For the first-order reaction `T_(av)` (average life), `T_(50)` and `T_(75)` in the increasing order are :

Options

A. `T_(50)ltT_(av)ltT_(75)`
B. `T_(50)ltT_(75)ltT_(av)`
C. `T_(av)ltT_(50)ltT_(75)`
D. `T_(av)=T_(50)ltT_(75)`

Answer

To solve the problem of arranging \( T_{av} \) (average life), \( T_{50} \) (half-life), and \( T_{75} \) (time taken for 75% completion) for a first-order reaction in increasing order, we will follow these steps:

### Step 1: Understand the Definitions
- **Half-life (\( T_{50} \))**: This is the time required for the concentration of the reactant to decrease to half of its initial concentration. For a first-order reaction, it is given by:
\[ T_{50} = \frac{0.693}{k} \]
where \( k \) is the rate constant.

- **Time for 75% completion (\( T_{75} \))**: This is the time required for the concentration of the reactant to decrease to 25% of its initial concentration. For a first-order reaction, it can be derived as follows:
\[ \frac{C_0}{4} = C_0 e^{-kT_{75}} \implies e^{-kT_{75}} = \frac{1}{4} \implies -kT_{75} = \ln\left(\frac{1}{4}\right) = -1.386 \implies T_{75} = \frac{1.386}{k} \]

- **Average life (\( T_{av} \))**: This is the average time a molecule of reactant spends in the reaction before it is converted to products. For a first-order reaction, it is given by:
\[ T_{av} = \frac{1}{k} \]

### Step 2: Calculate \( T_{50} \), \( T_{75} \), and \( T_{av} \)
From the above definitions, we can summarize:
- \( T_{50} = \frac{0.693}{k} \)
- \( T_{75} = \frac{1.386}{k} \)
- \( T_{av} = \frac{1}{k} \)

### Step 3: Compare the Values
Now we can compare the values of \( T_{50} \), \( T_{75} \), and \( T_{av} \):
- \( T_{50} = \frac{0.693}{k} \)
- \( T_{75} = \frac{1.386}{k} \)
- \( T_{av} = \frac{1}{k} \)

### Step 4: Arrange in Increasing Order
To find the increasing order, we note:
- \( T_{50} < T_{75} < T_{av} \)

Thus, the increasing order is:
\[ T_{50} < T_{75} < T_{av} \]

### Final Answer
The correct option is \( T_{50} < T_{75} < T_{av} \).