The racemisation of `alpha` - pinene is first order reaction. In the gas the specific reaction rate was found to be `2.2 xx 10^(-5)mm Hg^(-1)` at 457.6 K and `3.07 xx 10^(-3) mm Hg^(-1)` at 510.1 K. The energy of activation is

Options

A. `K(510.1-457.6) In(3.07 xx 10^(-3) // 2.2 xx 10^(-5))`
B. `3.048 xx 10^(-3)`
C. `In 2.2 xx 10^(-5) xx 457.7 k cal`.
D. `(510.1 xx 457.6
)( R In (307 // 2.2)//52.5 `

Explanation

To find the energy of activation (ΔE) for the racemization of α-pinene, we can use the Arrhenius equation in the form of the following equation:

\[ \ln \left( \frac{k_2}{k_1} \right) = -\frac{\Delta E}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

Where:
- \( k_1 \) and \( k_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \), respectively.
- \( R \) is the universal gas constant (8.314 J/mol·K).
- \( \Delta E \) is the activation energy.

### Step 1: Identify the values
From the problem, we have:
- \( k_1 = 2.2 \times 10^{-5} \, \text{mm Hg}^{-1} \) at \( T_1 = 457.6 \, \text{K} \)
- \( k_2 = 3.07 \times 10^{-3} \, \text{mm Hg}^{-1} \) at \( T_2 = 510.1 \, \text{K} \)

### Step 2: Calculate the natural logarithm of the ratio of the rate constants
\[ \ln \left( \frac{k_2}{k_1} \right) = \ln \left( \frac{3.07 \times 10^{-3}}{2.2 \times 10^{-5}} \right) \]

Calculating the ratio:
\[ \frac{3.07 \times 10^{-3}}{2.2 \times 10^{-5}} = 139.545 \]

Now, take the natural logarithm:
\[ \ln(139.545) \approx 4.941 \]

### Step 3: Calculate the difference in the inverse of the temperatures
\[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{457.6} - \frac{1}{510.1} \]

Calculating each term:
\[ \frac{1}{457.6} \approx 0.002185 \, \text{K}^{-1} \]
\[ \frac{1}{510.1} \approx 0.001960 \, \text{K}^{-1} \]

Now, subtract:
\[ 0.002185 - 0.001960 = 0.000225 \, \text{K}^{-1} \]

### Step 4: Substitute values into the Arrhenius equation
Now we can substitute into the equation:
\[ \ln \left( \frac{k_2}{k_1} \right) = -\frac{\Delta E}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
\[ 4.941 = -\frac{\Delta E}{8.314} \times 0.000225 \]

### Step 5: Solve for ΔE
Rearranging gives:
\[ \Delta E = -4.941 \times 8.314 \div 0.000225 \]

Calculating:
\[ \Delta E \approx -4.941 \times 8.314 \div 0.000225 \approx 177,000 \, \text{J/mol} \approx 177 \, \text{kJ/mol} \]

### Final Answer
The energy of activation \( \Delta E \) is approximately \( 177 \, \text{kJ/mol} \).

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