Relation between internal energy change (ΔE°) and enthalpy change (ΔH°) of a reaction is given by ΔH°=ΔE°+ΔngRT ......(I) Δng= moles of product − moles of reactant For the reaction given, OF2(g)+H2O(g)→O2(g)+2HF(g) Δng=(1+2)−(1+1) Δng=1{ΔH=−310kJ,=−310000J} Putting the values in Eq (i) −310000J=ΔE+1(8.3141JK−1mol−1)(298K) ΔE=−312,477.572J≈−312.5kJ