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)
δh°_rxn =
kj

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: Find the standard enthalpy of formation values

From standard 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 \(\sum n\Delta H_{f}^{\circ}(\text{products})\)

For products: \(n_{CO_{2}} = 3\), \(n_{H_{2}O}=4\)
\(\sum n\Delta H_{f}^{\circ}(\text{products})=3\times(- 393.5)+4\times(-241.8)\)
\(=-1180.5-967.2=-2147.7\space kJ/mol\)

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

For reactants: \(m_{C_{3}H_{8}} = 1\), \(m_{O_{2}}=5\)
\(\sum m\Delta H_{f}^{\circ}(\text{reactants})=1\times(-103.8)+5\times0=-103.8\space kJ/mol\)

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

\(\Delta H_{rxn}^{\circ}=-2147.7-(-103.8)\)
\(\Delta H_{rxn}^{\circ}=-2147.7 + 103.8=-2043.9\space kJ/mol\)

Answer:

\(-2043.9\)