Question

C3h8(g) 5o2(g)→3co2(g) 4h2o(g). calculate δh∘rxn for this reaction using standard enthalpies of formation. (the standard enthalpy of formation of gaseous propane is -103.9 kj/mol .)

Answer 1

To calculate the standard enthalpy of reaction for the combustion of propane, the standard enthalpies of formation for the reactants and products must be summed and subtracted according to the reaction stoichiometry. Using the given standard enthalpies of formation for propane, CO2, and H2O, we can determine the enthalpy change for this exothermic reaction.

Step-by-step explanation:

The question at hand involves calculating the ∆H°rxn (standard enthalpy of reaction) for the combustion of propane (C3H8) using standard enthalpies of formation. The balanced equation for the combustion of propane is provided as:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

To calculate the ∆H°rxn, one must use the standard enthalpies of formation (∆H°f) for the reactants and products:

∆H°rxn = (Σ∆H°f (products)) - (Σ∆H°f (reactants))

The standard enthalpy of formation for propane, CO2(g), and H2O(g) are provided as -103.9 kJ/mol, -393.5 kJ/mol and -241.8 kJ/mol, respectively. Note that the enthalpy of formation for O2, as an element in its standard state, is 0 kJ/mol.

Substituting the given values into the equation:

∆H°rxn = [(3 × -393.5 kJ/mol) + (4 × -241.8 kJ/mol)] - [1 × -103.9 kJ/mol]

Performing the calculations gives us the ∆H°rxn for the combustion of propane.