A first order reaction is carried out starting with `10" mol L"^(-1)` of the reactant. It is `40%` complete in one hour. If the same reaction is carried out with an initial concentration of `5" mol L"^(-1)`, the percentage of the reaction that is completed in one hour will be

Options

A. `40%`
B. `80%`
C. `20%`
D. `60%`

Explanation

To solve the problem step by step, we will use the first-order reaction kinetics formula.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The initial concentration of the reactant, \( [A_0] = 10 \, \text{mol L}^{-1} \).
- The reaction is 40% complete in 1 hour.

2. **Calculate Remaining Concentration After 40% Completion**:
- If the reaction is 40% complete, then the amount of reactant that has reacted is:
\[ \text{Amount reacted} = 0.40 \times 10 \, \text{mol L}^{-1} = 4 \, \text{mol L}^{-1} \]
- Therefore, the remaining concentration after 1 hour is:
\[ [A] = [A_0] - \text{Amount reacted} = 10 - 4 = 6 \, \text{mol L}^{-1} \]

3. **Use the First-Order Rate Equation to Find the Rate Constant \( K \)**:
- The first-order rate equation is given by:
\[ K = \frac{2.303}{t} \log\left(\frac{[A_0]}{[A]}\right) \]
- Substituting the values into the equation (where \( t = 1 \, \text{hour} \)):
\[ K = \frac{2.303}{1} \log\left(\frac{10}{6}\right) \]
- Calculate \( \log\left(\frac{10}{6}\right) \):
\[ \log\left(\frac{10}{6}\right) = \log(1.6667) \approx 0.2198 \]
- Therefore, substituting back gives:
\[ K \approx 2.303 \times 0.2198 \approx 0.506 \]

4. **Now, Use the Same Rate Constant \( K \) for the New Initial Concentration**:
- The new initial concentration is \( [A_0] = 5 \, \text{mol L}^{-1} \).
- We will use the same rate constant \( K \) to find the concentration after 1 hour:
\[ K = \frac{2.303}{1} \log\left(\frac{5}{[A]}\right) \]
- Rearranging gives:
\[ [A] = 5 \times 10^{-K/2.303} \]
- We need to find \( [A] \) using the value of \( K \):
\[ K \approx 0.506 \]
- Thus,
\[ [A] = 5 \times 10^{-0.506/2.303} \approx 5 \times 0.661 = 3.305 \, \text{mol L}^{-1} \]

5. **Calculate the Amount Reacted**:
- The amount reacted in this case is:
\[ \text{Amount reacted} = [A_0] - [A] = 5 - 3.305 = 1.695 \, \text{mol L}^{-1} \]

6. **Calculate the Percentage Completion**:
- The percentage of the reaction that is completed is given by:
\[ \text{Percentage completed} = \left(\frac{\text{Amount reacted}}{[A_0]}\right) \times 100 = \left(\frac{1.695}{5}\right) \times 100 \approx 33.9\% \]

### Final Answer:
The percentage of the reaction that is completed in one hour with an initial concentration of \( 5 \, \text{mol L}^{-1} \) is approximately **33.9%**.