For a reaction X + Y rightleftharpoons 2 Z,1. | vedclass

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(C) The reaction is $X + Y \rightleftharpoons 2 Z$. Initial moles: $X = 1.0 \ mol, Y = 1.5 \ mol, Z = 0.5 \ mol$. Volume of vessel = $1 \ L$. At equil

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) The reaction is $X + Y \rightleftharpoons 2 Z$. Initial moles: $X = 1.0 \ mol, Y = 1.5 \ mol, Z = 0.5 \ mol$. Volume of vessel = $1 \ L$. At equilibrium,$[Z] = 1.0 \ mol \ L^{-1}$,so moles of $Z = 1.0 \ mol$. Change in moles of $Z = 1.0 - 0.5 = 0.5 \ mol$. According to stoichiometry,for $2 \ mol$ of $Z$ produced,$1 \ mol$ of $X$ and $Y$ are consumed. So,for $0.5 \ mol$ of $Z$ produced,$0.25 \ mol$ of $X$ and $Y$ are consumed. Equilibrium moles: $X = 1.0 - 0.25 = 0.75 \ mol$. $Y = 1.5 - 0.25 = 1.25 \ mol$. $Z = 1.0 \ mol$. $K_{eq} = \frac{[Z]^2}{[X][Y]} = \frac{(1.0)^2}{(0.75)(1.25)} = \frac{1}{0.9375} = \frac{1}{15/16} = \frac{16}{15}$. Given $K_{eq} = \frac{x}{15}$,therefore $x = 16$.

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