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: Understanding the Concept: 
The half-life of a reaction (\(t_{1/2}\)) is the time required for the concentration of the reactant to reduce to half its initial value. 
The relationship between half-life and initial concentration depends on the order of the reaction. 
In general: \(t_{1/2} \propto [A_0]^{1-n}\), where \(n\) is the order of the reaction. 
From the graph (a straight line passing through the origin), \(t_{1/2}\) is directly proportional to \([A_0]\), which means the exponent \(1 - n = 1\), giving \(n = 0\). 
Therefore, this is a zero-order reaction. 

Step 2: Key Formula or Approach: 
For zero-order reactions: \(t_{1/2} = \frac{[A_0]}{2k} \propto [A_0]\). 
Use the direct proportionality ratio method: 
\[ \frac{(t_{1/2})_1}{(t_{1/2})_2} = \frac{[A_0]_1}{[A_0]_2} \] 
Step 3: Detailed Explanation: 
Given from the graph: 
When \([A_0]_1 = 4 \times 10^{-3} \text{ M}\), \((t_{1/2})_1 = 120 \text{ sec}\). 
We need to find \(x = (t_{1/2})_2\) when \([A_0]_2 = 1.5 \times 10^{-3} \text{ M}\). 
Applying the proportionality: 
\[ \frac{120}{x} = \frac{4 \times 10^{-3}}{1.5 \times 10^{-3}} = \frac{4}{1.5} \] 
\[ x = \frac{120 \times 1.5}{4} = \frac{180}{4} = 45 \text{ sec} \] 
Step 4: Final Answer: 
The value of `x' is 45 sec.