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Ch.19 - Chemical Thermodynamics
Chapter 19, Problem 56

A certain reaction has ΔH° = +23.7 kJ and ΔS° = +52.4 J>K. (c) Calculate ΔG° for the reaction at 298 K. (d) Is the reaction spontaneous at 298 K under standard conditions?

Verified step by step guidance
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Step 1: Understand the relationship between Gibbs free energy (ΔG°), enthalpy (ΔH°), and entropy (ΔS°) using the equation: ΔG° = ΔH° - TΔS°, where T is the temperature in Kelvin.
Step 2: Convert the entropy change (ΔS°) from J/K to kJ/K to match the units of ΔH°. Since 1 kJ = 1000 J, divide ΔS° by 1000.
Step 3: Substitute the given values into the equation: ΔH° = +23.7 kJ, ΔS° (converted to kJ/K), and T = 298 K.
Step 4: Calculate the term TΔS° by multiplying the temperature (298 K) by the converted ΔS° value.
Step 5: Substitute the values of ΔH° and TΔS° into the equation ΔG° = ΔH° - TΔS° to find ΔG°. Determine if the reaction is spontaneous by checking if ΔG° is negative.