The half - life for a given reaction was halved as the initial concentration of a reactant was doubled . The order for this component is

Options

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

Answer

To determine the order of the reaction based on the information given about the half-life and the initial concentration of the reactant, we can follow these steps:

### Step 1: Understand the relationship between half-life and concentration
The half-life (T_half) of a reaction is related to the initial concentration ([A]) and the order of the reaction (n) by the formula:
\[ T_{1/2} \propto [A]^{1-n} \]
This means that the half-life is directly proportional to the concentration raised to the power of (1 - n).

### Step 2: Set up the initial condition
Let the initial concentration of the reactant be [A] = a. Therefore, the half-life at this concentration can be expressed as:
\[ T_{1/2} = k \cdot a^{1-n} \]
where k is a constant.

### Step 3: Analyze the change in concentration
When the initial concentration is doubled, [A] becomes 2a. The new half-life can be expressed as:
\[ T'_{1/2} = k \cdot (2a)^{1-n} \]

### Step 4: Relate the two half-lives
According to the problem, the new half-life is half of the original half-life:
\[ T'_{1/2} = \frac{1}{2} T_{1/2} \]

### Step 5: Substitute the expressions for half-lives
Substituting the expressions from Steps 2 and 3 into this equation gives:
\[ k \cdot (2a)^{1-n} = \frac{1}{2} (k \cdot a^{1-n}) \]

### Step 6: Simplify the equation
We can cancel k from both sides and simplify:
\[ (2a)^{1-n} = \frac{1}{2} a^{1-n} \]

### Step 7: Expand the left side
Expanding the left side gives:
\[ 2^{1-n} \cdot a^{1-n} = \frac{1}{2} a^{1-n} \]

### Step 8: Cancel a^(1-n)
Assuming a ≠ 0, we can cancel \( a^{1-n} \) from both sides:
\[ 2^{1-n} = \frac{1}{2} \]

### Step 9: Solve for n
Now, we can rewrite \( \frac{1}{2} \) as \( 2^{-1} \):
\[ 2^{1-n} = 2^{-1} \]

### Step 10: Set the exponents equal
Since the bases are the same, we can set the exponents equal to each other:
\[ 1 - n = -1 \]

### Step 11: Solve for n
Rearranging gives:
\[ n = 2 \]

### Conclusion
The order of the reaction with respect to the reactant is **2**.

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