Question

2. Predict whether a gas mixture consisting of 10% H2S and 90% H2O can convert solid Cu20 into solid Cu2S at 1200K. Cu2O(s) + H2S(g) = Cu2S(s) + H2O(g)

Answer 1

Cu₂O(s) + H₂S(g) $\leftrightharpoons$ Cu₂S(s) + H₂O(g)

Since the total pressure of gaseous mixture is not given , I am assuming the total pressure be P atm .

Mole fraction of H₂S(g) = 10/100 = 0.1

Mole fraction of H₂O(g) = 90/100 = 0.9

K_p = P_H2O/P_H2S = 0.9 × P /(0.1×P) = 9

∆G^{​​​​​​}° = - RT ln K_p = - 8.314×1200×ln 9 = -21921.27J/mol

= - 21.9 kJ/mol.

∆G° is negative , means the reaction is spontaneous in nature . The solid Cu₂S can be created at given reaction conditions .   (Answer)