The value of \Delta _{f}H^{\ominus } for NH_{3} is – 91.8 kJ mol–1. Calculate the enthalpy change for the following reaction :
2NH_{3}(g) \rightarrow N_{2}(g) + 3H_{2}(g)

Answers (1)

The enthalpy change for the given reaction can be calculated as: -

N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3} (g)

\Delta _{f} H = -91.8\; kg/mol
The enthalpy of the reaction is +91.8 kg/mol. This happens as with the reversal of reaction, the value of\Delta _{r} H also gets reversed.

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