∆H0 (enthalpy of combustion at reference temperature T0) for benzene

1.	∆H0 (enthalpy of combustion at reference temperature T0) for benzene vapour (C6H6) at 25°C is – 3301000 kJ/mole with the H2O in the liquid phase. Calculate ∆H0 for the H2O in the vapour phase.
Answer» If H₂O remains as a vapour the heat transferred to the surroundings will be less than when the vapour condenses by the amount due to the change in enthalpy of the vapour during condensation at the reference temperature. ∆H₀ (vapour) = ∆H₀ (liquid) + m_s h_fgo where, m_s = Mass of H₂O formed, and h_fgo = Change in enthalpy of steam between saturated liquid and saturated vapour at the reference temperature T₀ = 2441.8 kJ at 25°C For the reaction : C₆H₆ + 7.5O₂ → 6CO₂ + 3H₂O 3 moles of H₂O are formed on combustion of 1 mole of C₆H₆ ; 3 moles of H₂O = 3 × 18 = 54 kg H₂O ∆H₀ (vapour) = – 3301000 + 54 × 2441.8 = – 3169143 kJ/mole.

∆H0 (enthalpy of combustion at reference temperature T0) for benzene vapour (C6H6) at 25°C is – 3301000 kJ/mole with the H2O in the liquid phase. Calculate ∆H0 for the H2O in the vapour phase.

Answer»

If H₂O remains as a vapour the heat transferred to the surroundings will be less than when the vapour condenses by the amount due to the change in enthalpy of the vapour during condensation at the reference temperature.

∆H₀ (vapour) = ∆H₀ (liquid) + m_s h_fgo

where, m_s = Mass of H₂O formed, and

h_fgo = Change in enthalpy of steam between saturated liquid and saturated vapour at the reference temperature T₀

= 2441.8 kJ at 25°C

For the reaction :

C₆H₆ + 7.5O₂ → 6CO₂ + 3H₂O

3 moles of H₂O are formed on combustion of 1 mole of C₆H₆ ; 3 moles of H₂O

= 3 × 18 = 54 kg H₂O

∆H₀ (vapour) = – 3301000 + 54 × 2441.8

= – 3169143 kJ/mole.

Discussion

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