Which of the following expression is correct for first order reaction? Ao refers to initial concentration of reactant

Options

A. `t_(1//2) prop A_(@)`
B. `t_(1//2) prop A_@^(-1)`
C. `t_(1//2) prop A_@^(-2)`
D. `t_(1//2) prop (A_@)^0`

Answer

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

### Step 1: Understand the first-order reaction rate equation
For a first-order reaction, the rate equation can be expressed as:
\[ k = \frac{1}{t} \ln \left( \frac{[A_0]}{[A]} \right) \]
where:
- \( k \) is the rate constant,
- \( [A_0] \) is the initial concentration of the reactant,
- \( [A] \) is the concentration of the reactant at time \( t \).

### Step 2: Define the half-life
The half-life (\( t_{1/2} \)) is the time required for the concentration of the reactant to decrease to half of its initial value. Therefore, at \( t = t_{1/2} \):
\[ [A] = \frac{[A_0]}{2} \]

### Step 3: Substitute into the rate equation
Substituting \( [A] = \frac{[A_0]}{2} \) into the rate equation gives:
\[ k = \frac{1}{t_{1/2}} \ln \left( \frac{[A_0]}{\frac{[A_0]}{2}} \right) \]

### Step 4: Simplify the logarithm
This simplifies to:
\[ k = \frac{1}{t_{1/2}} \ln \left( 2 \right) \]

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

### Step 6: Conclusion
From this expression, we can conclude that the half-life of a first-order reaction is independent of the initial concentration \( [A_0] \). Therefore, the correct expression for the half-life of a first-order reaction is:
\[ t_{1/2} = \frac{\ln(2)}{k} \]