Sr2+(aq) + F-(aq) ⇄ SrF+(aq) At 25°C, the equilibrium constant for the reaction represented above ha — Thermodynamics Chemistry Question

STEPS:

1. Analyze the temperature dependence of the equilibrium constant ($K_{eq}$): At $25^\circ\text{C}$, $K_{eq} = 1.3$, and at $50^\circ\text{C}$, $K_{eq} < 1.3$. This demonstrates that as the temperature increases, the value of the equilibrium constant decreases.
2. Relate $K_{eq}$ shift to Le Chatelier's principle: A decrease in $K_{eq}$ means that at a higher temperature, the equilibrium shifts to favor the reactants (to the left). Under Le Chatelier's principle, increasing the temperature of a system at equilibrium shifts the reaction in the endothermic direction to absorb the added thermal energy. Since the reaction shifts toward the reactants when heated, the reverse reaction must be endothermic, meaning the forward reaction must release heat (exothermic).
3. Determine the sign of $\Delta H^\circ$: Because the forward reaction is exothermic, the standard enthalpy change ($\Delta H^\circ$) at $25^\circ\text{C}$ must be negative ($\Delta H^\circ < 0$). This directly confirms that Option E is correct.
4. Determine the thermodynamical favorability ($\Delta G^\circ$): Using the equation $\Delta G^\circ = -RT\ln K$, we can evaluate standard free energy. At $25^\circ\text{C}$ ($298\text{ K}$), $K_{eq} = 1.3$. Since $K_{eq} > 1$, the value of $\ln(1.3)$ is positive. Because both the gas constant $R$ and temperature $T$ are positive, $\Delta G^\circ$ must be negative ($\Delta G^\circ < 0$), indicating that the reaction is thermodynamically favorable under standard conditions at $25^\circ\text{C}$.

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WHY_OTHERS_WRONG:

• A is incorrect: Increasing the temperature of a system increases the average kinetic energy of the particles, leading to a higher frequency of effective collisions and almost always increasing the reaction rate, regardless of whether the reaction is endothermic or exothermic.
• B is incorrect: Since $K_{eq} = 1.3 > 1$ at $25^\circ\text{C}$, the standard Gibbs free energy ($\Delta G^\circ$) is negative, meaning the reaction is already thermodynamically favorable at $25^\circ\text{C}$. As temperature increases, $K_{eq}$ decreases, making the reaction *less* thermodynamically favorable at higher temperatures.
• C is incorrect: Because $K_{eq} = 1.3 > 1$ at $25^\circ\text{C}$, the value of $\ln K_{eq}$ is positive, which makes $\Delta G^\circ$ negative ($\Delta G^\circ = -RT\ln K_{eq}$).
• D is incorrect: The formation of $\text{SrF}^+(aq)$ from its constituent aqueous ions ($\text{Sr}^{2+}$ and $\text{F}^-$) combines two separate particles into a single complex ion. This reduction in the number of independent dissolved particles restricts their translational freedom, which typically corresponds to a decrease in entropy, making $\Delta S^\circ$ negative rather than positive.