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Chemistry Question on Thermodynamics

Formation of ammonia is shown by the reaction, N2(g)+3H2(g)→2NH3(g)N_{2\left(g\right)}+3H_{2\left(g\right)} \to 2NH_{3\left(g\right)}; ΔrH∘=−91.8 kJ mol−1\Delta_{r}H^{\circ} = -91.8\,kJ\,mol^{-1} What will be the enthalpy of reaction for the decomposition of NH3NH_3 according to the reaction? 2NH3(g)→N2(g)+3H2(g);ΔrH∘=?2NH_{3\left(g\right)}\to N_{2\left(g\right)} + 3H_{2\left(g\right)}; \Delta_{r}H^{\circ} = ?

A

−91.8 kj mol−1-91.8 \,kj \,mol^{-1}

B

+91.8 kj mol−1+ 91.8\,kj \,mol^{-1}

C

−45.9 kj mol−1-45.9\,kj \,mol^{-1}

D

+45.9 kj mol−1+ 45.9\,kj \,mol^{-1}

Answer

+91.8 kj mol−1+ 91.8\,kj \,mol^{-1}

Explanation

Solution

The sign of ΔH\Delta H is reversed.