For a first order reaction, if ‘a’ is the initial concentration of reactant, then the half life time is

Options

A. independent of a
B. `propto a`
C. `propto a^(2)`
D. `propto a^(3)`

Explanation

To determine the half-life of a first-order reaction, we can follow these steps:

### Step 1: Understand the concept of half-life
The half-life (t₁/₂) of a reaction is the time required for the concentration of a reactant to decrease to half of its initial concentration.

### Step 2: Write the integrated rate law for a first-order reaction
For a first-order reaction, the integrated rate law can be expressed as:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]
Where:
- \([A]_0\) is the initial concentration of the reactant,
- \([A]\) is the concentration at time \(t\),
- \(k\) is the rate constant,
- \(t\) is the time.

### Step 3: Set up the equation for half-life
At half-life, the concentration of the reactant will be half of the initial concentration:
\[ [A] = \frac{[A]_0}{2} \]

Substituting this into the integrated rate law gives:
\[ \ln \left( \frac{[A]_0}{\frac{[A]_0}{2}} \right) = kt_{1/2} \]

### Step 4: Simplify the equation
This simplifies to:
\[ \ln(2) = kt_{1/2} \]

### Step 5: Solve for half-life
Rearranging the equation to solve for \(t_{1/2}\) gives:
\[ t_{1/2} = \frac{\ln(2)}{k} \]

### Step 6: Substitute the value of \(\ln(2)\)
The value of \(\ln(2)\) is approximately 0.693. Therefore, we can express the half-life as:
\[ t_{1/2} = \frac{0.693}{k} \]

### Conclusion
The half-life of a first-order reaction is independent of the initial concentration and is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]