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Q6.8 The reaction of cyanamide, NH2CN (s), with dioxygen was carried
out in a bomb calorimeter, and ΔU 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)
Dear Sha,
Enthalpy change for a reaction (ΔH) is given by the expression,
ΔH = ΔU + ΔngRT
Where,
ΔU = change in internal energy
Δng = change in number of moles
For the given reaction,
Δng = ∑ng (products) – ∑ng (reactants)
= (2 – 2.5) moles
Δng = –0.5 moles
And,
ΔU = –742.7 kJ mol–1
T = 298 K
R = 8.314 × 10–3 kJ mol–1 K–1
Substituting the values in the expression of ΔH:
ΔH = (–742.7 kJ mol–1) + (–0.5 mol) (298 K) (8.314 × 10–3 kJ mol–1 K–1)
= –742.7 – 1.2
ΔH = –743.9 kJ mol–1
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