For a reaction X + Y ⇌ 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 L–1. The equilibrium constant of reaction is x \over 15 . The value of x is \underline\hspace2cm .

Chemical Equilibrium PYQ — JEE Main 2020 · 5 Sep · Shift 2 · Q5 with Solution

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Correct answer: 16

$X + Y \rightleftharpoons 2Z$

Initial concentrations in a $1\,\text{L}$ vessel are same as initial moles:

$[X]_0 = 1.0, \quad [Y]_0 = 1.5, \quad [Z]_0 = 0.5$

From stoichiometry:

$X + Y \rightleftharpoons 2Z$

Change in concentrations:

$[X] = 1.0 - \xi$ $[Y] = 1.5 - \xi$ $[Z] = 0.5 + 2\xi$

Given:

$[Z]_{eq} = 1.0$

So,

$0.5 + 2\xi = 1.0$

$2\xi = 0.5$

$\xi = 0.25$

$[X]_{eq} = 1.0 - 0.25 = 0.75$ $[Y]_{eq} = 1.5 - 0.25 = 1.25$ $[Z]_{eq} = 1.0$

For

$X + Y \rightleftharpoons 2Z$

$K_c = \frac{[Z]^2}{[X][Y]}$

Substitute equilibrium values:

$K_c = \frac{(1.0)^2}{(0.75)(1.25)}$

Convert to fractions:

$0.75 = \frac{3}{4}, \qquad 1.25 = \frac{5}{4}$

Thus,

= \frac{1}{\frac{15}{16}} = \frac{16}{15}$$ 6. **Compare with given form** Given: $$K_c = \frac{x}{15}$$ So, $$\frac{x}{15} = \frac{16}{15} \implies x = 16$$ ## Final Answer $$\boxed{16}$$