The standard enthalpy of decomposition of the yellow complex H 3 NSO 2 into NH 3 and SO 2 is + 40 kJ mol –1 . Calculate the standard enthalpy of formation of H 3 NSO 2 . Δ H 0 f (NH 3 ) = – 46.17 kJ mol –1 , Δ H 0 f (SO 2 ) = –296.83
Text Solution
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(– 383 kJ mol –1 )
NH 3 (g) + SO 2 (g) ⎯→ NH 3 SO 2 (s) Δ Hº = – 40 kJ mol –1
Δ Hº = Δ Hº f (NH 3 SO 2 ,S) – Δ Hº f Δ Hº f (NH 3 ,g) – Δ H 0 f (SO 2 ,g)
Solving for Δ Hº f (NH 3 SO 2 ,S)
Δ Hº f = (NH 3 SO 2 ,S) – Δ Hº f (NH 3 ,g) + Δ Hº f (SO 2 ,g) + Δ Hº
= (– 46.17 – 296.83 – 40) KJ mol –1 = – 383 kJ mol –1
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