At 25°C and 1 atm pressure, the enthalpy of combustion of benzene (I) and acetylene (g) are - 3268 kJ mol⁻¹ and -1300 kJ mol⁻¹, respectively. The change in enthalpy for the reaction 3 C₂H₂(g) → C₆H₆ (I), is :

Solution

Chemistry ma’am · 3 steps

Combustion data run backwards: reactants' combustion enthalpy minus products' combustion enthalpy.

1. Build the cycle. Burn 3 mol of C₂H₂ to CO₂ and H₂O. Then run the combustion of benzene in reverse to turn those same products back into C₆H₆(l). The two steps add up to the target reaction.

   Δ H = Σ Δ_c H(reactants) - Σ Δ_c H(products)

2. Put in numbers. Burning 3 C₂H₂ gives 3 × (-1300) = -3900 kJ. Reversing benzene combustion gives +3268 kJ. So Δ H = -3900 + 3268 = -632 kJ mol⁻¹.
3. So. The trimerisation of acetylene to benzene releases heat: Δ H = -632 kJ mol⁻¹, option (c).

The idea to keep. With formation enthalpies it is products minus reactants. With combustion enthalpies it flips: reactants minus products, because you burn the reactants and un-burn the products.

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