For a certain chemical reaction, the standard Gibbs free energy change ΔG° is −45 kJ mol^-1 and the standard enthalpy change ΔH° is −90 kJ mol^-1 at 0 °C (273 K). Assuming that ΔH° and ΔS° are independent of temperature, at approximately what temperature (in kelvin) will ΔG° become zero, marking the boundary between spontaneous and non spontaneous behaviour?

Introduction / Context:

Gibbs free energy is the key criterion for spontaneity of chemical reactions at constant temperature and pressure. A reaction is spontaneous when the change in Gibbs free energy ΔG is negative, non spontaneous when ΔG is positive, and at equilibrium when ΔG is zero. This question gives you ΔG° and ΔH° at 0 °C and asks for the temperature at which ΔG° becomes zero, assuming both ΔH° and ΔS° do not change with temperature. This temperature marks the threshold between spontaneous and non spontaneous behaviour.

Given Data / Assumptions:

• ΔG° at T1 = 273 K is −45 kJ mol^-1.
• ΔH° is −90 kJ mol^-1 and taken as constant with temperature.
• ΔS° is assumed constant and can be found from ΔG° = ΔH° − T * ΔS°.
• We seek the temperature T at which ΔG° = 0.

Concept / Approach:

The basic relationship is ΔG° = ΔH° − T * ΔS°. At the given temperature T1 = 273 K, ΔG1° and ΔH° are known, so we can calculate ΔS° from the equation ΔS° = (ΔH° − ΔG°) / T1. Once ΔS° is known, we set ΔG° = 0 to find the threshold temperature T_eq at which ΔH° − T_eq * ΔS° = 0. Solving this gives T_eq = ΔH° / ΔS°. Because ΔH° is negative and ΔS° turns out to be negative here, the reaction is spontaneous (ΔG° < 0) at lower temperatures and becomes non spontaneous at sufficiently high temperatures.

Step-by-Step Solution:

Verification / Alternative check:

At T = 273 K, ΔG° is already negative (−45 kJ mol^-1), so the reaction is spontaneous at that temperature. Since both ΔH° and ΔS° are negative, the reaction is expected to be spontaneous only at sufficiently low temperatures. As T increases, the term −T * ΔS° becomes positive enough to cancel the negative ΔH°, making ΔG° zero and then positive. The calculated threshold temperature around 546 K fits this qualitative behaviour and matches the given option.

Why Other Options Are Wrong:

Option A: 273 K is the temperature at which ΔG° is given as −45 kJ mol^-1, not zero; the reaction is clearly spontaneous here, not at the threshold.

Option B: 298 K is close to room temperature and would not produce ΔG° = 0 according to the calculation using ΔG° = ΔH° − T * ΔS°.

Option D: 596 K is higher than the calculated threshold and would make ΔG° positive (non spontaneous), not zero.

Common Pitfalls:

Students often forget to convert Celsius to kelvin or mishandle signs in the Gibbs free energy equation. Another frequent error is to misinterpret the temperature at which ΔG° changes sign as a minimum temperature for spontaneity, when with negative ΔH° and negative ΔS° the reaction is actually spontaneous below a certain maximum temperature. Always calculate ΔS° carefully and then use ΔG° = 0 to find the threshold temperature.

Final Answer:

The reaction has ΔG° = 0 at approximately 546 K.