Question

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

• A. +184 kJ
• B. +23 kJ
• C. +92 kJ
• D. +46 kJ

Hint:

Reverse reaction = Sign change of $\Delta H$; Double the moles = Double the $\Delta H$.

Answer 1

The Correct Option is C

Step 1: Reaction Analysis
The target reaction is the reverse of the formation reaction of 2 moles of $NH_{3}$.
Step 2: Law of Thermochemistry
$\Delta H_{\text{reaction}} = -(2 \times \text{Enthalpy of formation of } NH_{3})$.
Step 3: Calculation
$\Delta H = -(2 \times -46) = +92$ kJ.
Final Answer: (c)