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Characteristics of Equilibrium Constant

For the react...

Question

For the reaction, $N_{2} O_{4} ⇌ 2 {N O}_{2}$ , if degree of dissociation of $N_{2} O_{4}$ are $25$ %, $50$ %, $75$ % and $100$ %, the gradation of observed vapour densities is:

d_{1} > d_{2} > d_{3} > d_{4}

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d_{4} > d_{3} > d_{2} > d_{1}

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d_{1} = d_{2} = d_{3} = d_{4}

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None of the above

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Solution

The correct option is C $d_{1} > d_{2} > d_{3} > d_{4}$
$E x p l a n a t i o n$

$N_{2} O_{4} ⇌ 2 N O_{2}$
$1$ $0$ At time $t = 0$
$1 - α$ $2 α$ At equilibrium

There is a relationship between the degree of dissociation and vapour density:

$α = \frac{(D - d)}{(n - 1) d} \to (1)$

$α =$ Degree of dissociation
$d =$ Vapor density
$n =$ No. of moles of gaseous product for $1$ mole of reactant

From the given equation $n = 2$ substitute in equation $1$
then we have:

$α = \frac{D - d}{d} \Rightarrow \frac{D}{d} - 1$

$\Rightarrow 1 + α = \frac{D}{d}$

$\Rightarrow d = \frac{D}{1 + α}$

If $D \sim 1$ then $d = \frac{1}{1 + α}$

$∴$ The higher the $α$ lower the $d$ value.

$∵ α_{1} < α_{2} < α_{3} < α_{4}$

$∴ d_{1} > d_{2} > d_{3} > d_{4}$

Hence the correct answer is option $A$ .

Suggest Corrections

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