For the reaction A rightleftharpoons frac{1}{ | vedclass

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(D) The reaction is $A \rightleftharpoons \frac{1}{2} B + C$. Initial moles: $1$ at $t=0$,$0$ at $t=0$,$0$ at $t=0$. Moles at equilibrium: $(1 - \alph

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$\frac{2(D_t - D_o)}{D_o}$

Vedclass · Answer

(D) The reaction is $A \rightleftharpoons \frac{1}{2} B + C$. Initial moles: $1$ at $t=0$,$0$ at $t=0$,$0$ at $t=0$. Moles at equilibrium: $(1 - \alpha)$ at $t=eq$,$\frac{\alpha}{2}$ at $t=eq$,$\alpha$ at $t=eq$. Total moles at equilibrium = $(1 - \alpha) + \frac{\alpha}{2} + \alpha = 1 + \frac{\alpha}{2}$. Let $n$ be the number of moles of products formed from $1$ mole of reactant,so $n = \frac{1}{2} + 1 = 1.5$. The formula for degree of dissociation is $\alpha = \frac{D_t - D_o}{D_o(n - 1)}$. Substituting $n = 1.5$: $\alpha = \frac{D_t - D_o}{D_o(1.5 - 1)} = \frac{D_t - D_o}{0.5 D_o} = \frac{2(D_t - D_o)}{D_o}$.

For the reaction $A \rightleftharpoons \frac{1}{2} B + C$,the degree of dissociation $\alpha$ in terms of vapour density $D_t$ (theoretical) and $D_o$ (observed) is given by:

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