Question

Half-life of zero order reaction \( A \to \) product is 1 hour, when initial concentration of reactant is 2.0 mol L\(^{-1}\). The time required to decrease concentration of A from 0.50 to 0.25 mol L\(^{-1}\) is:

• A. 0.5 hour
• B. 4 hour
• C. 15 min
• D. 60 min

Hint:

In zero-order reactions, equal amounts of reactant are consumed in equal intervals of time. If it takes 60 min to consume 1.0 mol, it will take 15 min to consume 0.25 mol.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a zero-order reaction, the rate is independent of the concentration of the reactant. The half-life (\(t_{1/2}\)) is directly proportional to the initial concentration (\([A]_0\)).

Step 2: Key Formula or Approach:
1. Zero order half-life: \(t_{1/2} = \frac{[A]_0}{2k}\)
2. Integrated rate law: \([A]_t = [A]_0 - kt \implies t = \frac{[A]_0 - [A]_t}{k}\)

Step 3: Detailed Explanation:
1. Find the rate constant (\(k\)): Using \(t_{1/2} = 1\text{ hr}\) and \([A]_0 = 2.0\text{ mol L}^{-1}\): \[ 1 = \frac{2.0}{2k} \implies k = 1.0\text{ mol L}^{-1}\text{ hr}^{-1} \] 2. Calculate the time (\(t\)) for concentration change: Initial concentration (\([A]_i\)) = \(0.50\text{ mol L}^{-1}\) Final concentration (\([A]_f\)) = \(0.25\text{ mol L}^{-1}\) \[ t = \frac{0.50 - 0.25}{1.0} = 0.25\text{ hours} \] 3. Convert to minutes: \[ 0.25\text{ hours} \times 60\text{ min/hr} = 15\text{ minutes} \]

Step 4: Final Answer
The time required is 15 min.