For a partiocular reaction with initial conc. Of the rectents as `a_(1)` and `a_(2)`, the half-life period are `t_(1)` and `t_(2)` respectively. The order of the reaction (n) is given by :

Options

A. `n=1+(log(t_(2)//t_(1)))/(log(a_(2)//a_(1))`
B. `n=(log(t_(2)//t_(1)))/(log(a_(2)//a_(1))`
C. `n=1+(log(t_(1)//t_(2)))/(log(a_(2)//a_(1))`
D. None of these

Answer

To determine the order of a reaction (n) based on the initial concentrations of the reactants and their respective half-lives, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Half-Life Expression**:
The half-life (t_half) of a reaction is related to the initial concentration (A) and the rate constant (k) by the general formula:
\[ t_{1/2} = \frac{2^{(n-1)}}{(n-1) \cdot k \cdot A^{(n-1)}} \]
Here, n is the order of the reaction.

2. **Expressing Half-Life in Terms of Concentration**:
From the above expression, we can see that the half-life is inversely proportional to the initial concentration raised to the power of (n-1):
\[ t_{1/2} \propto A^{(1-n)} \]
This indicates that as the concentration increases, the half-life decreases for reactions of order greater than 1.

3. **Setting up the Ratios**:
For two different initial concentrations \(A_1\) and \(A_2\) with corresponding half-lives \(t_1\) and \(t_2\), we can write:
\[ \frac{t_1}{t_2} = \frac{A_1^{(1-n)}}{A_2^{(1-n)}} \]

4. **Removing Proportionality**:
To eliminate the proportionality constant (k), we can rearrange the equation:
\[ \frac{t_1}{t_2} = \left(\frac{A_1}{A_2}\right)^{(1-n)} \]

5. **Taking Logarithms**:
Taking the logarithm of both sides gives:
\[ \log\left(\frac{t_1}{t_2}\right) = (1-n) \cdot \log\left(\frac{A_1}{A_2}\right) \]

6. **Isolating (1-n)**:
Rearranging this equation allows us to isolate \(1-n\):
\[ 1-n = \frac{\log\left(\frac{t_1}{t_2}\right)}{\log\left(\frac{A_1}{A_2}\right)} \]

7. **Finding n**:
Finally, we can solve for n:
\[ n = 1 - \frac{\log\left(\frac{t_1}{t_2}\right)}{\log\left(\frac{A_1}{A_2}\right)} \]

8. **Using the Reciprocal Property of Logarithms**:
Since \(\log\left(\frac{A_1}{A_2}\right) = -\log\left(\frac{A_2}{A_1}\right)\), we can rewrite n as:
\[ n = 1 + \frac{\log\left(\frac{t_1}{t_2}\right)}{\log\left(\frac{A_2}{A_1}\right)} \]

### Final Expression:
Thus, the order of the reaction (n) is given by:
\[ n = 1 + \frac{\log(t_1/t_2)}{\log(A_2/A_1)} \]