Given are the following data at 298 K. ΔfH∘SO2,g=−296.8kJmol−1  and  ΔfH∘SO3,g=−395.7kJmol−1 The value of Li/.!° for the reaction SO2(g)+12O2(g)⟶SO3(g) is

# Given are the following data at 298 K.  The value of Li/.!° for the reaction

1. A

2. B

3. C

4. D

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### Solution:

The required reaction ${\mathrm{SO}}_{2}\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_{2}\left(\mathrm{g}\right)⟶{\mathrm{SO}}_{3}\left(\mathrm{g}\right)$ is obtained as Eq.(ii) - Eq. (i) Hence, ${\mathrm{\Delta }}_{r}{H}^{\circ }=\left[-395.7-\left(-296.8\right)\right]\mathrm{kJ}{\mathrm{mol}}^{-1}=-98.9\mathrm{kJ}{\mathrm{mol}}^{-1}$ For the given reaction ,$\mathrm{\Delta }{V}_{\mathrm{g}}=-1/2$. Hence $\begin{array}{r}{\mathrm{\Delta }}_{\mathrm{r}}{U}^{\circ }={\mathrm{\Delta }}_{\mathrm{r}}{H}^{\circ }-\left(\mathrm{\Delta }{v}_{\mathrm{g}}\right)RT=-98.9\mathrm{kJ}{\mathrm{mol}}^{-1}-\left(-1/2\right)\left(8.314×{10}^{-3}{\mathrm{kJK}}^{-1}{\mathrm{mol}}^{-1}\left(298\mathrm{K}\right)\\ =-\left(98.9+1.24\right){\mathrm{kJmol}}^{-1}=-97.66\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$  