Question

From the given reaction, 
\(\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)},\ \Delta H = -92.6\,\text{kJ}, the enthalpy of formation of \mathrm{NH_3} is\)

• A. \(-92.6\,\text{kJ}\)
• B. \(-138.9\,\text{kJ}\)
• C. \(-185.2\,\text{kJ}\)
• D. \(-46.3\,\text{kJ}\)

Hint:

Always divide the given enthalpy change by the number of moles of product formed to get enthalpy of formation.

Answer 1

The Correct Option is D

Step 1: Understand enthalpy of formation.
Enthalpy of formation is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states.
Step 2: Analyze the given reaction.
The given reaction shows formation of \(2\) moles of \(\mathrm{NH_3}\) with enthalpy change \(-92.6\,\text{kJ}\).
Step 3: Calculate enthalpy of formation for one mole.
\[ \Delta H_f = \frac{-92.6}{2} = -46.3\,\text{kJ} \]
Step 4: Conclusion.
Thus, the enthalpy of formation of \(\mathrm{NH_3}\) is \(-46.3\,\text{kJ}\).