Question:medium
The enthalpy of formation of $NH_{3}$ is -46 kJ mol$^{-1}$. The enthalpy change for the reaction $2NH_{3}(g) \longrightarrow N_{2}(g) + 3H_{2}(g)$ is
The enthalpy of formation of $NH_{3}$ is -46 kJ mol$^{-1}$. The enthalpy change for the reaction $2NH_{3}(g) \longrightarrow N_{2}(g) + 3H_{2}(g)$ is
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Reverse reaction = Sign change of $\Delta H$; Double the moles = Double the $\Delta H$.
Updated On: Apr 10, 2026
- +184 kJ
- +23 kJ
- +92 kJ
- +46 kJ
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The Correct Option is C
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