In a first order reaction A `to` B, if k is rate constant and initial concentration of the reactant A is 2 M then the half life of the reaction is:

Options

A. `(log 2)/k`
B. `( log 2)/(ksqrt0.5)`
C. `(l n 2)/k`
D. `0.693/0.5k`

Answer

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

### Step-by-Step Solution:

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.

2. **Identify the Initial Concentration**:
- The initial concentration of reactant A (A₀) is given as 2 M.

3. **Determine the Concentration at Half-Life**:
- At half-life, the concentration of A (Aₜ) will be half of A₀:
\[ Aₜ = \frac{A₀}{2} = \frac{2 \, \text{M}}{2} = 1 \, \text{M} \]

4. **Use the First-Order Reaction Formula**:
- For a first-order reaction, the half-life is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
- Here, \( k \) is the rate constant.

5. **Substitute the Values**:
- Since we are not given the value of \( k \), we will express the half-life in terms of \( k \):
\[ t_{1/2} = \frac{0.693}{k} \]

6. **Select the Correct Option**:
- From the options provided, we see that \( \frac{\ln 2}{k} \) is equivalent to \( \frac{0.693}{k} \) since \( \ln 2 \approx 0.693 \).
- Therefore, the correct answer is:
\[ t_{1/2} = \frac{\ln 2}{k} \]

### Final Answer:
The half-life of the reaction is \( \frac{\ln 2}{k} \).

---