For the reaction $$A \to p$$ at  $$T K$$, the half  life ($$t_{1/2}$$) is plotted as a function of initial concentration $$[A]_o$$ of  $$A$$ as give below.

The value of $$x$$ in the given figure is ______ s (Nearest integer) 

The graph of $$t_{1/2}$$ versus initial concentration $$[A]_0$$ is a straight line passing through the origin. Hence,

$$t_{1/2} \propto [A]_0$$

This relationship is characteristic of a zero-order reaction, for which

$$t_{1/2}=\frac{[A]_0}{2k}$$

Since the graph is linear and passes through the origin, the ratio $$\dfrac{t_{1/2}}{[A]_0}$$ remains constant.

From the graph:

• When $$[A]0=4\times10^{-3},\text{mol L}^{-1}$$, $$t{1/2}=240,\text{s}$$
• When $$[A]0=1.5\times10^{-3},\text{mol L}^{-1}$$, $$t{1/2}=x,\text{s}$$

Therefore,

$$\frac{240}{4\times10^{-3}}=\frac{x}{1.5\times10^{-3}}$$

On solving,

$$x=\frac{240\times(1.5\times10^{-3})}{4\times10^{-3}}$$

Cancelling $$10^{-3}$$ from numerator and denominator,

$$x=\frac{240\times1.5}{4}$$

$$x=60\times1.5$$

$$x=90,\text{s}$$

Hence, the value of $$x$$ is

$$\boxed{90}$$