2 FeO(s) ⇌ 2 Fe(s) + O2(g) Keq = 1 × 10−6 at 1000 K CO2(g) ⇌ C(s) + O2(g) Keq = 1 × 10−32 at 1000 K — Equilibrium Chemistry Question

STEPS:

1. Identify the target reaction and the two given reactions at 1000 K:
The goal is to determine the equilibrium constant ($K_{\text{eq}}$) and the sign of the standard free energy change ($\Delta G^\circ$) for the target reaction:
$$2\text{ FeO}(s) + \text{C}(s) \rightleftharpoons 2\text{ Fe}(s) + \text{CO}_2(g) \quad \text{}$$
We are given two individual reactions at the same temperature:
* Reaction 1: $2\text{ FeO}(s) \rightleftharpoons 2\text{ Fe}(s) + \text{O}_2(g)$ with $K_{\text{eq}1} = 1 \times 10^{-6}$.
* Reaction 2: $\text{CO}_2(g) \rightleftharpoons \text{C}(s) + \text{O}_2(g)$ with $K_{\text{eq}2} = 1 \times 10^{-32}$.

2. Determine the mathematical manipulations needed for the equations:
To obtain the target reaction by adding the given equations, we must arrange the reactants and products appropriately:
* Reaction 1 is kept exactly as written, meaning its equilibrium constant remains unchanged:
$$K_a = K_{\text{eq}1} = 1 \times 10^{-6} \quad \text{}$$
* Reaction 2 must be reversed to place solid carbon ($\text{C}(s)$) on the reactant side and carbon dioxide ($\text{CO}_2(g)$) on the product side. When a chemical equation is reversed, its equilibrium constant is inverted:
$$K_b = \frac{1}{K_{\text{eq}2}} = \frac{1}{1 \times 10^{-32}} = 1 \times 10^{32} \quad \text{}$$

3. Verify the sum of the manipulated equations:
Adding Reaction 1 and the reversed Reaction 2 yields the target equation because the oxygen gas ($\text{O}_2(g)$) appears equally on both sides and cancels out completely:
$$2\text{ FeO}(s) + \text{C}(s) + \text{O}_2(g) \rightleftharpoons 2\text{ Fe}(s) + \text{O}_2(g) + \text{CO}_2(g) \quad \text{}$$
$$2\text{ FeO}(s) + \text{C}(s) \rightleftharpoons 2\text{ Fe}(s) + \text{CO}_2(g) \quad \text{}$$

4. Calculate the overall equilibrium constant ($K_{\text{eq}}$):
When adding chemical equations to find a net reaction, their individual equilibrium constants are multiplied.
$$K_{\text{eq}} = K_a \times K_b \quad$$
$$K_{\text{eq}} = (1 \times 10^{-6}) \times (1 \times 10^{32}) = \mathbf{1 \times 10^{26}} \quad \text{}$$

5. Determine the sign of the standard free energy change ($\Delta G^\circ$):
The thermodynamic relationship between Gibbs free energy and the equilibrium constant is defined by:
$$\Delta G^\circ = -RT \ln K \quad \text{}$$
Because the calculated equilibrium constant ($K_{\text{eq}} = 1 \times 10^{26}$) is significantly greater than $1.0$, the forward reaction is highly favored at equilibrium. Mathematically, when $K > 1$, $\ln K$ is positive, which makes $\Delta G^\circ$ negative. This corresponds directly to Option D.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: It lists an incorrect $K_{\text{eq}}$ value of $1 \times 10^{38}$, which is obtained if a student divides the reversed Reaction 1 constant by Reaction 2's constant instead of multiplying. It also claims $\Delta G^\circ$ is positive, which describes a thermodynamically unfavorable reaction at equilibrium, mathematically contradicting any large $K_{\text{eq}} > 1$.
• Option B is incorrect: While it correctly identifies that $\Delta G^\circ$ is negative, it displays the incorrect $K_{\text{eq}}$ value of $1 \times 10^{38}$ due to a mathematical division error instead of multiplying the correct manipulated constants.
• Option C is incorrect: It correctly calculates $K_{\text{eq}} = 1 \times 10^{26}$, but incorrectly states that $\Delta G^\circ$ is positive. A reaction with a highly favorable equilibrium constant ($K \gg 1$) must have a negative $\Delta G^\circ$ value.