Question

Using the given data, calculate the change in Gibbs free energy for each of the following reactions. In each case indicate whether the reaction is spontaneous at 298 K under standard conditions.

Part A: 2Ag (s) + Cl2 (g) --> 2AgCl (s) Gibbs free energy for AgCl (s) is -109.70 kJ/mol

Part B: spontaneous or nonspontaneous

Part C: P4O10 (s) + 16H2 (g) --> 4PH3 (g) + 10H2O (g)

Gibbs free energy for P4O10 (s) is -2675.2 kJ/mol

Gibbs free energy for PH3 (g) is 13.4 kJ/mol

Gibbs free energy for H2O (g) is -228.57 kJ/mol

Part D: spontaneous or nonspontaneous

Part E: CH4 (g) + 4F2 (g) --> CF4 (g) + 4HF (g)

Gibbs free energy for CH4 (g) is -50.8 kJ/mol

Gibbs free energy for CF4 (g) is -635.1 kJ/mol

Gibbs free energy for HF (g) is -270.70 kJ/mol

Part F: spontaneous or nonspontaneous

Part G: 2H2O2 (l) --> 2H2O (l) + O2 (g)

Gibbs free energy for H2O2 (l) is -120.4 kJ/mol

Gibbs free energy for H2O (l) is -237.13 kJ/mol

Part H: spontaneous or nonspontaneous

Answer 1

Using the given data, calculate the change in Gibbs free energy for each of the following reactions. In each case indicate whether the reaction is spontaneous at 298 K under standard conditions.

Part A: 2Ag (s) + Cl2 (g) --> 2AgCl (s)

Gibbs free energy for AgCl (s) is -109.70 kJ/mol

At standard condition,

ΔG = Gibbs free energy per mol of products - Gibbs free energy per mol of reactants

ΔG = - 2*109.70 – 0 = - 219.4 kJ/mol.

In this case, the ΔG < 0, so the reaction will be spontaneous.

Part B: spontaneous

Part C: P4O10 (s) + 16H2 (g) --> 4PH3 (g) + 10H2O (g)

Gibbs free energy for P4O10 (s) is -2675.2 kJ/mol

Gibbs free energy for PH3 (g) is 13.4 kJ/mol

Gibbs free energy for H2O (g) is -228.57 kJ/mol

ΔG = [4*13.4 + 10*(-228.57)] – [(-2675.2) + 0] kJ/mol = 443.1

In this case, the ΔG > 0, so the reaction will be nonspontaneous.

Part D: nonspontaneous

Part E: CH4 (g) + 4F2 (g) --> CF4 (g) + 4HF (g)

Gibbs free energy for CH4 (g) is -50.8 kJ/mol

Gibbs free energy for CF4 (g) is -635.1 kJ/mol

Gibbs free energy for HF (g) is -270.70 kJ/mol

ΔG = [4*(-270.70) + (-635.1)] – [(- 50.8) + 0] = -1667.1 kJ/mol.

In this case, the ΔG < 0, so the reaction will be spontaneous.

Part F: spontaneous

Part G: 2H2O2 (l) --> 2H2O (l) + O2 (g)

Gibbs free energy for H2O2 (l) is -120.4 kJ/mol

Gibbs free energy for H2O (l) is -237.13 kJ/mol

ΔG = [2*(-237.13) + 0] – 2*(-120.4) = -233.46 kJ/mol

In this case, the ΔG < 0, so the reaction will be spontaneous.

Part H: spontaneous