Which of the following is (are) not correct for a first order reaction?

Options

A. `t_(1//2) prop a`
B. `t_(1//2) prop 1//a`
C. `t_(1//2) prop a^(0)`
D. `t_(1//2) prop a^(2)`

Explanation

To determine which statements are not correct for a first-order reaction, we need to analyze the properties and equations associated with first-order kinetics.

### Step-by-Step Solution:

1. **Understanding First-Order Reactions**:
- A first-order reaction is characterized by a rate that is directly proportional to the concentration of one reactant. The general form of the rate law for a first-order reaction can be expressed as:
\[ \text{Rate} = k[A] \]
where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.

2. **Half-Life of a First-Order Reaction**:
- The half-life (\( t_{1/2} \)) of a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
- Importantly, this expression shows that the half-life is independent of the initial concentration of the reactant.

3. **Analyzing the Options**:
- We need to evaluate the statements provided in the options regarding the half-life of a first-order reaction:
- **Option A**: \( t_{1/2} \) is directly proportional to \( [A]^1 \) (Incorrect)
- **Option B**: \( t_{1/2} \) is directly proportional to \( [A]^0 \) (Correct)
- **Option C**: \( t_{1/2} \) is independent of concentration (Correct)
- **Option D**: \( t_{1/2} \) is directly proportional to \( [A]^2 \) (Incorrect)

4. **Identifying Incorrect Statements**:
- From our analysis:
- Option A is incorrect because \( t_{1/2} \) is not proportional to \( [A]^1 \).
- Option B is correct because \( t_{1/2} \) is proportional to \( [A]^0 \) (which is 1).
- Option C is correct as \( t_{1/2} \) is independent of concentration.
- Option D is incorrect because \( t_{1/2} \) is not proportional to \( [A]^2 \).

5. **Conclusion**:
- The statements that are **not correct** for a first-order reaction are **A and D**.

### Final Answer:
The incorrect statements for a first-order reaction are **A and D**.

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