1. a) Calculate the value of standard molar Gibbs free energy change of chemical reaction at $298 \mathrm{~K}$ using the data given below:
Is the chemical reaction exothermic reaction or endothermic? WHY?
$$
4 / 3 \mathrm{Al}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 / 3 \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~g}) \text { at } 298 \mathrm{~K}
$$
b) Calculate the enthalpy and entropy of chemical reaction at $1000 \mathrm{~K}$. Leave calculations in the integral form after putting cp values. Organize them as we do it in the class.
$$
4 / 3 \mathrm{Al}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 / 3 \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~g}) \text { at } 1273 \mathrm{~K}
$$

DATA:
$\triangle \mathrm{H} \mathrm{Al}(\mathrm{s} \rightarrow \mathrm{l})=10500$ joules at $\mathbf{9 3 2} \mathrm{K}$,
Standard molar enthalpy formation and entropy at 298K:
$\Delta \mathrm{H}^{{ }^{\circ}} \mathrm{Al}_{2} \mathrm{O}_{3}=-1674000$ joules,
$$
\mathrm{S}^{\mathrm{o} \mathrm{Al}}(\mathrm{s})=28.3 \mathrm{~J} / \underline{\underline{\mathrm{K}},} \quad \mathrm{S}^{\mathrm{o}} \mathrm{Al}_{2} \mathrm{O}_{3}=59.6 \mathrm{~J} / \mathrm{K}, \quad \mathrm{S}^{\circ} \mathrm{O}_{2}=205 \mathrm{~J} / \mathrm{K}
$$

Question

1. a) Calculate the value of standard molar Gibbs free energy change of chemical reaction at $298 \mathrm{~K}$ using the data given below:
Is the chemical reaction exothermic reaction or endothermic? WHY?
$$
4 / 3 \mathrm{Al}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 / 3 \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~g}) \text { at } 298 \mathrm{~K}
$$
b) Calculate the enthalpy and entropy of chemical reaction at $1000 \mathrm{~K}$. Leave calculations in the integral form after putting cp values. Organize them as we do it in the class.
$$
4 / 3 \mathrm{Al}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 / 3 \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~g}) \text { at } 1273 \mathrm{~K}
$$

DATA:
$\triangle \mathrm{H} \mathrm{Al}(\mathrm{s} \rightarrow \mathrm{l})=10500$ joules at $\mathbf{9 3 2} \mathrm{K}$,
Standard molar enthalpy formation and entropy at 298K:
$\Delta \mathrm{H}^{{ }^{\circ}} \mathrm{Al}_{2} \mathrm{O}_{3}=-1674000$ joules,
$$
\mathrm{S}^{\mathrm{o} \mathrm{Al}}(\mathrm{s})=28.3 \mathrm{~J} / \underline{\underline{\mathrm{K}},} \quad \mathrm{S}^{\mathrm{o}} \mathrm{Al}_{2} \mathrm{O}_{3}=59.6 \mathrm{~J} / \mathrm{K}, \quad \mathrm{S}^{\circ} \mathrm{O}_{2}=205 \mathrm{~J} / \mathrm{K}
$$

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Write the balanced chemical equation**: The balanced chemical equation for the reaction given is: $4 \mathrm{Al}(s) + 3 \mathrm{O}_{2}(g) ightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)$. However, the equation provided in the question is not balanced correctly. To balance it, we need to multiply the entire reaction by 3 to get whole numbers for the stoichiometric coefficients: $4 \mathrm{Al}(s) + 3 \mathrm{O}_{2}(g) ightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)$. ‖ ‖ 2 ⋮ **Calculate the standard molar Gibbs free energy change ($\Delta G^\circ$)**: Use the formula $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. The standard molar enthalpy change ($\Delta H^\circ$) for the formation of $\mathrm{Al}_{2} \mathrm{O}_{3}$ is given as $-1674000$ joules. The standard molar entropy change ($\Delta S^\circ$) for the reaction can be calculated using the standard molar entropies of reactants and products: $\Delta S^\circ = (2 imes S^\circ_{\mathrm{Al}_{2} \mathrm{O}_{3}}) - (4 imes S^\circ_{\mathrm{Al}} + 3 imes S^\circ_{\mathrm{O}_{2}})$. Plugging in the values: $\Delta S^\circ = (2 imes 59.6 \mathrm{J/K}) - (4 imes 28.3 \mathrm{J/K} + 3 imes 205 \mathrm{J/K}) = -319.8 \mathrm{J/K}$. Now calculate $\Delta G^\circ$: $\Delta G^\circ = -1674000 \mathrm{J} - 298 \mathrm{K} imes (-319.8 \mathrm{J/K}) = -1674000 \mathrm{J} + 95364.4 \mathrm{J} = -1578635.6 \mathrm{J}$. ‖ ‖ 3 ⋮ **Determine if the reaction is exothermic or endothermic**: Since the standard molar enthalpy change ($\Delta H^\circ$) is negative, the reaction is exothermic. ‖ ∻Answer∻ ⚹ $\Delta G^\circ = -1578635.6 \mathrm{J}$, and the reaction is exothermic. ⚹ ∻Key Concept∻ ⚹ Gibbs free energy change and reaction heat ⚹ ∻Explanation∻ ⚹ The Gibbs free energy change indicates the spontaneity of a reaction, and a negative enthalpy change indicates an exothermic reaction. ⚹ ∻Solution∻ ‖ 1 ⋮ **Calculate the enthalpy change at 1000 K**: The enthalpy change at 1000 K can be calculated using the heat capacities and the enthalpy of phase transition. Since the reaction involves aluminum in the liquid state at 1273 K, we must account for the phase transition from solid to liquid at 932 K. The enthalpy change at 1000 K is given by $\Delta H = \Delta H^\circ + \int_{298}^{1000} \Delta C_p \, dT$, where $\Delta H^\circ$ is the standard enthalpy change and $\Delta C_p$ is the change in heat capacity. ‖ ‖ 2 ⋮ **Calculate the entropy change at 1000 K**: The entropy change at 1000 K can be calculated using the standard molar entropies and the temperature dependence of entropy. The entropy change at 1000 K is given by $\Delta S = \Delta S^\circ + \int_{298}^{1000} \Delta \left(\frac{C_p}{T} ight) \, dT$, where $\Delta S^\circ$ is the standard entropy change and $\Delta \left(\frac{C_p}{T} ight)$ is the change in the temperature-dependent part of entropy. ‖ ∻Answer∻ ⚹ The enthalpy and entropy changes at 1000 K are given by the integrals $\Delta H = \Delta H^\circ + \int_{298}^{1000} \Delta C_p \, dT$ and $\Delta S = \Delta S^\circ + \int_{298}^{1000} \Delta \left(\frac{C_p}{T} ight) \, dT$, respectively. The exact values depend on the heat capacities, which are not provided. ⚹ ∻Key Concept∻ ⚹ Temperature dependence of enthalpy and entropy ⚹ ∻Explanation∻ ⚹ The enthalpy and entropy of a reaction can change with temperature, and these changes can be calculated using integrals involving heat capacities. ⚹◊What is the formula to calculate the standard Gibbs free energy change of a chemical reaction?⍭ Generate me a similar question◊

Sia