The equilibrium constant for the reaction $\mathrm{W}+\mathrm{X} ightleftharpoons \mathrm{Y}+\mathrm{Z}$ is…

The equilibrium constant for the reaction $\mathrm{W}+\mathrm{X} ightleftharpoons \mathrm{Y}+\mathrm{Z}$ is $9 .$ If one mole of each of $\mathrm{W}$ and $\mathrm{X}$ are mixed and there is no change in volume, the number of moles of Y formed is
  1. $0.10$
  2. $0.50$
  3. $0.75$
  4. $0.54$

Solution



$$
\begin{array}{l}
\mathrm{K}_{\mathrm{eq}}=9=\frac{\alpha^{2}}{(1-\alpha)^{2}} \\
\therefore \alpha=0.75 \\
\text { Moles of } \mathrm{Y}=075
\end{array}
$$ .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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