Question

Half-life of zero-order reaction $ A \to $ product is 1 hour, when initial concentration of reaction 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:

For zero-order reactions, the concentration of reactant decreases linearly with time, and the rate constant \( k \) is directly related to the half-life.

Answer 1

The Correct Option is C

For zero-order reaction: The half-life is given by: \[ \text{Half life} = \frac{A_0}{2k} \] Given, \( \text{half-life} = 1 \) hour and the initial concentration \( A_0 = 2.0 \, \text{mol/L} \), we can write: \[ 60 \, \text{min} = \frac{2}{2k} \] Solving for \( k \): \[ k = \frac{1}{60} \, \text{M/min} \] Now, using the formula for zero-order reaction: \[ A_t = A_0 - kt \] \[ t = \frac{A_0 - A_t}{k} \] Substitute the values: \[ t = \frac{0.5 - 0.25}{\frac{1}{60}} = 0.25 \times 60 = 15 \, \text{min} \]
Thus, the time required to decrease the concentration of A from 0.50 to 0.25 mol L\(^{-1}\) is 15 minutes.