In a first order reaction, `75%` of the reactants disappeared in `1.386 hr`. What is the rate constant ?

Options

A. `3.6 xx 10^(-3) s^(-1)`
B. `2.7 xx 10^(-4) s^(-1)`
C. `72 xx 10^(-3) s^(-1)`
D. `1.8 xx 10^(-3) s^(-1)`

Explanation

To find the rate constant \( k \) for a first-order reaction where 75% of the reactants have disappeared in 1.386 hours, we can follow these steps:

### Step 1: Understand the relationship between percentage disappearance and half-life
In a first-order reaction, when 75% of the reactants have disappeared, 25% of the reactants remain. The time taken for 75% disappearance is related to the half-life of the reaction. The relationship is given by:
\[ T_{75\%} = T_{1/2} \times 2 \]
This means that the time taken for 75% of the reactants to disappear is twice the half-life.

### Step 2: Calculate the half-life
Given that \( T_{75\%} = 1.386 \) hours, we can find the half-life \( T_{1/2} \):
\[ T_{1/2} = \frac{T_{75\%}}{2} = \frac{1.386 \text{ hours}}{2} = 0.693 \text{ hours} \]

### Step 3: Use the half-life to find the rate constant
For a first-order reaction, the relationship between the half-life and the rate constant \( k \) is given by:
\[ T_{1/2} = \frac{\ln 2}{k} \]
Rearranging this equation to solve for \( k \):
\[ k = \frac{\ln 2}{T_{1/2}} \]

### Step 4: Substitute the values
We know that \( \ln 2 \approx 0.693 \) and \( T_{1/2} = 0.693 \) hours. Substituting these values into the equation gives:
\[ k = \frac{0.693}{0.693 \text{ hours}} = 1 \text{ hour}^{-1} \]

### Step 5: Convert the rate constant to seconds
Since the answer is typically required in seconds, we convert hours to seconds:
\[ 1 \text{ hour} = 3600 \text{ seconds} \]
Thus,
\[ k = \frac{1}{3600} \text{ seconds}^{-1} \approx 2.77 \times 10^{-4} \text{ seconds}^{-1} \]

### Final Answer
The rate constant \( k \) for the reaction is approximately \( 2.77 \times 10^{-4} \text{ seconds}^{-1} \).

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