TS EAMCET · Chemistry · Thermodynamics (C)
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 \(-74 \mathrm{~kJ}\)
- B +98 kJ
- C +89 kJ
- D -47 kJ
Answer & Solution
Correct Answer
(B) +98 kJ
Step-by-step Solution
Detailed explanation
\[ \Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\left[2 \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{NO})\right]-\left[\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{N}_2 \mathrm{O}\right)+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{O}_2\right)\right] \]…
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