Question

Given the data:

PbO(s), D_f H^q = -217.3 kJ mol^–1, S^q = +68.7 J mol^–1 K^–1
PbO₂(s), D_f H^q = -277.0 kJ mol^–1, S^q = +68.6 J mol^–1 K^–1
O₂(g), D_f H^q= 0.00 kJ mol^–1, S^q = +205 J mol^–1 K^–1

calculate the standard free energy change, D_rG^q, for the reaction:
PbO(s) + ½ O₂(g) ® PbO₂(s)

Select one:

a. +1.45 kJ mol^–1

b. –29.1 kJ mol^–1

c. –68.3 kJ mol^–1

d. –90.3 kJ m

Answer 1

answer : a.) +1.45 kJ mol^–1

solution:

PbO(s) + ½ O₂(g) ------------------------> PbO₂(s)

H = Hproducts - Hreactants

        = (-277.0 - (-217.3 +0))

       = -59.7 kJ/mol

S = Sproducts - Sreactants

        =(68.6 - (68.7 + 205))

         = -205.1 J / K

         = -0.205.1 kJ /K mol

G = H - T S

         = -59.7 - 298 x -0.205.1

        = 1.45 kJ/mol