Using the given data, calculate the change in Gibbs free energy for each of the

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

Explanation / Answer

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

Answer:

At standard condition,

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

G = - 2*109.70 – 0 = - 219.4 kJ/mol.

In this case, the G < 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

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

In this case, the G > 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

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

In this case, the G < 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

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

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

Part H: spontaneous

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