The half life of a first order reaction is 30 min and the initial concentration of the reactant is 0.1 M . If the initial concentration of reactant is doubled ,then the half life of the reaction will be

Options

A. 1800S
B. 60 min
C. 15 min
D. 900s

Answer

To solve the problem, we need to understand the relationship between the half-life of a first-order reaction and the concentration of the reactant.

### Step-by-Step Solution:

1. **Understand the Half-Life Formula for First-Order Reactions**:
The half-life (T_half) of a first-order reaction is given by the formula:
\[ T_{half} = \frac{0.693}{k} \]
where \( k \) is the rate constant of the reaction.

2. **Given Information**:
- The initial half-life (T_half) is 30 minutes.
- The initial concentration of the reactant is 0.1 M.

3. **Determine the Rate Constant (k)**:
Since we know the half-life, we can calculate the rate constant \( k \):
\[ T_{half} = 30 \text{ min} = 30 \times 60 \text{ seconds} = 1800 \text{ seconds} \]
Rearranging the half-life formula to find \( k \):
\[ k = \frac{0.693}{T_{half}} = \frac{0.693}{1800 \text{ seconds}} \approx 0.000385 \text{ s}^{-1} \]

4. **Doubling the Initial Concentration**:
The problem states that the initial concentration of the reactant is doubled to 0.2 M. However, for first-order reactions, the half-life is independent of the initial concentration.

5. **Conclusion**:
Since the half-life of a first-order reaction does not depend on the concentration of the reactant, the half-life remains the same:
\[ T_{half} = 30 \text{ minutes} = 1800 \text{ seconds} \]

### Final Answer:
The half-life of the reaction remains 30 minutes or 1800 seconds, even when the initial concentration is doubled.

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