Calculate the standard internal energy charge for the reaction; OF2(g)+H2O(g)→O2(g)+2HF(g)at 298 K. The standard enthalpies of formation of OF2(g), H2O(g),HF(g) are +20,+250 , and −270kJmol−1 respectivley?

ΔHReaction ∘=∑HProducts ∘−∑HReactants ∘=2×HHF∘(g)+H∘O2(g)−H∘OF2(g)+HH2O∘(g)∴H∘ of free elements =0∴HO2∘=0∴ΔHf∘=H∘, i.e., standard heat of formation = standard heat enthalpy of a compoundAlso, ΔHf∘ for F2O is +ve;ΔHf∘ for H2O is -ve,  because heat of combustions are exothermic.ΔHR∘=[2×(−270)]+0−[20+(−250)]=−310kJ Now ΔH∘=ΔU∘+ΔnRT∴ΔHR∘=−310×103J;Δn=3−2=1R=8.314J;T=298K∴ −310×103=ΔU∘+1×8.314×298∴ ΔU∘=−312477.5joule∴ ΔU∘=−312.4775kJ