Heat of formation of H₂O(g) at 25⁰C is -243 kJ,ΔE for the reaction H₂(g) + 1/2 O₂(g) → H₂O(g) at 25⁰C is

Options

(a) -243 kJ
(b) -241.8 kJ
(c) 241.8 kJ
(d) 243 kJ

Correct Answer:

-241.8 kJ

Explanation:

H₂(g) + 1/2 O₂(g) → H₂O(g),
ΔHf = -243 kJ, ΔH = ΔE + Δn(g)RT,
where ΔH = enthalpy change of reaction = -243 kJ,
ΔE = internal energy change of reaction, Δn(g) = number of gaseous product – number of gaseous reactant = 1-(1+ 1/2) = – 1/2. ⇒ ΔE = ΔH – Δn(g)RT = -243000 + 0.5 x 8.314 x 298 = -241.76 kJ.