If D_T and D_0 are the theoretical and observ | vedclass

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Question

(C) Consider the dissociation reaction: $A \rightleftharpoons nB$
Initial moles: $1 \quad 0$
Moles at equilibrium: $1 - \alpha \quad n\alpha$
Total

Vedclass · Accepted Answer

$\alpha = \frac{D_T - D_0}{(n - 1)D_0}$

Vedclass · Answer

(C) Consider the dissociation reaction: $A ightleftharpoons nB$
Initial moles: $1 \quad 0$
Moles at equilibrium: $1 - \alpha \quad n\alpha$
Total moles at equilibrium: $1 - \alpha + n\alpha = 1 + \alpha(n - 1)$
Since vapour density is inversely proportional to the number of moles,we have: $\frac{D_T}{D_0} = \frac{	ext{Total moles at equilibrium}}{	ext{Initial moles}} = \frac{1 + \alpha(n - 1)}{1}$
Rearranging for $\alpha$: $\frac{D_T}{D_0} = 1 + \alpha(n - 1)$
$\frac{D_T}{D_0} - 1 = \alpha(n - 1)$
$\frac{D_T - D_0}{D_0} = \alpha(n - 1)$
$\alpha = \frac{D_T - D_0}{(n - 1)D_0}$

If $D_T$ and $D_0$ are the theoretical and observed vapour densities at a definite temperature and $\alpha$ is the degree of dissociation of a substance,then $\alpha$ in terms of $D_0, D_T$ and $n$ (number of moles of product formed from $1 \, \text{mole}$ of reactant) is calculated by the formula:

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