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

We have the gaseous equilibrium reaction $$X + Y \rightleftharpoons 2Z$$ carried out in a vessel of volume $$1\ \text{L}$$, so numerical moles are equal to numerical molar concentrations.

Initially, the number of moles (and hence initial concentrations) are

$$[X]_0 = 1.0\ \text{mol L}^{-1},\qquad [Y]_0 = 1.5\ \text{mol L}^{-1},\qquad [Z]_0 = 0.5\ \text{mol L}^{-1}.$$

Let the extent of reaction be such that $$x\ \text{mol L}^{-1}$$ of $$X$$ and $$Y$$ get converted. According to the stoichiometry $$X + Y \to 2Z$$, the change and equilibrium concentrations are:

$$\begin{aligned} [X]_{\text{eq}} &= 1.0 - x,\\[4pt] [Y]_{\text{eq}} &= 1.5 - x,\\[4pt] [Z]_{\text{eq}} &= 0.5 + 2x. \end{aligned}$$

We are told that at equilibrium $$[Z]_{\text{eq}} = 1.0\ \text{mol L}^{-1}.$$ Substituting in the third expression,

$$0.5 + 2x = 1.0.$$

Now, solving for $$x$$ step by step,

$$2x = 1.0 - 0.5 = 0.5,$$

$$x = \frac{0.5}{2} = 0.25.$$

Using this value of $$x$$, we obtain the equilibrium concentrations of all species:

$$[X]_{\text{eq}} = 1.0 - 0.25 = 0.75\ \text{mol L}^{-1},$$

$$[Y]_{\text{eq}} = 1.5 - 0.25 = 1.25\ \text{mol L}^{-1},$$

$$[Z]_{\text{eq}} = 1.0\ \text{mol L}^{-1}.$$

The equilibrium constant expression for the reaction $$X + Y \rightleftharpoons 2Z$$ is, by definition,

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

Substituting the equilibrium concentrations, we have

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

Multiplying the denominator first,

$$0.75 \times 1.25 = 0.9375.$$

Thus,

$$K_c = \frac{1}{0.9375}.$$

Recognising that $$0.9375 = \frac{15}{16},$$ the reciprocal is

$$\frac{1}{0.9375} = \frac{16}{15}.$$

According to the statement of the problem, $$K_c = \dfrac{x}{15}.$$ Setting the two values equal gives

$$\frac{x}{15} = \frac{16}{15}.$$

Because the denominators are identical, the numerators must also be equal:

$$x = 16.$$

So, the answer is $$16$$.