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 completion and half-life
In a first-order reaction, when 75% of the reactants have disappeared, 25% of the reactants remain. The time taken to reach 75% completion is related to the half-life \( t_{1/2} \) of the reaction. Specifically, the time for 75% completion is twice the half-life:
\[ t_{75\%} = 2 \times t_{1/2} \]

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

### 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:
\[ t_{1/2} = \frac{0.693}{k} \]
Rearranging this equation to solve for \( k \):
\[ k = \frac{0.693}{t_{1/2}} \]

### Step 4: Substitute the half-life value
Now, substituting the value of \( t_{1/2} \):
\[ k = \frac{0.693}{0.693 \text{ hr}} = 1 \text{ hr}^{-1} \]

### Step 5: Convert the rate constant to seconds
Since we want the rate constant in seconds, we need to convert hours to seconds. There are 3600 seconds in an hour:
\[ k = 1 \text{ hr}^{-1} = \frac{1}{3600} \text{ s}^{-1} \approx 2.77 \times 10^{-4} \text{ s}^{-1} \]

### Final Answer
Thus, the rate constant \( k \) is approximately:
\[ k \approx 2.77 \times 10^{-4} \text{ s}^{-1} \]

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