Q29 / 150 · MHT CET 2025 (21 Apr Shift 1) · Chemistry

Find standard enthalpy of formation of ammonia from following reaction. $\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NH}_{3(\mathrm{~g})} ; \Delta_{\mathrm{r}} \mathrm{H}^0=-92.0 \mathrm{~kJ}$

  • A
    $-92.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • B
    $-69.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • Correct answer
    C
    $-46.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • D
    $-184.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

The reaction $\mathrm{N}_{2(\mathrm{g})} + 3\mathrm{H}_{2(\mathrm{g})} \rightarrow 2\mathrm{NH}_{3(\mathrm{g})}$ has an enthalpy change $\Delta_{\mathrm{r}} \mathrm{H}^0 = -92.0\ \mathrm{kJ}$ for the formation of 2 moles of ammonia from its elements in standard states.

Standard enthalpy of formation represents the enthalpy change per mole formed, so $\Delta_{\mathrm{f}} \mathrm{H}^0 (\mathrm{NH}_3) = \frac{-92.0\ \mathrm{kJ}}{2\ \mathrm{mol}} = -46.0\ \mathrm{kJ\cdot mol^{-1}}$.

Final answer: $\boxed{\text{C}}$

Hence, the correct answer is option: 'C'.

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