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

A
-76.5 kJ mol^-1
B
76.5 mol^-1
C
97.5 kJ mol^-1
D
-97.5 kJ mol^-1

Solution:

$(i) H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ_{f} H^{\circ} = - 286 kJ {mol}^{- 1} (ii) \frac{1}{2} N_{2} (g) + \frac{3}{2} O_{2} (g) + \frac{1}{2} H_{2} (g) ⟶ {HNO}_{3} (l); Δ_{f} H^{\circ} = - 174 kJ {mol}^{- 1} (iii) N_{2} (g) + \frac{5}{2} O_{2} (g) ⟶ N_{2} O_{5} (g); {ΔH}^{\circ} = - 14.5 {kJmol}^{- 1}$

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

Solution:

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