Questions 42-43 refer to the following. H2(g) + Cl2(g) <=> 2 HCl(g) K_p = 2 x 10^30 at 298 K HCl(g) — Equilibrium Chemistry Question

STEPS:

1. Understand the thermodynamic relationship between the equilibrium constant ($K_p$) and Gibbs free energy ($\Delta G^\circ$):
The equilibrium constant of a reaction is directly related to the standard Gibbs free energy change by the equation:
$$\Delta G^\circ = -RT \ln K_p$$
Alternatively, this can be written as:
$$K_p = e^{-\Delta G^\circ / RT}$$
For a reaction to have a massive equilibrium constant ($K_p = 2 \times 10^{30}$), the value of $\Delta G^\circ$ must be highly negative (highly spontaneous).

2. Connect $\Delta G^\circ$ to enthalpy ($\Delta H^\circ$) and entropy ($\Delta S^\circ$):
The Gibbs free energy change is determined by both the enthalpy and entropy changes of the reaction:
$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$
To make $\Delta G^\circ$ highly negative at a standard temperature of $298\text{ K}$, we need a highly negative $\Delta H^\circ$ (highly exothermic) and/or a highly positive $\Delta S^\circ$ (highly increased disorder).

3. Calculate the overall reaction enthalpy ($\Delta H^\circ$) using Hess's Law:
Since enthalpy is a state function, the overall enthalpy change of a multi-step reaction is the sum of the enthalpy changes of its elementary steps:
* Step 1: $\Delta H^\circ_1 = +242\text{ kJ/mol}_{\text{rxn}}$
* Step 2: $\Delta H^\circ_2 = +4\text{ kJ/mol}_{\text{rxn}}$
* Step 3: $\Delta H^\circ_3 = -432\text{ kJ/mol}_{\text{rxn}}$
$$\Delta H^\circ_{\text{overall}} = 242\text{ kJ/mol} + 4\text{ kJ/mol} + (-432\text{ kJ/mol}) = \mathbf{-186\text{ kJ/mol}_{\text{rxn}}}$$
The overall enthalpy of the reaction is large and negative (strongly exothermic).

4. Evaluate the contribution of entropy ($\Delta S^\circ$) to the reaction:
Look at the states of matter in the balanced overall equation:
$$\text{H}_2(g) + \text{Cl}_2(g) \rightleftharpoons 2\ \text{HCl}(g)$$
There are exactly $2\text{ moles}$ of gas on the reactant side and $2\text{ moles}$ of gas on the product side. Because there is no change in the number of moles of gas, the overall change in entropy ($\Delta S^\circ$) will be very close to zero (small and positive/negative). It does not contribute significantly to the massive value of $K_p$.

5. Synthesize the findings:
Since $\Delta S^\circ$ is tiny, the thermodynamic driving force that drives $\Delta G^\circ$ to be highly negative (making $K_p$ extremely large) is almost entirely due to the fact that $\Delta H^\circ$ is large and negative ($-186\text{ kJ/mol}_{\text{rxn}}$). This matches Option D.

*

WHY_OTHERS_WRONG:

• Options A and B are incorrect: The activation energy ($E_a$) is a kinetic parameter that determines the *rate* of a reaction (how fast it occurs). The equilibrium constant ($K_p$) is a thermodynamic parameter that describes the relative concentrations of reactants and products at equilibrium. Because thermodynamics and kinetics are independent of one another, changing or having a high activation energy does not influence the overall value of the equilibrium constant.
• Option C is incorrect: Although a positive entropy change ($\Delta S^\circ > 0$) thermodynamically favors a reaction, the entropy change for this reaction is extremely small because there are equal moles of gas on both sides of the equation. A small entropy value multiplied by standard temperature ($T\Delta S^\circ$) is far too small to account for a massive equilibrium constant of $2 \times 10^{30}$.