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 half-life for the reaction is:

Options

A. `2.772xx10^(-24)s`
B. `2.772xx10^(-12)s`
C. `1.386xx10^(-24)s`
D. `1.386xx10^(-12)s`

Explanation

To solve the problem, we need to determine the half-life of a first-order reaction given the rate constant and the information about the concentration of the reactant. Here’s a step-by-step solution:

### 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.

### Step 2: Use the First-Order Rate Constant
The rate constant \( k \) is given as \( 2.5 \times 10^{-15} \, \text{s}^{-1} \).

### Step 3: Recall the Half-Life Formula for First-Order Reactions
The half-life (\( t_{1/2} \)) for a first-order reaction is calculated using the formula:
\[ t_{1/2} = \frac{0.693}{k} \]

### Step 4: Substitute the Value of \( k \)
Substituting the value of \( k \) into the half-life formula:
\[ t_{1/2} = \frac{0.693}{2.5 \times 10^{-15}} \]

### Step 5: Perform the Calculation
Now, we will calculate the half-life:
\[ t_{1/2} = \frac{0.693}{2.5 \times 10^{-15}} = 2.772 \times 10^{14} \, \text{s} \]

### Step 6: Final Answer
Thus, the half-life for the reaction is:
\[ t_{1/2} \approx 2.772 \times 10^{14} \, \text{s} \]