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 the reaction (n) based on the initial concentrations of the reactants (A1 and A2) and their respective half-lives (T1 and T2), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Half-Life and Reaction Order**:
- The half-life of a reaction is the time required for the concentration of a reactant to decrease to half of its initial value.
- For a reaction of order n, the half-life (t) is related to the initial concentration ([A]) by the formula:
\[ t \propto [A]^{(n-1)} \]
- This means that the half-life is inversely proportional to the initial concentration raised to the power of (n - 1).

2. **Setting Up the Relationship**:
- For the two reactants with initial concentrations A1 and A2 and half-lives T1 and T2, we can express the relationship as:
\[ \frac{T1}{T2} = \left(\frac{A2}{A1}\right)^{(n-1)} \]

3. **Taking Logarithms**:
- To simplify the equation, take the logarithm of both sides:
\[ \log\left(\frac{T1}{T2}\right) = (n-1) \log\left(\frac{A2}{A1}\right) \]

4. **Rearranging the Equation**:
- Rearranging gives:
\[ n - 1 = \frac{\log\left(\frac{T1}{T2}\right)}{\log\left(\frac{A2}{A1}\right)} \]
- Adding 1 to both sides results in:
\[ n = 1 + \frac{\log\left(\frac{T1}{T2}\right)}{\log\left(\frac{A2}{A1}\right)} \]

5. **Final Expression for Order of Reaction**:
- Thus, the order of the reaction (n) can be expressed as:
\[ n = 1 + \frac{\log\left(T1\right) - \log\left(T2\right)}{\log\left(A2\right) - \log\left(A1\right)} \]
- This is the final formula to determine the order of the reaction based on the given half-lives and concentrations.