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 slope of the curve in the reaction is:

Options

A. K
B. `-K`
C. 2K
D. `-2K`

Explanation

To solve the question step by step, we will analyze the information provided and apply the concepts of chemical kinetics.

### 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 a first-order reaction. For a first-order reaction, the rate can be expressed as:
\[ \text{Rate} = k [C_2H_4] \]
where \( k \) is the rate constant and \([C_2H_4]\) is the concentration of ethene.

**Hint:** The rate of a reaction can indicate its order. If it is proportional to the concentration, it is likely first-order.

### Step 2: Identify the Rate Constant
The rate constant \( k \) is given as \( 2.5 \times 10^{-15} \, s^{-1} \). This unit confirms that the reaction is first-order, as the unit of \( k \) for first-order reactions is \( s^{-1} \).

**Hint:** The unit of the rate constant helps determine the order of the reaction.

### Step 3: Use the First-Order Reaction Equation
For a first-order reaction, the integrated rate law is given by:
\[ \ln \left( \frac{[R_0]}{[R]} \right) = k t \]
where:
- \([R_0]\) is the initial concentration,
- \([R]\) is the concentration at time \( t \),
- \( t \) is the time elapsed.

### Step 4: Determine Concentration Change
According to the problem, the concentration of the reactant reduces to one-third of its initial concentration in 5 minutes (300 seconds). Therefore:
\[ [R] = \frac{[R_0]}{3} \]

### Step 5: Substitute Values into the Integrated Rate Law
Substituting the values into the integrated rate law:
\[ \ln \left( \frac{[R_0]}{\frac{[R_0]}{3}} \right) = k \cdot 300 \]
This simplifies to:
\[ \ln(3) = k \cdot 300 \]

### Step 6: Solve for \( k \)
Now we can solve for \( k \):
\[ k = \frac{\ln(3)}{300} \]

### Step 7: Calculate the Slope
The slope of the plot of \(\ln[R]\) versus time \(t\) for a first-order reaction is equal to \(-k\). Therefore, the slope of the curve is:
\[ \text{Slope} = -k = -\left( \frac{\ln(3)}{300} \right) \]

### Conclusion
The slope of the curve in the reaction is:
\[ \text{Slope} = -\frac{\ln(3)}{300} \]

### Final Answer
The slope of the curve in the reaction is \(-k\), which is equal to \(-\frac{\ln(3)}{300}\).

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