The rate constant of a zero order reaction is `1.5 times 10^–2 mol l^–1 min^–1` at 0.5 M concentration of the reactant. The half life of the reaction is

Options

A. 0.8min
B. 23.1min
C. 16.6min
D. 7.53min

Answer

To find the half-life of a zero-order reaction, we can use the formula for half-life in zero-order kinetics:

\[ t_{1/2} = \frac{[A_0]}{2k} \]

Where:
- \( t_{1/2} \) is the half-life,
- \( [A_0] \) is the initial concentration of the reactant,
- \( k \) is the rate constant.

### Step-by-Step Solution:

1. **Identify the given values:**
- Rate constant \( k = 1.5 \times 10^{-2} \, \text{mol L}^{-1} \text{min}^{-1} \)
- Initial concentration \( [A_0] = 0.5 \, \text{M} \)

2. **Substitute the values into the half-life formula:**
\[ t_{1/2} = \frac{[A_0]}{2k} = \frac{0.5 \, \text{M}}{2 \times (1.5 \times 10^{-2} \, \text{mol L}^{-1} \text{min}^{-1})} \]

3. **Calculate the denominator:**
\[ 2k = 2 \times 1.5 \times 10^{-2} = 3.0 \times 10^{-2} \, \text{mol L}^{-1} \text{min}^{-1} \]

4. **Perform the division:**
\[ t_{1/2} = \frac{0.5}{3.0 \times 10^{-2}} = \frac{0.5}{0.03} = \frac{50}{3} \approx 16.67 \, \text{min} \]

5. **Round the result:**
- The half-life \( t_{1/2} \approx 16.67 \, \text{min} \) can be approximated to \( 16.6 \, \text{min} \) for the final answer.

### Conclusion:
The half-life of the reaction is approximately \( 16.6 \, \text{min} \).

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