What is the $\Delta G^{\circ}$ for the following reaction from the equilibrium constant at the temperature given?
$$
\mathrm{CaCO}_{3} 3(\mathrm{~s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \mathrm{T}=900^{\circ} \mathrm{C} \mathrm{Kp}=1.04
$$

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∻Solution∻ ‖ 1 ⋮ **Identify the reaction**: The given reaction is $\mathrm{CaCO}_{3}(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g})$ at $T=900^{\circ} \mathrm{C}$ with $\mathrm{Kp}=1.04$. ‖ ‖ 2 ⋮ **Use the relationship between $\Delta G^{\circ}$ and $K_p$**: The standard Gibbs free energy change ($\Delta G^{\circ}$) for a reaction can be calculated using the equation: $$ \Delta G^{\circ} = -RT \ln K_p $$ where $R$ is the gas constant ($8.314 \, \mathrm{J \, mol^{-1} \, K^{-1}}$) and $T$ is the temperature in Kelvin. ‖ ‖ 3 ⋮ **Convert temperature to Kelvin**: $$ T = 900^{\circ} \mathrm{C} + 273.15 = 1173.15 \, \mathrm{K} $$ ‖ ‖ 4 ⋮ **Calculate $\Delta G^{\circ}$**: $$ \Delta G^{\circ} = - (8.314 \, \mathrm{J \, mol^{-1} \, K^{-1}}) \times (1173.15 \, \mathrm{K}) \times \ln(1.04) $$ $$ \Delta G^{\circ} \approx - (8.314 \times 1173.15) \times 0.0392 $$ $$ \Delta G^{\circ} \approx - 384.1 \, \mathrm{J \, mol^{-1}} $$ ‖ ∻Answer∻ ⚹ $\Delta G^{\circ} \approx - 384.1 \, \mathrm{J \, mol^{-1}}$ ⚹ ∻Key Concept∻ ⚹ Relationship between Gibbs free energy change and equilibrium constant ⚹ ∻Explanation∻ ⚹ The standard Gibbs free energy change ($\Delta G^{\circ}$) for a reaction can be determined from the equilibrium constant ($K_p$) using the equation $\Delta G^{\circ} = -RT \ln K_p$. This relationship shows how the spontaneity of a reaction is related to its equilibrium position. ⚹◊What is the relationship between equilibrium constant $K_p$ and standard Gibbs free energy change $\Delta G^\circ$ for a chemical reaction?⍭ Generate me a similar question◊

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