Answer to Question #150909 in General Chemistry for Lyna Moinifar

Question

What is ΔH° rxn (in kilojoules) for the reaction below?
3 Fe 2O 3(s) + CO(g) ↔ 2 Fe 3O 4(s) + CO 2(g)

Fe 2O 3(s) ΔH f o = -824.2 kJ/mol
CO(g) ΔH f o = -110.5 kJ/mol.
Fe 3O 4(s) ΔH f o = -1118.4 kJ/mol.
CO 2(g) ΔH f o = -393.5 kJ/mol.

Answer 1

Q150909

What is ΔH° rxn (in kilojoules) for the reaction below?

3 Fe₂O₃(s) + CO(g) ↔ 2 Fe₃O₄(s) + CO₂(g)

Fe₂O₃(s) ΔH_f⁰ = -824.2 kJ/mol

CO(g) ΔH_f⁰ = -110.5 kJ/mol.

Fe₃O₄(s) ΔH_f⁰ = = -1118.4 kJ/mol.

CO₂(g) ΔH_f⁰ = = -393.5 kJ/mol.

Solution:

The given reaction is already balanced for all the element, so there is no need for balancing the

reaction.

The heat of reaction / Enthalpy of reaction: Heat of reaction is equal to the difference between the sum of enthalpies of products and the sum of enthalpies of reactants.

ΔH° _rxn = Sum of enthalpies of products - Sum of enthalpies of reactants

ΔH° _rxn = ΣaH⁰_products - ΣbH⁰ _reactants

a and b are the coefficients of the reactants and products in the balanced equation.

ΔH° _rxn = ( 2 * ΔH_F of Fe₃O₄ + 1 * Δ H_F of CO₂ )– ( 3 * ΔH_F of Fe₂O₃ + 1 * Δ H_F of CO )

ΔH° _rxn = ( 2 * – 1118.4 kJ/mol + 1 * – 393.5 kJ/mol ) – ( 3 * – 824.2 kJ/mol + 1 * – 110.5 kJ/mol )

ΔH° _rxn = ( – 2236.8 – 393.5 ) – ( – 2472.6 – 110.5)

ΔH° _rxn = (– 2630.3)– (– 2583.1)

ΔH° _rxn = -2630.3 + 2583.1 = -47.2 kJ ;

Hence ΔH° rxn (in kilojoules) for the given reaction is -47.2 kilojoules.