From different sets of data of `t_(1//2)` at diifferent initial concentrations say 'a' for a given reaction, the `[t_(1//2)xxqa]` is found to be constant. The order or reaction is :

Options

A. zero
B. 1
C. 2
D. 3

Answer

To determine the order of the reaction based on the given information about the half-life (\(t_{1/2}\)) being constant at different initial concentrations, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Half-Life**: The half-life (\(t_{1/2}\)) of a reaction is the time required for the concentration of a reactant to reduce to half of its initial concentration. The relationship between half-life and initial concentration varies depending on the order of the reaction.

2. **Given Condition**: We are given that \(t_{1/2} \times [A]\) (where \([A]\) is the initial concentration) is constant. This implies that as the concentration changes, the product of half-life and concentration remains the same.

3. **Relationship of Half-Life with Concentration**: For a reaction of order \(n\), the half-life can be expressed as:
\[ t_{1/2} \propto [A]^{(1-n)} \]
This means that the half-life is inversely related to the concentration raised to the power of \(n-1\).

4. **Setting Up the Equation**: Since \(t_{1/2} \times [A]\) is constant, we can express this mathematically:
\[ t_{1/2} \times [A] = k \quad \text{(constant)} \]
Rearranging gives:
\[ t_{1/2} = \frac{k}{[A]} \]

5. **Comparing the Two Relationships**: From the relationship we derived earlier:
\[ t_{1/2} \propto [A]^{(1-n)} \]
We can equate this with our expression for \(t_{1/2}\):
\[ \frac{k}{[A]} \propto [A]^{(1-n)} \]

6. **Solving for \(n\)**: If we assume \(t_{1/2} \times [A]\) is constant, we can write:
\[ t_{1/2} \times [A]^{n-1} = \text{constant} \]
This implies that \(n - 1 = 1\) (to satisfy the condition of constancy). Therefore:
\[ n = 2 \]

7. **Conclusion**: The order of the reaction is 2.

### Final Answer:
The order of the reaction is 2.

---