Question

Use standard entropies and heats of formation to calculate \(\Delta G_{f}^{\circ}\) at \(25^{\circ} \mathrm{C}\) for (a) cadmium(II) chloride (s). (b) methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH}(l)\). (c) copper(I) sulfide \((s)\).

Short answer

Question: Calculate the Gibbs free energy of formation (\(\Delta G_{f}^{\circ}\)) at \(25^{\circ} \mathrm{C}\) for the following compounds: (a) cadmium(II) chloride (s), (b) methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH}(l)\), and (c) copper(I) sulfide \((s)\). Answer: (a) -681.6 kJ/mol, (b) -269.8 kJ/mol, (c) -183.3 kJ/mol

Step-by-step solution

Convert the temperature to Kelvin

To convert the temperature from Celsius to Kelvin, add 273.15 to the Celsius temperature. \(T = 25^{\circ} \mathrm{C} + 273.15 = 298.15\,\mathrm{K}\)

Calculate Gibbs free energy for cadmium(II) chloride(s)

For CdCl2 (s), we are given the following values: \(\Delta H_{f}^{\circ} = -645.2 \,\mathrm{kJ/mol}\) (standard heat of formation) \(\Delta S^{\circ} = 123.51 \,\mathrm{J/mol \cdot K}\) (standard entropy) Now, we can use the equation: \(\Delta G_{f}^{\circ} = \Delta H_{f}^{\circ} - T \Delta S^{\circ} = -645.2 \,\mathrm{kJ/mol} - (298.15 \,\mathrm{K})(123.51 \,\mathrm{J/mol \cdot K})\) Note that 1000 J = 1 kJ, so we need to convert J to kJ: \(= -645.2 - (298.15)(0.12351) \,\mathrm{kJ/mol} = -681.6 \,\mathrm{kJ/mol}\)

Calculate Gibbs free energy for methyl alcohol (CH3OH(l))

For CH3OH(l), we are given the following values: \(\Delta H_{f}^{\circ} = -238.6 \,\mathrm{kJ/mol}\) \(\Delta S^{\circ} = 126.8 \,\mathrm{J/mol \cdot K}\) Now, we can use the equation: \(\Delta G_{f}^{\circ} = \Delta H_{f}^{\circ} - T \Delta S^{\circ} = -238.6 \,\mathrm{kJ/mol} - (298.15 \,\mathrm{K})(126.8 \,\mathrm{J/mol \cdot K})\) Convert J to kJ: \(= -238.6 - (298.15)(0.1268) \,\mathrm{kJ/mol} = -269.8 \,\mathrm{kJ/mol}\)

Calculate Gibbs free energy for copper(I) sulfide (s)

For Cu2S(s), we are given the following values: \(\Delta H_{f}^{\circ} = -158.6 \,\mathrm{kJ/mol}\) \(\Delta S^{\circ} = 85.29 \,\mathrm{J/mol \cdot K}\) Now, we can use the equation: \(\Delta G_{f}^{\circ} = \Delta H_{f}^{\circ} - T \Delta S^{\circ} = -158.6 \,\mathrm{kJ/mol} - (298.15 \,\mathrm{K})(85.29 \,\mathrm{J/mol \cdot K})\) Convert J to kJ: \(= -158.6 - (298.15)(0.08529) \,\mathrm{kJ/mol} = -183.3 \,\mathrm{kJ/mol}\) The calculated Gibbs free energy of formation \(\Delta G_{f}^{\circ}\) at \(25^{\circ} \mathrm{C}\) for the three compounds are: (a) cadmium(II) chloride (s): -681.6 kJ/mol (b) methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH}(l)\): -269.8 kJ/mol (c) copper(I) sulfide \((s)\): -183.3 kJ/mol