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Question

The enthalpy of formation of $NH_3$ is $-46 \, \text{kJ mol}^{-1}$. The enthalpy change for the reaction: $2 \, NH_3 (g) \rightarrow N_2 (g) + 3 \, H_2 (g)$ is

+92 kJ

+46 kJ

+184 kJ

+23 kJ

Solution

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