Consider the reaction,
$4 {N O}_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (g), Δ_{r} H = - 111 k J .$
If $N_{2} O_{5} (s)$ is formed instead of $N_{2} O_{5} (g)$ in the above reaction, the $Δ_{r} H$ value will be
( Given, $Δ H$ of sublimation for $N_{2} O_{5}$ is $54 k J {m o l}^{- 1}$ )

- 165 k J

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+ 54 k J

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+ 219 k J

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- 219 k J

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Solution

The correct option is A $- 219 k J$
As $2$ $m o l e s$ of $N_{2} O_{5}$ is formed net sublimation energy $= 108$ $k J$ .
$N O_{2} + O_{2} (g) ⟶ 2 N_{2} O_{5} (s) Δ H_{r} = ?$
Adding
$2 N_{2} O_{5} (s) Δ ⟶ 2 N_{2} O_{5} (g) Δ H = + 108$ $k J$
$4 N O_{2} + O_{2} ⟶ 2 N_{2} O_{5} (g) Δ H_{n e t} = - 111$ $k J$
$⟹ Δ H_{r} + Δ H = Δ H_{n e t}$
$⟹ Δ H_{r} = - 111 - 108$
$⟹ Δ H_{r} = - 219$ $k J$

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Similar questions

Consider the reaction:

$4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (g), Δ_{r} H = - 111 k J$ . If $N_{2} O_{5} (s)$ is formed instead of $N_{2} O_{5} (g)$ in the above reaction, the $Δ_{r} H$ value will be:

(given, $Δ H$ of sublimation for $N_{2} O_{5}$ is $54 k J m o l^{- 1}$ )

Consider the reaction:

$4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (g), Δ_{r} H = - 111 k J$ . If $N_{2} O_{5} (s)$ is formed instead of $N_{2} O_{5} (g)$ in the above reaction, the $Δ_{r} H$ value will be: