The rate of the reaction is proportional to the concentration of the reactant. Hydrogenation of ethene results in the formation of ethane. The rate constant, k for the reaction was found to be `2.5xx10^(-15)s^(-1)`. The concentration of the reactant reduces to one-third of the initial concentration in 5 minutes.
Q. The rate constant of the reaction after 5 minutes is:

Options

A. 0.4290 `"min"^(-1)`
B. 0.1297 `"min"^(-1)`
C. 0.2197 `"min"^(-1)`
D. 0.6591 `"min"^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction Order
The problem states that the rate of the reaction is proportional to the concentration of the reactant. This indicates that the reaction is first-order with respect to the reactant.

### Step 2: Identify Given Data
- Initial rate constant, \( k = 2.5 \times 10^{-15} \, s^{-1} \)
- The concentration of the reactant reduces to one-third of the initial concentration in \( t = 5 \) minutes.

### Step 3: Write the First-Order Rate Equation
For a first-order reaction, the relationship between the concentration and time can be expressed as:
\[ k = \frac{2.303}{t} \log \left( \frac{[R_0]}{[R]} \right) \]
where:
- \( [R_0] \) is the initial concentration,
- \( [R] \) is the concentration after time \( t \).

### Step 4: Substitute Known Values
Since the concentration reduces to one-third, we can express this as:
\[ [R] = \frac{[R_0]}{3} \]
Substituting this into the equation gives:
\[ k = \frac{2.303}{5 \, \text{minutes}} \log \left( \frac{[R_0]}{\frac{[R_0]}{3}} \right) \]
This simplifies to:
\[ k = \frac{2.303}{5} \log(3) \]

### Step 5: Calculate the Logarithm
Using the logarithmic value:
\[ \log(3) \approx 0.477 \]
Now substituting this value into the equation:
\[ k = \frac{2.303}{5} \times 0.477 \]

### Step 6: Perform the Calculation
Calculating the above expression:
\[ k \approx \frac{2.303 \times 0.477}{5} \approx \frac{1.098351}{5} \approx 0.21967 \, \text{min}^{-1} \]

### Step 7: Convert Units if Necessary
Since we calculated the rate constant in minutes, we can express it as:
\[ k \approx 0.2197 \, \text{min}^{-1} \]

### Conclusion
The rate constant of the reaction after 5 minutes is approximately \( 0.2197 \, \text{min}^{-1} \).

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