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. 0
B. 1
C. 2
D. 3

Answer

To determine the order of the reaction based on the given information about the half-life (\(t_{1/2}\)) and the initial concentration (\(a\)), we will follow these steps:

### Step 1: Understand the relationship between half-life and concentration
The half-life of a reaction is related to the initial concentration of the reactants. For a reaction of order \(n\), the relationship can be expressed as:
\[ t_{1/2} \propto a^{(1-n)} \]
where \(t_{1/2}\) is the half-life and \(a\) is the initial concentration.

### Step 2: Rearranging the relationship
From the proportionality, we can express it in terms of a constant:
\[ t_{1/2} \cdot a^{(n-1)} = \text{constant} \]
This means that if we multiply the half-life by the initial concentration raised to the power of \(n-1\), we will get a constant value.

### Step 3: Analyze the given information
According to the problem, it is stated that \(t_{1/2} \cdot a\) is a constant. This implies:
\[ t_{1/2} \cdot a^{(n-1)} = \text{constant} \]
Comparing this with the given information, we can set:
\[ n - 1 = 1 \]

### Step 4: Solve for \(n\)
From the equation \(n - 1 = 1\), we can solve for \(n\):
\[ n = 2 \]
This indicates that the reaction is of second order.

### Conclusion
The order of the reaction is **2**.

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