Question

Use the information below to answer the question that follows.

N2(g)+3H2(g)⟶2NH3(g)Bond energy of(N≡N)=941 kJBond energy of(H−H)=436 kJBond energy of(N−H)=391 kJ


What is the enthalpy of the reaction using bond energies, in kilojoules?

Answer 1

The enthalpy of the reaction is -97 kJ/mol, calculated using bond energies by subtracting the energy to break reactant bonds from the energy released in forming product bonds, indicating an exothermic process.

Step-by-step explanation:

The enthalpy of the reaction using bond energies is calculated by taking the sum of the energies needed to break the bonds on the reactant side and subtracting the energy released by forming the bonds on the product side.

Reactants:

• 1 mole N≡N bonds: 941 kJ/mol
• 3 moles H−H bonds: 3 × 436 kJ/mol = 1308 kJ/mol
• Total energy to break bonds: 941 kJ/mol + 1308 kJ/mol = 2249 kJ/mol

Products:

6 moles N−H bonds: 6 × 391 kJ/mol = 2346 kJ/mol

• Total energy released in bond formation: 2346 kJ/mol

Enthalpy change (ΔH) is calculated by subtracting the energy of bond formation from the energy required to break the bonds:

ΔH = Total energy released in bond formation - Total energy to break bonds

ΔH = 2346 kJ/mol - 2249 kJ/mol = -97 kJ/mol

Therefore, the enthalpy of the reaction is -97 kJ/mol, indicating that it is an exothermic reaction, releasing energy to the surroundings.