The half-life periofs of a reaction at initial concentration `0.1 mol L^(-1)` and `0.5 mol L^(-1)` are `200s` and `40 s`, respectively. The order of the reaction is

Options

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

Explanation

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

### Step 1: Write the relationship between half-life and concentration
The half-life (\(t_{1/2}\)) of a reaction is inversely proportional to the concentration raised to the power of \(n - 1\), where \(n\) is the order of the reaction. This can be expressed as:
\[ t_{1/2} \propto \frac{1}{[A]^{n-1}} \]

### Step 2: Set up the equations for the two cases
For the first case with initial concentration \(A_1 = 0.1 \, \text{mol L}^{-1}\) and half-life \(t_{1/2} = 200 \, \text{s}\):
\[ t_{1/21} = k \cdot \frac{1}{(0.1)^{n-1}} \quad \text{(Equation 1)} \]

For the second case with initial concentration \(A_2 = 0.5 \, \text{mol L}^{-1}\) and half-life \(t_{1/2} = 40 \, \text{s}\):
\[ t_{1/22} = k \cdot \frac{1}{(0.5)^{n-1}} \quad \text{(Equation 2)} \]

### Step 3: Divide the two equations
To eliminate \(k\), divide Equation 2 by Equation 1:
\[ \frac{t_{1/22}}{t_{1/21}} = \frac{(0.1)^{n-1}}{(0.5)^{n-1}} \]

Substituting the values of \(t_{1/21}\) and \(t_{1/22}\):
\[ \frac{40}{200} = \frac{(0.1)^{n-1}}{(0.5)^{n-1}} \]

### Step 4: Simplify the equation
This simplifies to:
\[ \frac{1}{5} = \frac{(0.1)^{n-1}}{(0.5)^{n-1}} \]

### Step 5: Rewrite the right side
Rewrite the right side:
\[ \frac{1}{5} = \left(\frac{0.1}{0.5}\right)^{n-1} \]
\[ \frac{0.1}{0.5} = \frac{1}{5} \]
Thus, we have:
\[ \frac{1}{5} = \left(\frac{1}{5}\right)^{n-1} \]

### Step 6: Equate the exponents
Since the bases are the same, we can equate the exponents:
\[ 1 = n - 1 \]

### Step 7: Solve for \(n\)
Solving for \(n\):
\[ n = 2 \]

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

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