After how many seconds will the concentration of the reactant in a first order reaction be halved if the rate constant is `1.155xx10^(-3)s^(-1)`?

Options

A. 100 sec
B. 200 sec
C. 400 sec
D. 600 sec

Answer

To solve the problem of determining the time it takes for the concentration of a reactant in a first-order reaction to be halved, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formula for Half-Life of a First-Order Reaction**:
The half-life (\(t_{1/2}\)) for a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \(k\) is the rate constant.

2. **Substitute the Given Rate Constant**:
The rate constant \(k\) is provided as \(1.155 \times 10^{-3} \, s^{-1}\). We substitute this value into the half-life formula:
\[ t_{1/2} = \frac{0.693}{1.155 \times 10^{-3}} \]

3. **Perform the Calculation**:
Now, we perform the division:
\[ t_{1/2} = \frac{0.693}{1.155 \times 10^{-3}} \approx 600 \, seconds \]

4. **Conclusion**:
Therefore, the time after which the concentration of the reactant will be halved is approximately 600 seconds.

### Final Answer:
The concentration of the reactant in a first-order reaction will be halved after **600 seconds**.

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