Given are the following data at 298 K.ΔfH∘H2O,l=−286kJmol−1 ΔfH∘HNO3,1=−174kJmol−1; ΔfH∘N2O5,g=14.5kJmol−1, The ΔrH∘ for the reaction N2O5(g)+H2O(l)⟶2HNO3(l) is

# Given are the following data at 298 K.${\mathrm{\Delta }}_{\mathrm{f}}{H}^{\circ }\left({\mathrm{H}}_{2}\mathrm{O},\mathrm{l}\right)=-286\mathrm{kJ}{\mathrm{mol}}^{-1}$ , The ${\mathrm{\Delta }}_{\mathrm{r}}{H}^{\circ }$ for the reaction ${\mathrm{N}}_{2}{\mathrm{O}}_{5}\left(\mathrm{g}\right)+{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)⟶2{\mathrm{HNO}}_{3}\left(\mathrm{l}\right)$ is

1. A

-76.5 kJ mol-1

2. B

76.5 mol-1

3. C

97.5 kJ mol-1

4. D

-97.5 kJ mol-1

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### Solution:

The required reaction ${\mathrm{N}}_{2}{\mathrm{O}}_{5}\left(\mathrm{g}\right)+{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)⟶2{\mathrm{NHO}}_{3}\left(\mathrm{l}\right)$ is obtained as -Eq.(iii) - Eq. (i) + 2 Eq. (ii) Hence, ${\mathrm{\Delta }}_{1}{H}^{\circ }=\left[-14.5-\left(-286\right)\right)+2\left(-174\right)]\mathrm{kJ}{\mathrm{mol}}^{-1}=-76.5\mathrm{kJ}{\mathrm{mol}}^{-1}$  