Q8 / 160 · TS EAMCET 2023 (12 May Shift 2) · Chemistry

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 $\mathrm{N}_2 \mathrm{O}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow$ $2 \mathrm{NO}(\mathrm{g})$ is

  • A
    $-74 \mathrm{~kJ}$
  • Correct answer
    B
    +98 kJ
  • C
    +89 kJ
  • D
    -47 kJ

Solution

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

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

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