The half-life for a reaction at initial concentration of `0.5` and `1.0 "mole litre"^(-1)` are `200 sec` and `100 sec` respectively. The order of the reaction is

Options

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

Explanation

To determine the order of the reaction based on the given half-lives at different initial concentrations, we can follow these steps:

### Step 1: Write down the relationship for half-life
For a reaction of order \( n \), the half-life \( T_{1/2} \) is given by the formula:
\[ T_{1/2} \propto \frac{1}{[A]^{n-1}} \]
where \( [A] \) is the initial concentration of the reactant.

### Step 2: Set up the equations for the two cases
For the two given cases:
1. When \( [A] = 0.5 \, \text{mol L}^{-1} \), \( T_{1/2} = 200 \, \text{sec} \)
2. When \( [A] = 1.0 \, \text{mol L}^{-1} \), \( T_{1/2} = 100 \, \text{sec} \)

We can express these relationships as:
\[ T_{1/2,1} = k \cdot \frac{1}{(0.5)^{n-1}} \quad \text{(Equation 1)} \]
\[ T_{1/2,2} = k \cdot \frac{1}{(1)^{n-1}} \quad \text{(Equation 2)} \]
where \( k \) is a proportionality constant.

### Step 3: Substitute the values into the equations
Substituting the half-life values into the equations gives us:
\[ 200 = k \cdot \frac{1}{(0.5)^{n-1}} \quad \text{(1)} \]
\[ 100 = k \cdot \frac{1}{(1)^{n-1}} \quad \text{(2)} \]

### Step 4: Divide the two equations
Dividing Equation 1 by Equation 2:
\[ \frac{200}{100} = \frac{k \cdot \frac{1}{(0.5)^{n-1}}}{k \cdot \frac{1}{(1)^{n-1}}} \]
This simplifies to:
\[ 2 = \frac{1}{(0.5)^{n-1}} \quad \text{(since } (1)^{n-1} = 1\text{)} \]

### Step 5: Solve for \( n \)
Rearranging the equation gives:
\[ 2 = (0.5)^{-(n-1)} \]
This can be rewritten as:
\[ 2 = 2^{(n-1)} \]
Taking logarithm base 2 on both sides:
\[ 1 = n - 1 \]
Thus, solving for \( n \):
\[ n = 2 \]

### Conclusion
The order of the reaction is \( n = 2 \).

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