Question

For a reaction $X+Y\rightleftharpoons 2Z$ 1.0 mol of X, 1.5 mol of Y and 0.5 mol of Z were taken in a 1 L vessel and allowed to react.
At equilibrum, the concentration of Z was $1.0mol{L}^{-1}$
The equilibrium constant of the reaction is $\frac{x}{15}$. The value of x is

Answer 1

$x+y\rightleftharpoons 2Z$

$t=01\frac{3}{2}\frac{1}{2}\text{initial moles}$

$t_{eq}--1mol(2x=\frac{1}{2})$

$t_{eq}(1-x)(\frac{3}{2}-x)\frac{1}{2}2x(x=\frac{1}{4})$

$t_{eq}\frac{3}{4}\frac{5}{4}1$

$K_{eq}=\frac{(1{)}^2}{\frac{5}{4}\times \frac{3}{4}}=\frac{16}{15}$

$K_{eq}=16$