Rate constant for first order reaction is `5.78xx10^(-5)S^(-1)`. What percentage of initial reactant will react in 10 hours ?

Options

A. `12. 5%`
B. `25%`
C. `87. 5%`
D. `75%`

Explanation

To solve the problem, we need to find out what percentage of the initial reactant will react in 10 hours for a first-order reaction given the rate constant \( k = 5.78 \times 10^{-5} \, \text{s}^{-1} \).

### Step-by-Step Solution:

1. **Convert Time to Seconds**:
Since the rate constant is given in seconds, we need to convert 10 hours into seconds.
\[ 10 \, \text{hours} = 10 \times 3600 \, \text{seconds} = 36000 \, \text{seconds} \]

2. **Use the First-Order Reaction Formula**:
The integrated rate law for a first-order reaction is given by:
\[ k = \frac{2.303}{t} \log\left(\frac{A_0}{A}\right) \]
Where:
- \( k \) = rate constant
- \( t \) = time
- \( A_0 \) = initial concentration
- \( A \) = concentration at time \( t \)

3. **Substitute the Known Values**:
Substitute \( k = 5.78 \times 10^{-5} \, \text{s}^{-1} \) and \( t = 36000 \, \text{s} \) into the equation:
\[ 5.78 \times 10^{-5} = \frac{2.303}{36000} \log\left(\frac{A_0}{A}\right) \]

4. **Rearranging the Equation**:
Rearranging the equation to isolate the logarithm:
\[ \log\left(\frac{A_0}{A}\right) = 5.78 \times 10^{-5} \times 36000 / 2.303 \]

5. **Calculate the Right Side**:
Calculate the value on the right side:
\[ \log\left(\frac{A_0}{A}\right) \approx 0.9635 \]

6. **Take Antilog**:
Now, take the antilogarithm to find \( \frac{A_0}{A} \):
\[ \frac{A_0}{A} = 10^{0.9635} \approx 9.14 \]

7. **Calculate \( A \)**:
From \( \frac{A_0}{A} = 9.14 \), we can express \( A \):
\[ A = \frac{A_0}{9.14} \]

8. **Calculate the Percentage of Reactant that has reacted**:
The fraction of the reactant that has reacted is:
\[ \text{Fraction reacted} = 1 - \frac{A}{A_0} = 1 - \frac{1}{9.14} \approx 0.891 \]
To find the percentage:
\[ \text{Percentage reacted} = 0.891 \times 100 \approx 89.1\% \]

### Final Answer:
Approximately **89.1%** of the initial reactant will react in 10 hours.