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Consider the reaction,
$4 {NO}_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (g), ∆_{r} H = - 111 kJ . If N_{2} O_{5} (s)$ is formed instead of $N_{2} O_{5} (g)$ in the above reaction, the $∆_{r} H$ value will be
(Given, $∆ H$ of sublimation for $N_{2} O_{5} is 54 kJ {mol}^{- 1})$

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$- 165 kJ$

+54 kJ

+219 kJ

$- 219 kJ$

answer is A.

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Detailed Solution

From Hess's law,

$∆ H_{f} + ∆ H_{sub} = ∆ H_{reaction} ∆ H_{f} = ∆ H_{reaction} - ∆ H_{sub} = - 111 kJ - (54 kJ) = - 111 - 54 kJ = - 165 kJ$

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