If doubting the initial concentrations of a reactant doubles `t_(1/2)` of the reaction , the order of the reaction is

Options

A. 3
B. 2
C. 1
D. 0

Answer

To determine the order of the reaction given that doubling the initial concentration of a reactant doubles the half-life (\(t_{1/2}\)) of the reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The half-life of a reaction (\(t_{1/2}\)) is related to the initial concentration (\(a\)) and the order of the reaction (\(n\)). The general relationship can be expressed as:
\[ t_{1/2} \propto a^{(1-n)} \]

2. **Setting Up the Initial Conditions**:
Let the initial concentration be \(a\). According to the problem, when the concentration is doubled, it becomes \(2a\).

3. **Expressing the Half-Life for Both Concentrations**:
For the initial concentration \(a\):
\[ t_{1/2} = k \cdot a^{(1-n)} \]
For the doubled concentration \(2a\):
\[ t'_{1/2} = k \cdot (2a)^{(1-n)} \]

4. **Substituting the New Concentration**:
Expanding the expression for \(t'_{1/2}\):
\[ t'_{1/2} = k \cdot (2^{(1-n)} \cdot a^{(1-n)}) = 2^{(1-n)} \cdot k \cdot a^{(1-n)} \]
Since \(t_{1/2} = k \cdot a^{(1-n)}\), we can rewrite \(t'_{1/2}\) as:
\[ t'_{1/2} = 2^{(1-n)} \cdot t_{1/2} \]

5. **Using the Given Condition**:
According to the problem, doubling the initial concentration doubles the half-life:
\[ t'_{1/2} = 2 \cdot t_{1/2} \]

6. **Setting Up the Equation**:
We can now equate the two expressions for \(t'_{1/2}\):
\[ 2^{(1-n)} \cdot t_{1/2} = 2 \cdot t_{1/2} \]

7. **Cancelling \(t_{1/2}\)**:
Assuming \(t_{1/2} \neq 0\), we can cancel it out:
\[ 2^{(1-n)} = 2 \]

8. **Solving for \(n\)**:
Since the bases are the same, we can equate the exponents:
\[ 1 - n = 1 \]
Rearranging gives:
\[ n = 0 \]

### Conclusion:
The order of the reaction is \(n = 0\).