JEE PYQ: Ionic Equilibrium Question 14
Question 14 - 2020 (05 Sep Shift 2)
Type: Numerical
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 equilibrium, the concentration of $Z$ was 1.0 mol $\text{L}^{-1}$.
The equilibrium constant of the reaction is $\frac{x}{15}$. The value of $x$ is ____.
Show Answer
Answer: 16
Solution
$X + Y \rightleftharpoons 2Z$
Initial: 1 mol, 1.5 mol, 0.5 mol
At equilibrium: $Z = 1.0$ mol, so change $= 0.5 + 2a = 1$, $a = 0.25$
Equilibrium: $X = 0.75$, $Y = 1.25$, $Z = 1$
$K_{eq} = \frac{[Z]^2}{[X][Y]} = \frac{1}{0.75 \times 1.25} = \frac{x}{15}$
$x = \frac{15}{0.75 \times 1.25} = 16$
BETA
