For the reaction 2A rightleftharpoons B + 2C, | vedclass

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Question

(B) For the reaction $2A \rightleftharpoons B + 2C$,the initial moles are $1$ and at equilibrium,moles are $(1-x)$,$(x/2)$,and $x$ respectively.Total

Vedclass · Accepted Answer

$x = 2 \left( \frac{D_t - D_0}{D_0(3 - 1)} \right)$

Vedclass · Answer

(B) For the reaction $2A \rightleftharpoons B + 2C$,the initial moles are $1$ and at equilibrium,moles are $(1-x)$,$(x/2)$,and $x$ respectively.Total moles at equilibrium = $(1-x) + x/2 + x = 1 + x/2$.The relation between degree of dissociation $(x)$,theoretical vapour density $(D_t)$,and observed vapour density $(D_0)$ is given by $x = \frac{D_t - D_0}{D_0(n - 1)}$,where $n$ is the number of moles of products formed from $1$ mole of reactant.Here,$n = (1 + 2)/2 = 1.5$.Substituting $n = 1.5$ into the formula: $x = \frac{D_t - D_0}{D_0(1.5 - 1)} = \frac{D_t - D_0}{0.5 D_0} = 2 \left( \frac{D_t - D_0}{D_0} \right)$.

For the reaction $2A \rightleftharpoons B + 2C$,the degree of dissociation $(x)$ in terms of vapour density is:

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