For a first order reaction, `t_(1//2) = underset(k)ul(...........)`.

Answer

To find the expression for the half-life (\(t_{1/2}\)) of a first-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definition of Half-Life**:
The half-life of a reaction is the time required for the concentration of a reactant to decrease to half of its initial concentration.

2. **Write the Rate Law for a First-Order Reaction**:
For a first-order reaction, the rate constant \(k\) is related to the concentration of the reactants by the equation:
\[ k = \frac{1}{t} \ln\left(\frac{A_0}{A_t}\right) \]
where \(A_0\) is the initial concentration and \(A_t\) is the concentration at time \(t\).

3. **Substitute for Half-Life**:
At half-life (\(t = t_{1/2}\)), the concentration \(A_t\) is half of the initial concentration:
\[ A_t = \frac{A_0}{2} \]

4. **Plug in the Half-Life Condition**:
Substitute \(A_t\) into the rate law:
\[ k = \frac{1}{t_{1/2}} \ln\left(\frac{A_0}{\frac{A_0}{2}}\right) \]

5. **Simplify the Logarithm**:
The logarithm simplifies as follows:
\[ \ln\left(\frac{A_0}{\frac{A_0}{2}}\right) = \ln(2) \]
Therefore, we can rewrite the equation as:
\[ k = \frac{1}{t_{1/2}} \ln(2) \]

6. **Rearranging to Solve for Half-Life**:
Rearranging the equation to solve for \(t_{1/2}\):
\[ t_{1/2} = \frac{\ln(2)}{k} \]

7. **Substituting the Value of \(\ln(2)\)**:
The value of \(\ln(2)\) is approximately \(0.693\). Thus, we can express the half-life as:
\[ t_{1/2} = \frac{0.693}{k} \]

### Final Expression:
The final expression for the half-life of a first-order reaction is:
\[ t_{1/2} = \frac{0.693}{k} \]