The rate coefficient (k) for a particular reactions is `1.3 xx 10^(-4)M^(-1)s^(-1)` at `100^(@)C`, and `1.3 xx 10^(-3)M^(-1)s^(-1)` at `150^(@)C`. What is the energy of activation (`E_(a)`) (in kJ) for this reaction? (R=molar gas constant = `8.314 JK^(-1) mol^(-1)`)

Options

A. 16
B. 60
C. 99
D. 132

Explanation

To find the energy of activation (Ea) for the given reaction, we can use the Arrhenius equation in its logarithmic form:

\[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

Where:
- \( k_1 \) and \( k_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \) respectively.
- \( R \) is the universal gas constant, which is \( 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \).
- \( T_1 \) and \( T_2 \) are the absolute temperatures in Kelvin.

### Step 1: Convert temperatures from Celsius to Kelvin
- \( T_1 = 100^\circ C = 100 + 273 = 373 \, K \)
- \( T_2 = 150^\circ C = 150 + 273 = 423 \, K \)

### Step 2: Identify the rate constants
- \( k_1 = 1.3 \times 10^{-4} \, M^{-1} s^{-1} \)
- \( k_2 = 1.3 \times 10^{-3} \, M^{-1} s^{-1} \)

### Step 3: Substitute values into the Arrhenius equation
We can now substitute the values into the logarithmic form of the Arrhenius equation:

\[ \log \left( \frac{1.3 \times 10^{-3}}{1.3 \times 10^{-4}} \right) = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{373} - \frac{1}{423} \right) \]

### Step 4: Simplify the left side
The left side simplifies as follows:

\[ \log \left( \frac{1.3 \times 10^{-3}}{1.3 \times 10^{-4}} \right) = \log \left( 10 \right) = 1 \]

### Step 5: Calculate the right side
Now we need to calculate the right side. First, calculate \( \frac{1}{373} - \frac{1}{423} \):

\[ \frac{1}{373} \approx 0.002684 \, K^{-1} \]
\[ \frac{1}{423} \approx 0.002365 \, K^{-1} \]
\[ \frac{1}{373} - \frac{1}{423} \approx 0.002684 - 0.002365 = 0.000319 \, K^{-1} \]

### Step 6: Substitute and solve for \( E_a \)
Now substitute this value back into the equation:

\[ 1 = \frac{E_a}{2.303 \times 8.314} \times 0.000319 \]

Now, calculate \( 2.303 \times 8.314 \):

\[ 2.303 \times 8.314 \approx 19.1 \]

Now we can solve for \( E_a \):

\[ 1 = \frac{E_a \times 0.000319}{19.1} \]

Rearranging gives:

\[ E_a = \frac{19.1}{0.000319} \]

Calculating \( E_a \):

\[ E_a \approx 59700 \, J/mol = 59.7 \, kJ/mol \]

### Final Answer
Thus, the energy of activation \( E_a \) for the reaction is approximately:

\[ E_a \approx 60 \, kJ/mol \]