Question

One mole of ethane was burnt in oxygen at
constant volume to give CO2 at 298 K. The heat
evolved was found to be 1554 kJ. Calculate the
heat of reaction at constant pressure.​

Answer 1

Heat of reaction of the ethane is - 1504.53 KJ

Given,

one mole of ethane = C₂H₈

temperature (T) = 298 K

heat evolved (ΔE) = 1554 KJ

universal gas constant (R) = 8.3 J mol⁻¹ K⁻¹

C₂H₈  + 4 O₂ →  2 Co₂ + 4 H₂O

change in no of moles (Δn)  = gaseous moles in product - gaseous moles in reactants

Δn = 2 - 4 = -2

Δn = -2

heat of the reaction (ΔH) = ΔE + ΔnRT

ΔH = 1554 × 10³ J + (-2) (8.3 J mol⁻¹ K⁻¹) (298 K)

ΔH  = - 1504.53 KJ