Question

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

Explanation:

Step1: Identify reactants and products

Reactants: \( C_3H_8(g) \) (\( \Delta H_f = -103.8 \, \text{kJ/mol} \)) and \( O_2(g) \) (standard enthalpy of formation for elements in standard state is \( 0 \, \text{kJ/mol} \)).
Products: \( 3CO_2(g) \) (\( \Delta H_f = -393.5 \, \text{kJ/mol} \)) and \( 4H_2O(g) \) (\( \Delta H_f = -241.82 \, \text{kJ/mol} \)).

Step2: Apply the enthalpy change formula

$$ \Delta H_{rxn} = \sum (\Delta H_{f, \text{products}}) - \sum (\Delta H_{f, \text{reactants}}) $$

Calculate \( \sum (\Delta H_{f, \text{products}}) \):

$$ 3 \times (-393.5) + 4 \times (-241.82) = -1180.5 - 967.28 = -2147.78 \, \text{kJ/mol} $$

Calculate \( \sum (\Delta H_{f, \text{reactants}}) \):

$$ -103.8 + 5 \times 0 = -103.8 \, \text{kJ/mol} $$

Step3: Compute \( \Delta H_{rxn} \)

$$ \Delta H_{rxn} = -2147.78 - (-103.8) = -2147.78 + 103.8 = -2043.98 \approx -2044.0 \, \text{kJ/mol} $$

Answer:

-2,044.0 kJ/mol