Calculate and report ΔrG°T for the dissolution of Ca(OH)2(s) in water at 25°C, given that ΔH° = -985.2 kJ/mol and ΔS° = -0.353 kJ/mol·K.

• A. -879.9 kJ/mol
• B. -990.1 kJ/mol
• C. -987.5 kJ/mol
• D. -882.3 kJ/mol

Answer: (D)

Verified step by step guidance

1. Understand the relationship between Gibbs free energy change (ΔrG°), enthalpy change (ΔH°), and entropy change (ΔS°) using the equation: ΔrG° = ΔH° - TΔS°. This equation helps us determine the spontaneity of a reaction at a given temperature.
2. Identify the given values: ΔH° = -985.2 kJ/mol, ΔS° = -0.353 kJ/mol·K, and the temperature T = 25°C. Convert the temperature to Kelvin by adding 273.15 to the Celsius temperature: T = 25 + 273.15 = 298.15 K.
3. Substitute the values into the Gibbs free energy equation: ΔrG° = ΔH° - TΔS°. Ensure that the units are consistent. Here, both ΔH° and ΔS° are in kJ, so no conversion is needed.
4. Calculate the term TΔS° by multiplying the temperature in Kelvin (298.15 K) by the entropy change (ΔS° = -0.353 kJ/mol·K): TΔS° = 298.15 K * (-0.353 kJ/mol·K).
5. Substitute the calculated TΔS° value back into the Gibbs free energy equation: ΔrG° = -985.2 kJ/mol - TΔS°. Perform the subtraction to find the Gibbs free energy change for the dissolution of Ca(OH)2(s) in water at 25°C.