Combustion of 1 mole of benzene is expressed at
C₆ H₆(l)+15/2 O₂( g) → 6 CO₂( g)+3 H₂ O(l) .
The standard enthalpy of combustion of 2 mol of benzene is -′ x′ kJ. x= __________.
Given :
1. standard Enthalpy of formation of 1 mol of C₆ H₆(l), for the reaction 6 C (graphite) +3 H₂(g) → C₆ H₆(l) is 48.5 kJ mol⁻¹.
2. Standard Enthalpy of formation of 1 mol of CO₂(g), for the reaction C (graphite) +O₂(g) → CO₂(g) is -393.5 kJ mol⁻¹.
3. Standard and Enthalpy of formation of 1 mol of H₂ O(l), for the reaction H₂(g)+1/2 O₂(g) → H₂ O(l) is -286 kJ mol⁻¹.

Solution

Chemistry ma’am · 4 steps

Products minus reactants with formation enthalpies, then double it, because the question asks for 2 mol.

1. Use formation enthalpies. By Hess's law, the enthalpy of any reaction is the sum of the formation enthalpies of the products minus those of the reactants. O₂ is an element in its standard state, so its Δ_f H° is zero.

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

2. Put in numbers. Products: 6 × (-393.5) + 3 × (-286) = -2361 - 858 = -3219 kJ. Subtract benzene's +48.5 kJ to get Δ_c H° = -3267.5 kJ for 1 mol of benzene.
3. Scale to 2 mol. Enthalpy change grows in step with the amount that reacts, so for 2 mol it is 2 × (-3267.5) = -6535 kJ.
4. So. The enthalpy of combustion of 2 mol of benzene is -6535 kJ, so x = 6535.

The common slip: 3267.5. Stopping at the per-mole value gives 3267.5, not 6535. The question asks for 2 mol of benzene, so double it.

The idea to keep. Reaction enthalpy is the sum of Δ_f H° of the products minus that of the reactants, each times its coefficient. Elements count as zero. Always check how many moles the question asks for.

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