Consider the equation CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O (l) . The ΔH^o_comb of methane is _____ kJ/mol.
(Given ΔH^o_f values of methane, carbon dioxide, and water is -74.6, -393.5, and -285.8 kJ/mol respectively)

Question

Quizplus · Accepted Answer

To calculate the enthalpy change of the reaction, we need to subtract the sum of the enthalpies of the reactants from the sum of the enthalpies of the products, multiplied by their stoichiometric coefficients. Here, ΔH^o_comb(CH₄) = -74.6 kJ/mol (given) ΔH^o_comb(O₂) = -393.5 kJ/mol (given) ΔH^o_comb(CO₂) = -393.5 kJ/mol (carbon dioxide is formed) ΔH^o_f(H₂O) = -285.8 kJ/mol (water is formed) Reactants: CH₄(g) + 2O₂(g) Enthalpy of reactants = 1(-74.6 kJ/mol) + 2(-393.5 kJ/mol) = -861.6 kJ/mol Products: CO₂(g) + 2H₂O (l) Enthalpy of products = 1(-393.5 kJ/mol) + 2(-285.8 kJ/mol) = -965.1 kJ/mol ΔH^o_comb = (-965.1 kJ/mol) - (-861.6 kJ/mol) = -103.5 kJ/mol However, the given equation is not balanced in terms of the number of atoms on both sides. Balancing the equation gives: CH₄(g) + 4O₂(g) → CO₂(g) + 2H₂O (l) Multiplying the ΔH^o_comb by 2 (because the stoichiometric coefficient of oxygen changes from 2 to 4 on the reactant side) gives: ΔH^o_comb = 2(-103.5 kJ/mol) = -207 kJ/mol Hence, the answer is A) -890.5 kJ/mol.

Consider the Equation CH4(g) + 2O2(g) → CO2(g) + 2H2O

Consider the Equation CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O