Calculate the enthalpy and entropy of chemical reaction at 1000K. Leave calculations in the integral form after putting cp values. Organize them as we do it in the class. $4 / 3 \mathrm{Al}(\mathrm{l})+\mathrm{O} 2(\mathrm{~g}) \boxtimes 2 / 3 \mathrm{Al} 2 \mathrm{O} 3(\mathrm{~g})$ at $1273 \mathrm{~K}$ DATA: $\boxtimes \mathrm{H}$ Al (s『l)=10500 joules at $932 \mathrm{~K}$, Standard molar enthalpy formation and entropy at 298K: $ riangle \mathrm{HofAl} 2 \mathrm{O}=$ -1674000 joules, SoAl(s)=28.3 J/K, So Al2O3=59.6 J/K, SoO2=205 J/K

Question

Calculate the enthalpy and entropy of chemical reaction at 1000K. Leave calculations in the integral form after putting cp values. Organize them as we do it in the class. $4 / 3 \mathrm{Al}(\mathrm{l})+\mathrm{O} 2(\mathrm{~g}) \boxtimes 2 / 3 \mathrm{Al} 2 \mathrm{O} 3(\mathrm{~g})$ at $1273 \mathrm{~K}$ DATA: $\boxtimes \mathrm{H}$ Al (s『l)=10500 joules at $932 \mathrm{~K}$, Standard molar enthalpy formation and entropy at 298K: $	riangle \mathrm{HofAl} 2 \mathrm{O}=$ -1674000 joules, SoAl(s)=28.3 J/K, So Al2O3=59.6 J/K, SoO2=205 J/K

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Write the balanced chemical equation**: The balanced chemical equation for the reaction given is $4/3 \mathrm{Al(s)} + \mathrm{O}_2(g) ightarrow 2/3 \mathrm{Al}_2\mathrm{O}_3(g)$. ‖ ‖ 2 ⋮ **Calculate the standard enthalpy change for the reaction at 298 K**: The standard enthalpy change for the reaction ($\Delta H^\circ_{ ext{rxn}}$) at 298 K can be calculated using the standard enthalpies of formation ($\Delta H^\circ_{ ext{f}}$) of the products and reactants. The formula is $\Delta H^\circ_{ ext{rxn}} = \sum (\Delta H^\circ_{ ext{f, products}}) - \sum (\Delta H^\circ_{ ext{f, reactants}})$. For the given reaction, $\Delta H^\circ_{ ext{rxn}} = \frac{2}{3} \Delta H^\circ_{ ext{f, Al}_2\mathrm{O}_3} - \frac{4}{3} \Delta H^\circ_{ ext{f, Al}} - \Delta H^\circ_{ ext{f, O}_2}$. Since $\Delta H^\circ_{ ext{f, O}_2}$ is zero for elements in their standard state, we have $\Delta H^\circ_{ ext{rxn}} = \frac{2}{3}(-1674000 ext{ J}) - \frac{4}{3}(0 ext{ J})$. ‖ ‖ 3 ⋮ **Calculate the standard entropy change for the reaction at 298 K**: The standard entropy change for the reaction ($\Delta S^\circ_{ ext{rxn}}$) at 298 K can be calculated using the standard molar entropies ($S^\circ$) of the products and reactants. The formula is $\Delta S^\circ_{ ext{rxn}} = \sum (S^\circ_{ ext{products}}) - \sum (S^\circ_{ ext{reactants}})$. For the given reaction, $\Delta S^\circ_{ ext{rxn}} = \frac{2}{3} S^\circ_{ ext{Al}_2\mathrm{O}_3} - \frac{4}{3} S^\circ_{ ext{Al}} - S^\circ_{ ext{O}_2}$. Plugging in the values, we get $\Delta S^\circ_{ ext{rxn}} = \frac{2}{3}(59.6 ext{ J/K}) - \frac{4}{3}(28.3 ext{ J/K}) - 205 ext{ J/K}$. ‖ ‖ 4 ⋮ **Adjust enthalpy and entropy to 1000 K**: To adjust the enthalpy and entropy to 1000 K, we need to integrate the heat capacities ($C_p$) from 298 K to 1000 K. The change in enthalpy ($\Delta H$) and entropy ($\Delta S$) can be calculated using the formulas $\Delta H = \Delta H^\circ_{ ext{rxn}} + \int_{298}^{1000} C_p \, dT$ and $\Delta S = \Delta S^\circ_{ ext{rxn}} + \int_{298}^{1000} \frac{C_p}{T} \, dT$. Since the heat capacities are not provided, we leave the integrals in their general form. ‖ ∻Answer∻ ⚹ The standard enthalpy change for the reaction at 298 K is $\Delta H^\circ_{ ext{rxn}} = \frac{2}{3}(-1674000 ext{ J}) = -1116000 ext{ J}$. The standard entropy change for the reaction at 298 K is $\Delta S^\circ_{ ext{rxn}} = \frac{2}{3}(59.6 ext{ J/K}) - \frac{4}{3}(28.3 ext{ J/K}) - 205 ext{ J/K} = -149.1 ext{ J/K}$. The changes in enthalpy and entropy at 1000 K cannot be calculated without the heat capacities and are left in integral form. ⚹ ∻Key Concept∻ ⚹ The standard enthalpy and entropy changes of a reaction at a given temperature can be calculated using the standard enthalpies of formation and standard molar entropies of the reactants and products. ⚹ ∻Explanation∻ ⚹ The enthalpy change at a different temperature requires the integration of heat capacities over the temperature range, and the entropy change also requires integration of heat capacities divided by temperature. Without specific $C_p$ values, these calculations remain in integral form. ⚹◊What is the formula to calculate the enthalpy change of the chemical reaction at 1000K using the provided data and the heat capacities? ⍭ Generate me a similar question◊

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