3. $\mathrm{Fe}$ and $\mathrm{FeO}$ are in equilibrium with gas mixture of composition $71.8 \% \mathrm{CO}-28.2 \% \mathrm{CO}_{2}$ at $1273 \mathrm{~K}$. Which of the two solid phases disappears if the composition of the gas is held constant and temperature of the system is increased?
$$
\begin{array}{l}
\mathrm{Fe}(\mathrm{s})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})=\mathrm{FeO}(\mathrm{s}) \Delta \mathrm{G}^{\mathrm{o}}=-263700+64.35 \mathrm{~T} \
\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})=\mathrm{CO}_{2}(\mathrm{~g}) \quad \Delta \mathrm{G}^{0}=-394100-0.84 \mathrm{~T} \
\mathrm{C}(\mathrm{s})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})=\mathrm{CO} \quad \Delta \mathrm{G}^{0}=-111700-87.65 \mathrm{~T}
\end{array}
$$

Question

3. $\mathrm{Fe}$ and $\mathrm{FeO}$ are in equilibrium with gas mixture of composition $71.8 \% \mathrm{CO}-28.2 \% \mathrm{CO}_{2}$ at $1273 \mathrm{~K}$. Which of the two solid phases disappears if the composition of the gas is held constant and temperature of the system is increased?
$$
\begin{array}{l}
\mathrm{Fe}(\mathrm{s})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})=\mathrm{FeO}(\mathrm{s}) \Delta \mathrm{G}^{\mathrm{o}}=-263700+64.35 \mathrm{~T} \
\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})=\mathrm{CO}_{2}(\mathrm{~g}) \quad \Delta \mathrm{G}^{0}=-394100-0.84 \mathrm{~T} \
\mathrm{C}(\mathrm{s})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})=\mathrm{CO} \quad \Delta \mathrm{G}^{0}=-111700-87.65 \mathrm{~T}
\end{array}
$$

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Determine the equilibrium constant for the reaction**: The equilibrium constant $K$ for the reaction $\mathrm{Fe}(s) + \frac{1}{2} \mathrm{O}_2(g) = \mathrm{FeO}(s)$ can be calculated using the Gibbs free energy change $\Delta G^{\circ}$ at standard conditions. The relationship between $\Delta G^{\circ}$ and $K$ is given by $\Delta G^{\circ} = -RT \ln K$, where $R$ is the gas constant and $T$ is the temperature in Kelvin. ‖ ‖ 2 ⋮ **Calculate the change in Gibbs free energy with temperature**: The change in Gibbs free energy $\Delta G$ for the reaction at temperature $T$ can be calculated using the equation $\Delta G = \Delta G^{\circ} + 64.35T$ for the formation of $\mathrm{FeO}$ from $\mathrm{Fe}$ and $\mathrm{O}_2$. ‖ ‖ 3 ⋮ **Analyze the effect of temperature on the equilibrium**: As the temperature increases, the value of $\Delta G$ for the formation of $\mathrm{FeO}$ becomes less negative (since $64.35T$ is positive and increases with $T$). This implies that the reaction becomes less spontaneous at higher temperatures, and the equilibrium will shift to favor the reactants, $\mathrm{Fe}(s)$ and $\mathrm{O}_2(g)$. ‖ ‖ 4 ⋮ **Determine which solid phase disappears**: Since the equilibrium shifts to favor the reactants at higher temperatures, $\mathrm{FeO}(s)$ will tend to decompose back into $\mathrm{Fe}(s)$ and $\mathrm{O}_2(g)$. Therefore, the solid phase $\mathrm{FeO}$ will disappear if the temperature is increased while the gas composition is held constant. ‖ ∻Answer∻ ⚹ The solid phase $\mathrm{FeO}$ will disappear upon increasing the temperature. ⚹ ∻Key Concept∻ ⚹ Le Chatelier's principle and the temperature dependence of Gibbs free energy ⚹ ∻Explanation∻ ⚹ According to Le Chatelier's principle, increasing the temperature of an exothermic reaction will shift the equilibrium to favor the reactants. The temperature dependence of Gibbs free energy for the formation of $\mathrm{FeO}$ indicates that the reaction becomes less favorable at higher temperatures, leading to the disappearance of $\mathrm{FeO}$. ⚹◊What is the equilibrium constant expression for the reaction involving Fe and FeO at 1273 K?⍭ Generate me a similar question◊

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