FeF2(s) ⇄ Fe2+(aq) + 2 F−(aq) K1 = 2 × 10−6 F−(aq) + H+(aq) ⇄ HF(aq) K2 = 1 × 103 FeF2(s) + 2 H+(aq) — Equilibrium Chemistry Question

STEPS:

1. Analyze the target reaction: The question asks us to evaluate the thermodynamic favorability of the coupled dissolution reaction:
$$\text{FeF}_2(s) + 2\text{ H}^+(aq) \rightleftharpoons \text{Fe}^{2+}(aq) + 2\text{ HF}(aq) \quad K_3 = \ ?$$
2. Utilize Hess's Law for equilibrium constants: We can construct this target equation by combining the two individual elementary equilibrium steps provided:
* Step 1 (Dissolution): $\text{FeF}_2(s) \rightleftharpoons \text{Fe}^{2+}(aq) + 2\text{ F}^-(aq) \quad K_1 = 2 \times 10^{-6}$
* Step 2 (Protonation): $\text{F}^-(aq) + \text{H}^+(aq) \rightleftharpoons \text{HF}(aq) \quad K_2 = 1 \times 10^3$
3. Manipulate the equations to match the stoichiometry:
To match the target equation, we need 2 moles of $\text{HF}$ as a product and 2 moles of $\text{H}^+$ as a reactant. Therefore, we must multiply the second reaction by a factor of 2:
$$2\text{ F}^-(aq) + 2\text{ H}^+(aq) \rightleftharpoons 2\text{ HF}(aq)$$
4. Determine the new equilibrium constant for the manipulated step:
When a chemical equation is multiplied by a stoichiometric coefficient $n$, its corresponding equilibrium constant is raised to the power of $n$:
$$K_{\text{new}} = (K_2)^2 = (1 \times 10^3)^2 = \mathbf{1 \times 10^6} \quad$$
5. Add the chemical equations together:
Adding the dissolution step to the doubled protonation step allows the intermediate fluoride ions ($2\text{ F}^-$) on the reactant and product sides to cancel out, yielding the net equation:
$$\text{FeF}_2(s) \rightleftharpoons \text{Fe}^{2+}(aq) + \bcancel{2\text{ F}^-(aq)}$$
$$\bcancel{2\text{ F}^-(aq)} + 2\text{ H}^+(aq) \rightleftharpoons 2\text{ HF}(aq)$$
$$\text{Rule Net:} \quad \text{FeF}_2(s) + 2\text{ H}^+(aq) \rightleftharpoons \text{Fe}^{2+}(aq) + 2\text{ HF}(aq)$$
6. Calculate the overall equilibrium constant ($K_3$):
When individual chemical equations are added together to produce a net equation, their respective equilibrium constants are multiplied to find the overall constant:
$$K_3 = K_1 \times (K_2)^2 = (2 \times 10^{-6}) \times (1 \times 10^6) = \mathbf{2} \quad$$
7. Relate the equilibrium constant to thermodynamic favorability:
The standard Gibbs free energy change ($\Delta G^\circ$) and the equilibrium constant ($K$) are related by the equation $\Delta G^\circ = -RT \ln K$.
* If $K > 1$, then $\ln K$ is positive, making $\Delta G^\circ$ negative ($\Delta G^\circ < 0$), which indicates that the process is thermodynamically favorable under standard conditions.
* Because our calculated value for $K_3$ is 2 (which is strictly greater than 1), the overall dissolution process is thermodynamically favorable. This points directly to Option B as the correct answer.

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WHY_OTHERS_WRONG:

• A is incorrect: While $K_2 > 1$ indicates that the formation of $\text{HF}$ from its constituent ions is thermodynamically favorable on its own, this single constant does not account for the substantial energy barrier required to break apart the ionic crystal lattice of solid $\text{FeF}_2$. The favorability of the coupled system must be evaluated using the overall equilibrium constant, $K_3$.
• C is incorrect: Although $K_1 < 1$ indicates that the dissolution of $\text{FeF}_2(s)$ in *pure water* is highly unfavorable, it does not represent the system in an *acidic* environment. In acid, the $\text{H}^+$ ions react with and consume the basic fluoride ions to form stable $\text{HF}$ molecules. This continually draws the dissolution forward (applying Le Chatelier's Principle), making the coupled reaction thermodynamically favorable.
• D is incorrect: While this option correctly identifies that $K_3$ is the governing constant for evaluating the coupled process, it asserts that $K_3 < 1$ and that the reaction is not favorable. This is mathematically disproven by the calculation showing $K_3 = 2$.