For a chemical reaction at `27^(@)C`, the activation energy is 600R. The ratio of the rate constants at `327^(@)C` to that of at `27^(@)C` will be

Options

A. 2
B. 40
C. e
D. `e^(2)`

Explanation

To solve the problem of finding the ratio of the rate constants \( K_2 \) at \( 327^\circ C \) to \( K_1 \) at \( 27^\circ C \), we will use the Arrhenius equation. The Arrhenius equation relates the rate constant \( K \) to the temperature \( T \) and the activation energy \( E_a \):

\[ K = A e^{-\frac{E_a}{RT}} \]

Where:
- \( K \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin.

### Step 1: Convert temperatures to Kelvin
First, we need to convert the given temperatures from Celsius to Kelvin.

- For \( T_1 = 27^\circ C \):
\[ T_1 = 27 + 273 = 300 \, K \]

- For \( T_2 = 327^\circ C \):
\[ T_2 = 327 + 273 = 600 \, K \]

### Step 2: Write the Arrhenius equation for both temperatures
Using the Arrhenius equation, we can express the rate constants at both temperatures:

\[ K_1 = A e^{-\frac{E_a}{RT_1}} \quad \text{and} \quad K_2 = A e^{-\frac{E_a}{RT_2}} \]

### Step 3: Find the ratio of rate constants
To find the ratio \( \frac{K_2}{K_1} \):

\[ \frac{K_2}{K_1} = \frac{A e^{-\frac{E_a}{RT_2}}}{A e^{-\frac{E_a}{RT_1}}} \]

The \( A \) cancels out:

\[ \frac{K_2}{K_1} = e^{-\frac{E_a}{RT_2} + \frac{E_a}{RT_1}} = e^{E_a \left( \frac{1}{RT_1} - \frac{1}{RT_2} \right)} \]

### Step 4: Substitute the values
Given \( E_a = 600R \):

\[ \frac{K_2}{K_1} = e^{600R \left( \frac{1}{R \cdot 300} - \frac{1}{R \cdot 600} \right)} \]

This simplifies to:

\[ \frac{K_2}{K_1} = e^{600 \left( \frac{1}{300} - \frac{1}{600} \right)} \]

Calculating the expression inside the exponent:

\[ \frac{1}{300} - \frac{1}{600} = \frac{2 - 1}{600} = \frac{1}{600} \]

Thus, we have:

\[ \frac{K_2}{K_1} = e^{600 \cdot \frac{1}{600}} = e^1 = e \]

### Final Answer
The ratio of the rate constants at \( 327^\circ C \) to that at \( 27^\circ C \) is:

\[ \frac{K_2}{K_1} = e \]