The enthalpies of formation of gaseous $\mathrm{N}_2 \mathrm{O}$ and $\mathrm{NO}$ at $298 \mathrm{~K}$ are…

The enthalpies of formation of gaseous $\mathrm{N}_2 \mathrm{O}$ and $\mathrm{NO}$ at $298 \mathrm{~K}$ are 82.0 and $90.0 \mathrm{KJ} \mathrm{mol}^{-1}$ respectively. The enthalpy change of the reaction $\mathrm{N}_2 \mathrm{O}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) ightarrow$ $2 \mathrm{NO}(\mathrm{g})$ is
  1. $-74 \mathrm{~kJ}$
  2. +98 kJ
  3. +89 kJ
  4. -47 kJ

Solution

$$ \Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\left[2 \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{NO})ight]-\left[\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{N}_2 \mathrm{O}ight)+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{O}_2ight)ight] $$ $\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{O}_2ight)=0$ (standard state of an element) $$ \begin{gathered} \Rightarrow \Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}=[2 \times 90.0]-[82.0+0.0] \\ \quad=180.0-82.0=+98.0 \mathrm{~kJ} . \end{gathered} $$

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