A student observes that the equilibrium constant for a reaction is greater than 1.0 at temperatures — Thermodynamics Chemistry Question

STEPS:

1. Relate the equilibrium constant ($K$) to thermodynamic favorability ($\Delta G^\circ$):
The relationship between the standard Gibbs free energy change and the equilibrium constant is given by the equation:
$$\Delta G^\circ = -RT \ln K \quad \text{}$$
* When $K > 1.0$ (at temperatures below $500\text{ K}$), the term $\ln K$ is positive, which makes $\Delta G^\circ < 0$. This indicates that the reaction is thermodynamically favorable at lower temperatures.
* When $K < 1.0$ (at temperatures above $500\text{ K}$), the term $\ln K$ is negative, which makes $\Delta G^\circ > 0$. This indicates that the reaction is thermodynamically unfavorable at higher temperatures.

2. Recall the temperature dependence of Gibbs free energy:
The relationship between Gibbs free energy, enthalpy, and entropy is represented by the formula:
$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \quad \text{}$$

3. Determine the sign of the standard enthalpy change ($\Delta H^\circ$):
* Since the reaction is thermodynamically favorable ($\Delta G^\circ < 0$) at low temperatures, the enthalpy change must drive the favorability of the process.
* For $\Delta G^\circ$ to be negative when the temperature $T$ approaches absolute zero, the enthalpy change must be negative ($\Delta H^\circ < 0$), indicating an exothermic reaction.

4. Determine the sign of the standard entropy change ($\Delta S^\circ$):
* Since the reaction becomes thermodynamically unfavorable ($\Delta G^\circ > 0$) at high temperatures, the entropy term ($-T\Delta S^\circ$) must work against favorability as $T$ increases.
* If $\Delta S^\circ$ is negative ($\Delta S^\circ < 0$), then the term $-T\Delta S^\circ$ becomes a large positive value at high temperatures.
* At temperatures above $500\text{ K}$, this positive entropy term dominates and overcomes the favorable negative enthalpy ($\Delta H^\circ$), causing $\Delta G^\circ$ to become positive.

5. Conclude the signs of both parameters:
* Therefore, the reaction must have $\Delta H^\circ < 0$ and $\Delta S^\circ < 0$ to be favorable only at low temperatures. This matches Option D.

*

WHY_OTHERS_WRONG:

• Option A is incorrect ($\Delta H^\circ > 0$ and $\Delta S^\circ > 0$): A reaction with these signs is endothermic and entropy-favored. It is thermodynamically unfavorable at low temperatures but becomes favorable at high temperatures, which is the exact opposite of the student's observations.
• Option B is incorrect ($\Delta H^\circ > 0$ and $\Delta S^\circ < 0$): A reaction with these signs has both unfavorable enthalpy ($\Delta H^\circ > 0$) and unfavorable entropy ($\Delta S^\circ < 0$). As a result, $\Delta G^\circ$ is positive and $K < 1.0$ at all temperatures.
• Option C is incorrect ($\Delta H^\circ < 0$ and $\Delta S^\circ > 0$): A reaction with these signs has both favorable enthalpy ($\Delta H^\circ < 0$) and favorable entropy ($\Delta S^\circ > 0$). Consequently, $\Delta G^\circ$ is negative and $K > 1.0$ at all temperatures.