Question

calculate the standard enthalpy change for the reaction at 25°c. standard enthalpy of formation values can be found in this list of thermodynamic properties.

c₃h₈(g) + 5o₂(g) → 3co₂(g) + 4h₂o(g)

δhᵣₓₙ° =

Explanation:

Step1: Recall the formula for standard enthalpy change of reaction

The formula is \(\Delta H_{rxn}^{\circ}=\sum n\Delta H_{f}^{\circ}(\text{products})-\sum m\Delta H_{f}^{\circ}(\text{reactants})\), where \(n\) and \(m\) are the stoichiometric coefficients.

Step2: Look up the standard enthalpy of formation values

From thermodynamic tables: \(\Delta H_{f}^{\circ}(C_{3}H_{8}(g))=-103.8\space kJ/mol\), \(\Delta H_{f}^{\circ}(O_{2}(g)) = 0\space kJ/mol\), \(\Delta H_{f}^{\circ}(CO_{2}(g))=-393.5\space kJ/mol\), \(\Delta H_{f}^{\circ}(H_{2}O(g))=-241.8\space kJ/mol\)

Step3: Calculate \(\sum n\Delta H_{f}^{\circ}(\text{products})\)

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Step4: Calculate \(\sum m\Delta H_{f}^{\circ}(\text{reactants})\)

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Step5: Calculate \(\Delta H_{rxn}^{\circ}\)

$$
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Answer:

\(-2043.9\)