The rate constant of a reaction is `0.01S^-1`, how much time does it take for 2.4` mol L^-1` concentration of reactant reduced to 0.3 `mol L^-1`?

Options

A. `108.3S^-1`
B. `207.9S^-1`
C. `248.2S^-1`
D. `164.3S^-1`

Answer

To solve the problem, we need to determine how much time it takes for the concentration of a reactant to decrease from 2.4 mol L^-1 to 0.3 mol L^-1, given that the rate constant (k) of the reaction is 0.01 s^-1. Since the rate constant is in s^-1, we can conclude that this is a first-order reaction.

### Step-by-Step Solution:

1. **Identify the Reaction Order**:
- The rate constant (k) is given as 0.01 s^-1, indicating that this is a first-order reaction.

2. **Use the First-Order Reaction Formula**:
- For a first-order reaction, the relationship between the initial concentration \([R_0]\), the final concentration \([R]\), and time (t) can be expressed using the equation:
\[ \ln\left(\frac{[R_0]}{[R]}\right) = kt \]
- Here, \([R_0] = 2.4 \, \text{mol L}^{-1}\) and \([R] = 0.3 \, \text{mol L}^{-1}\).

3. **Substitute the Values**:
- Substitute the known values into the equation:
\[ \ln\left(\frac{2.4}{0.3}\right) = 0.01 \cdot t \]

4. **Calculate the Ratio**:
- Calculate the ratio:
\[ \frac{2.4}{0.3} = 8 \]

5. **Calculate the Natural Logarithm**:
- Now, calculate the natural logarithm:
\[ \ln(8) \approx 2.079 \]

6. **Solve for Time (t)**:
- Now, substitute back into the equation:
\[ 2.079 = 0.01 \cdot t \]
- Rearranging gives:
\[ t = \frac{2.079}{0.01} = 207.9 \, \text{s} \]

### Final Answer:
The time required for the concentration of the reactant to decrease from 2.4 mol L^-1 to 0.3 mol L^-1 is approximately **207.9 seconds**.

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