Consider the following data :
Heat of formation of CO₂(g)=-393.5 kJ mol ⁻¹
Heat of formation of H₂ O(l)=-286.0 kJ mol ⁻¹
Heat of combustion of benzene =-3267.0 kJ mol⁻¹
The heat of formation of benzene is __________ kJ mol⁻¹. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Write the combustion equation, then use products minus reactants with benzene as the only unknown.

1. Combustion equation. Burning one mole of benzene gives 6 mol of CO₂ and 3 mol of liquid water: C₆H₆ + 15/2 O₂ → 6CO₂ + 3H₂O. The heat of formation of O₂, an element in its standard state, is zero.

   Δ_c H = [6 Δ_f H(CO₂) + 3 Δ_f H(H₂O)] - Δ_f H(C₆H₆)

2. Products side. 6 × (-393.5) = -2361 kJ and 3 × (-286.0) = -858 kJ. Together the products give -3219 kJ.
3. Solve for benzene. -3267 = -3219 - Δ_f H(C₆H₆), so Δ_f H(C₆H₆) = -3219 + 3267 = +48 kJ mol⁻¹.
4. So. The heat of formation of benzene is 48 kJ mol⁻¹. It is positive, so benzene is less stable than the carbon and hydrogen it is formed from.

The common slip: -48. Subtracting the wrong way round, Δ_c H - (-3219), flips the sign and gives -48 instead of +48.

The idea to keep. Heat of combustion equals the formation enthalpies of the products minus those of the reactants. Put elements at zero, and the compound being burnt is the only unknown.

Answer from the NTA final answer key. Solution by PrepWiser.