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 of reaction is constant and independent of the reactant concentration. The half-life of a zero-order reaction is directly proportional to its initial concentration. 
Step 2: Key Formula or Approach: 
The half-life ($t_{1/2}$) for a zero-order reaction is given by: \[ t_{1/2} = \frac{[A]_0}{2k} \] The integrated rate law for a zero-order reaction is: \[ [A]_t = [A]_0 - kt \] where $k$ is the rate constant. 
Step 3: Detailed Explanation: 
First, calculate the rate constant $k$ using the given half-life information: \[ 1 \text{ hour} = \frac{2.0 \text{ mol L}^{-1}}{2k} \] \[ k = \frac{2.0}{2 \times 1} = 1.0 \text{ mol L}^{-1} \text{ h}^{-1} \] Next, find the time required to decrease the concentration from $0.50 \text{ mol L}^{-1}$ to $0.25 \text{ mol L}^{-1}$. 
Here, the starting concentration for this interval is $[A]_0 = 0.50 \text{ mol L}^{-1}$ and the final concentration is $[A]_t = 0.25 \text{ mol L}^{-1}$. Using the integrated rate law: \[ 0.25 = 0.50 - (1.0) \times t \] \[ 1.0 \times t = 0.50 - 0.25 = 0.25 \text{ hours} \] Convert the time into minutes: \[ t = 0.25 \times 60 \text{ min} = 15 \text{ min} \] Step 4: Final Answer: 
The time required is 15 minutes.