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in Thermodynamics by (48.1k points)

The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be -742.7 kJ mol-1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH2CN(g) + 3/2O2(g) ➝ N2(g) + CO2(g) + H2O(l)

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

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