Question

For the reaction \(A \longrightarrow \text{Product}\), the following graph is observed between half life \((t_{1/2})\) and initial concentration of \(A\). Then find the value of \(x\).

Hint:

For a zero order reaction, half-life depends on the initial concentration and is given by \(t_{1/2} = \frac{[A_0]}{2k}\). That is why the graph of \(t_{1/2}\) versus \([A_0]\) is a straight line through the origin.

Answer 1

Correct Answer: 45

Step 1: Identify the order of the reaction from the graph.
The graph between half-life \((t_{1/2})\) and initial concentration \([A_0]\) is a straight line passing through the origin.
This means that half-life is directly proportional to the initial concentration:
\[ t_{1/2} \propto [A_0] \] Such a relation is valid for a zero order reaction, because for zero order kinetics:
\[ t_{1/2} = \frac{[A_0]}{2k} \]
Step 2: Write the proportional relation from the graph.
Since
\[ t_{1/2} \propto [A_0] \] we can write:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \frac{[A_0]_1}{[A_0]_2} \] From the graph,
when \([A_0] = 4 \times 10^{-3}\, \text{M}\), then
\[ t_{1/2} = 120 \text{ sec} \] And when \([A_0] = 1.5 \times 10^{-3}\, \text{M}\), then
\[ t_{1/2} = x \]
Step 3: Substitute the values into the proportionality relation.
So,
\[ \frac{x}{120} = \frac{1.5 \times 10^{-3}}{4 \times 10^{-3}} \]
Step 4: Simplify and calculate \(x\).
\[ \frac{x}{120} = \frac{1.5}{4} \] \[ \frac{x}{120} = 0.375 \] \[ x = 120 \times 0.375 \] \[ x = 45 \]
Step 5: State the final answer.
Hence, the value of \(x\) is:
\[ \boxed{45 \text{ sec}} \]