52) Consider the reaction: $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})$
$$
K_{\text {eq }}=0.1134 \text { at } 20^{\circ} \mathrm{C} . \Delta H^{\circ}{ }_{\text {nxn }}=58.03 \mathrm{~kJ} / \mathrm{mol}
$$

What is $\Delta G^{\circ}$ for this reaction at $20^{\circ} \mathrm{C}$ ?
A) $5.4 \mathrm{~kJ}$
B) $0.36 \mathrm{~kJ}$
C) $4.9 \mathrm{~kJ}$
D) $-5.4 \mathrm{~kJ}$
E) $5.3 \mathrm{~kJ}$

Question

52) Consider the reaction: $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})$
$$
K_{\text {eq }}=0.1134 \text { at } 20^{\circ} \mathrm{C} . \Delta H^{\circ}{ }_{\text {nxn }}=58.03 \mathrm{~kJ} / \mathrm{mol}
$$

What is $\Delta G^{\circ}$ for this reaction at $20^{\circ} \mathrm{C}$ ?
A) $5.4 \mathrm{~kJ}$
B) $0.36 \mathrm{~kJ}$
C) $4.9 \mathrm{~kJ}$
D) $-5.4 \mathrm{~kJ}$
E) $5.3 \mathrm{~kJ}$

Accepted Answer

***D*** ∻Key Concept∻ ⚹Gibbs Free Energy Equation⚹ ∻Explanation∻ ⚹The Gibbs Free Energy change ($\Delta G^\circ$) for a reaction at a constant temperature can be calculated using the equation $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. Here, $\Delta H^\circ$ is the enthalpy change, $T$ is the temperature in Kelvin, and $\Delta S^\circ$ is the entropy change. The entropy change can be derived from the equilibrium constant ($K_{eq}$) using the equation $\Delta S^\circ = -R \ln K_{eq}$, where $R$ is the gas constant (8.314 J/mol·K). By substituting the values and converting units appropriately, we can find the correct answer.⚹◊What is the formula to calculate Gibbs Free Energy Change (ΔG°) for a reaction at a given temperature?⍭ Generate me a similar question◊

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