Question

propane c₃h₈ (g), δh_f = 103.8 kj/mol reacts with oxygen to produce carbon dioxide co₂, δh_f = -393.5 kj/mol and water h₂o, δh_f = -241.82 kj/mol according to the equation c₃h₈ (g) + 5o₂ (g) → 3co₂ (g) + 4h₂o (g). what is the enthalpy of combustion (per mole) of c₃h₈ (g)? use δh_rxn = σ(δh_f,products) - σ(δh_f,reactants). 531.5 kj/mol -2,044.0 kj/mol 2,044.0 kj/mol -531.5 kj/mol

Explanation:

Step1: Identify reactants and products

Reactants: $C_3H_8(g)$; Products: $3CO_2(g)+4H_2O(g)$

Step2: Find $\Delta H_f$ values for products

For $CO_2$, $\Delta H_f=-393.5$ kJ/mol; for $H_2O$, $\Delta H_f = - 241.82$ kJ/mol.
$\sum(\Delta H_{f,products})=3\times(-393.5)+4\times(-241.82)$
$= - 1180.5-967.28=-2147.78$ kJ/mol

Step3: Find $\Delta H_f$ value for reactant

For $C_3H_8$, $\Delta H_f=-103.8$ kJ/mol

Step4: Calculate $\Delta H_{rxn}$

$\Delta H_{rxn}=\sum(\Delta H_{f,products})-\sum(\Delta H_{f,reactants})$
$=-2147.78-(-103.8)=-2043.98\approx - 2044.0$ kJ/mol

Answer:

$-2,044.0$ kJ/mol