSWER IS a)