2 NO2(g) <-> N2O4(g) (dark brown to colorless) The dimerization of NO2(g), an exothermic process, is — Thermodynamics Chemistry Question

STEPS:

1. Identify the thermodynamic criteria for chemical favorability:
A reaction is thermodynamically favored (spontaneous) when its standard Gibbs free energy change is negative ($\Delta G^\circ < 0$). The relationship between Gibbs free energy, enthalpy, and entropy is defined by the fundamental equation:
$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$
where $T$ represents the absolute temperature in Kelvin (which is always a positive value).

2. Determine the sign of the enthalpy change ($\Delta H^\circ$):
The problem states that the dimerization of $\text{NO}_2(g)$ is an exothermic process. Therefore, heat is released, and the standard enthalpy change is negative:
$$\Delta H^\circ < 0 \quad (\text{negative})$$

3. Determine the sign of the entropy change ($\Delta S^\circ$):
Examine the physical phases and stoichiometry of the system:
$$2\ \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)$$
* The reaction converts two moles of reactant gas into one mole of product gas.
* A decrease in the number of gaseous molecules reduces the positional disorder of the system.
* Thus, the standard entropy change for this process is negative:
$$\Delta S^\circ < 0 \quad (\text{negative})$$

4. Analyze the mathematical signs within the Gibbs free energy expression:
Substitute the signs of $\Delta H^\circ$ (negative) and $\Delta S^\circ$ (negative) into the equation:
$$\Delta G^\circ = (\text{negative}) - T(\text{negative})$$
$$\Delta G^\circ = -\Delta H^\circ + T|\Delta S^\circ|$$
* The enthalpy term ($-\Delta H^\circ$) is thermodynamically favorable and drives $\Delta G^\circ$ to be negative.
* The entropy term ($+ T|\Delta S^\circ|$) is thermodynamically unfavorable and drives $\Delta G^\circ$ to be positive.

5. Determine the temperature dependence of favorability:
* At low temperatures, the magnitude of $T$ is small, making the unfavorable term ($+ T|\Delta S^\circ|$) very small. Under these conditions, the favorable negative enthalpy term dominates, resulting in a negative free energy change ($\Delta G^\circ < 0$).
* At high temperatures, the magnitude of $T$ is large, meaning the unfavorable term ($+ T|\Delta S^\circ|$) increases and overcomes the enthalpy term, resulting in a positive free energy change ($\Delta G^\circ > 0$).
* Therefore, the forward reaction is thermodynamically favored at low temperatures only (Option B).

*

WHY_OTHERS_WRONG:

• Option A is incorrect: This reaction cannot be spontaneous at all temperatures because the unfavorable entropy change ($-\Delta S^\circ > 0$) will dominate at sufficiently high temperatures, making $\Delta G^\circ$ positive and the process non-spontaneous.
• Option C is incorrect: High temperatures increase the influence of the unfavorable negative entropy change (where $-T\Delta S^\circ$ becomes a large positive value), driving $\Delta G^\circ$ above zero and rendering the forward reaction thermodynamically unfavorable. High temperatures instead favor the reverse, endothermic decomposition reaction.
• Option D is incorrect: The reaction is indeed thermodynamically favored at low temperatures because the favorable enthalpy of the reaction ($\Delta H^\circ < 0$) easily overcomes the unfavorable entropy penalty ($T\Delta S^\circ < 0$) when the thermal energy $T$ is small.