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Electrolytes

At 340K one a...

Question

At 340K one atmospheric pressure $N_{2} O_{4}$ is 66% dissociated into $N O_{2}$ . What volume of log $N_{2} O_{4}$ occupy under these conditions?

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Solution

$N_{2} O_{4} ⇋ 2 H O_{2}$
$10.0 g$
$T = 340 K$
$p = 1 a t m$
We start with $10 g$ of $N_{2} O_{4}$ and $66 %$ of it decomposed , that will leave $3.4 g$ of $N_{2} O_{4}$ and make $66 g$ of $N O_{2}$ at equilibrium,
$3.4 g$ of $N_{2} O_{4} ⟷ 0.03696 m o l N_{2} O_{4}$
$6.6 g$ of $N_{2} O_{4} ⟷ 0.1435 m o l N O_{2}$
Total moles $= 0.1805$ moles of $N_{2} O_{4}$ and $N O_{2}$ , at a total pressure of $1 a t m$
$P V = n R T$
$V = \frac{n R T}{p} = \frac{(0.1805) (0.0821) \times (340)}{1}$
$= 5.04 L$
You will get $5.02 L$ if you use $0.18$ moles of gas.

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