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Given are the following data at 298 K. $Δ_{f} H^{\circ} (H_{2} O, l) = - 286 kJ {mol}^{- 1}$ $Δ_{f} H^{\circ} ({HNO}_{3}, 1) = - 174 kJ {mol}^{- 1}; Δ_{f} H^{\circ} (N_{2} O_{5}, g) = 14.5 kJ {mol}^{- 1}$ , The $Δ_{r} H^{\circ}$ for the reaction $N_{2} O_{5} (g) + H_{2} O (l) ⟶ 2 {HNO}_{3} (l)$ is

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a

-76.5 kJ mol^-1

b

76.5 mol^-1

c

97.5 kJ mol^-1

d

-97.5 kJ mol^-1

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detailed solution

Correct option is A

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

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, $Δ_{1} H^{\circ} = [- 14.5 - (- 286)) + 2 (- 174)] kJ {mol}^{- 1} = - 76.5 kJ {mol}^{- 1}$

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