The equation alpha = frac{D - d}{(n - 1)d} re | vedclass

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(B) The general formula for the degree of dissociation $\alpha$ is given by $\alpha = \frac{D - d}{(n - 1)d}$,where $n$ is the total number of moles o

Vedclass · Accepted Answer

$A \rightleftharpoons \frac{nB}{2} + \left( \frac{2n}{3} \right)C$

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(B) The general formula for the degree of dissociation $\alpha$ is given by $\alpha = \frac{D - d}{(n - 1)d}$,where $n$ is the total number of moles of products formed from $1$ mole of reactant.For the reaction $A \rightleftharpoons xB + yC$,the total number of moles of products is $n = x + y$.Substituting $n = x + y$ into the formula,we get $\alpha = \frac{D - d}{(x + y - 1)d}$.Checking option $B$: $A \rightleftharpoons \frac{n}{3}B + \frac{2n}{3}C$.Here,the sum of stoichiometric coefficients is $x + y = \frac{n}{3} + \frac{2n}{3} = \frac{3n}{3} = n$.Substituting this into the denominator $(x + y - 1)$,we get $(n - 1)$,which matches the given equation.

The equation $\alpha = \frac{D - d}{(n - 1)d}$ represents the degree of dissociation. For which of the following reactions is this equation valid,where $D$ is the theoretical vapour density and $d$ is the observed vapour density?

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