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

To solve the problem, let's consider the details of the reaction given:

The reaction is zero-order, denoted as \( A \to \text{product} \), where the rate is independent of the concentration of the reactant.

The formula for the half-life of a zero-order reaction is:

\(t_{1/2} = \frac{[A]_0}{2k}\)

where \([A]_0\) is the initial concentration and \(k\) is the rate constant.

Given that the half-life (\( t_{1/2} \)) is 1 hour and \([A]_0 = 2.0 \, \text{mol L}^{-1}\), we can calculate \(k\):

\(1 = \frac{2.0}{2k}\)

Solving for \( k \):

\(k = 1.0 \, \text{mol L}^{-1} \text{h}^{-1}\)

Next, we want to find the time required to decrease the concentration of \(A\) from 0.50 mol L^-1 to 0.25 mol L^-1. For a zero-order reaction, the time (\( t \)) for the concentration change from \([A]_0\) to \([A]\) is given by:

\(t = \frac{[A]_0 - [A]}{k}\)

Substituting the given values:

\(t = \frac{0.50 - 0.25}{1.0} = 0.25 \, \text{hours}\)

Since we need to find the time in minutes:

\(0.25 \, \text{hours} \times 60 \, \text{min/hour} = 15 \, \text{minutes}\)

Therefore, the time required to decrease the concentration of A from 0.50 mol L^-1 to 0.25 mol L^-1 is 15 minutes.

Hence, the correct answer is: 15 min.

Answer 2

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.