For a 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 given below. The value of $x$ in the given figure is ______ s (Nearest integer)}
Step 1: Understanding the Concept:
The graph of $t_{1/2}$ versus $[A]_o$ is a straight line passing through the origin. This implies that $t_{1/2} \propto [A]_o$, which is the defining characteristic of a zero-order reaction. : Key Formula or Approach:
For a zero-order reaction: $t_{1/2} = \frac{[A]_o}{2k}$.
Since the graph is a straight line through the origin, the slope is constant:
Slope = $\frac{t_{1/2, 1}}{[A]_{o, 1}} = \frac{t_{1/2, 2}}{[A]_{o, 2}}$. Step 2: Detailed Explanation:
We have two points from the graph:
Point 1: $[A]_{o, 1} = 4 \times 10^{-3} \text{ mol/L}, t_{1/2, 1} = 240 \text{ s}$.
Point 2: $[A]_{o, 2} = 1.5 \times 10^{-3} \text{ mol/L}, t_{1/2, 2} = x \text{ s}$.
Using the constant slope property:
$\frac{x}{1.5 \times 10^{-3}} = \frac{240}{4 \times 10^{-3}}$.
$x = \frac{240 \times 1.5 \times 10^{-3}}{4 \times 10^{-3}}$.
$x = \frac{240 \times 1.5}{4} = 60 \times 1.5 = 90 \text{ s}$. Step 3: Final Answer:
The value of $x$ is 90.
