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Thermodynamics — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryThermodynamics3 Marks
Question
Given$\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \xrightarrow{ \ \ \ \ \ \ }2\text{NH}_3(\text{g}) ; \Delta_{\text{r}}\text{H}^\ominus= –92.4\text{kJ} \ \text{mol}^{–1}$
What is the standard enthalpy of formation of $\text{NH}_{3}$ gas?
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Answer

Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state. Re-writing the given equation for 1 mole of $\text{NH}_{3(\text{g})}$,
$\frac{1}{2}\text{N}_{2(\text{g})} + \frac{3}{2}\text{H}_{2(\text{g})} \xrightarrow{ \ \ \ \ \ \ } \text{NH}_{3(\text{g})}$
$\therefore$ Standard enthalpy of formation of $\text{NH}_{3(\text{g})}$
$=\frac{1}{2}\Delta_{\text{r}}\text{H}^\ominus$
$=\frac{1}{2}(-92.4\text{kJ}) \ \text{mol}^{-1}$
$=-46.2\text{kJ} \ \text{mol}^{-1}$

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