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Chemistry Question on Thermodynamics terms

The reaction of cyanamide, NH2CN(s)NH_2CN(s), with dioxygen was carried out in a bomb calorimeter, and ∆U∆U was found to be –742.7 kJ mol–1–742.7\ kJ\ mol^{–1} at 298 K298\ K. Calculate enthalpy change for the reaction at 298 K298\ K.
NH2CN(g)+32O2(g)→N2(g)+CO2(g)+H2O(l)NH_2CN(g)+\frac 32O_2(g)→N_2(g)+CO_2(g)+H_2O(l)

Answer

Enthalpy change for a reaction (∆H∆H) is given by the expression,
∆H=∆U+∆ngRT∆H = ∆U + ∆n_gRT
Where,
∆U∆U = change in internal energy
∆ng∆n_g = change in number of moles
For the given reaction,
∆ng∆n_g = ∑ng∑n_g (products) – ∑ng∑n_g (reactants)
∆ng∆n_g = (2–2.5)(2 – 2.5) moles
∆ng∆n_g = –0.5–0.5 moles
And, ∆U=–742.7∆U = –742.7 kJmol–1kJ mol^{–1}
T=298 KT = 298\ K
R=8.314×10–3 kJmol–1K–1R = 8.314 × 10^{–3}\ kJ mol^{–1}K^{–1}
Substituting the values in the expression of ∆H∆H:
∆H=(–742.7 kJmol–1)+(–0.5 mol)(298 K)(8.314×10–3kJ mol–1K–1)∆H = (–742.7 \ kJ mol^{–1}) + (–0.5\ mol) (298\ K) (8.314 × 10^{–3} kJ\ mol^{–1} K^{–1})
∆H=–742.7–1.2∆H= –742.7 – 1.2
∆H=–743.9 kJ mol–1∆H = –743.9 \ kJ\ mol^{–1}