Consider reaction in which reactant `R` is converted into product `P`: `RtoP`. When the initial concentration of `R^(')` is `0.5M` The half life of the reaction is 20 minute. When the initial concentration is increased to 1.3 M, the half life decreases to 7.69 minute.
Q. What is the rate canstant of the reaction?

Options

A. `0.1M^(-1)min^(-1)`
B. `0.6M^(-1)min^(-1)`
C. `0.3M^(-1)min^(-1)`
D. `1.0M^(-1)min^(-1)`

Explanation

To solve the problem, we need to calculate the rate constant \( k \) for the reaction \( R \to P \) using the given half-life values at different initial concentrations. We will use the formula for the half-life of a second-order reaction:

\[ t_{1/2} = \frac{1}{k [R]_0} \]

where:
- \( t_{1/2} \) is the half-life,
- \( k \) is the rate constant,
- \( [R]_0 \) is the initial concentration of the reactant.

### Step 1: Calculate \( k \) for the first concentration

Given:
- Initial concentration \( [R]_0 = 0.5 \, M \)
- Half-life \( t_{1/2} = 20 \, \text{minutes} \)

Using the formula:

\[ 20 = \frac{1}{k \cdot 0.5} \]

Rearranging to solve for \( k \):

\[ k = \frac{1}{20 \cdot 0.5} = \frac{1}{10} = 0.1 \, \text{M}^{-1} \text{min}^{-1} \]

### Step 2: Calculate \( k \) for the second concentration

Given:
- Initial concentration \( [R]_0 = 1.3 \, M \)
- Half-life \( t_{1/2} = 7.69 \, \text{minutes} \)

Using the same formula:

\[ 7.69 = \frac{1}{k \cdot 1.3} \]

Rearranging to solve for \( k \):

\[ k = \frac{1}{7.69 \cdot 1.3} \]

Calculating \( k \):

\[ k = \frac{1}{10.0} \approx 0.1 \, \text{M}^{-1} \text{min}^{-1} \]

### Conclusion

The rate constant \( k \) for the reaction is approximately \( 0.1 \, \text{M}^{-1} \text{min}^{-1} \).