The rate constant of a reaction is 6 × `10^-3 s^-1` at 50° and 9 × `10^-3 s^-1` at 100° C. Calculate the energy of activation of the reaction.

Options

A. 6.123 kJ `mol^-1`
B. 8.124 kJ `mol^-1`
C. 12.357 kJ `mol^-1`
D. 18.256 kJ `mol^-1`

Answer

To calculate the energy of activation (Ea) for the reaction given the rate constants at two different temperatures, 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:
- \( k_1 = 6 \times 10^{-3} \, \text{s}^{-1} \) at \( T_1 = 50^\circ C \)
- \( k_2 = 9 \times 10^{-3} \, \text{s}^{-1} \) at \( T_2 = 100^\circ C \)
- \( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \)

### Step 1: Convert temperatures to Kelvin
First, we need to convert the temperatures from Celsius to Kelvin:
\[ T_1 = 50 + 273 = 323 \, \text{K} \]
\[ T_2 = 100 + 273 = 373 \, \text{K} \]

### Step 2: Calculate the ratio of rate constants
Next, we calculate the ratio of the rate constants:
\[ \frac{k_2}{k_1} = \frac{9 \times 10^{-3}}{6 \times 10^{-3}} = \frac{9}{6} = 1.5 \]

### Step 3: Calculate the logarithm of the ratio
Now, we take the logarithm (base 10) of the ratio:
\[ \log \left( \frac{k_2}{k_1} \right) = \log(1.5) \approx 0.1761 \]

### Step 4: Substitute values into the Arrhenius equation
Now we can substitute the values into the Arrhenius equation:
\[ 0.1761 = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{323} - \frac{1}{373} \right) \]

### Step 5: Calculate the difference in the reciprocal of temperatures
Calculate \( \frac{1}{T_1} - \frac{1}{T_2} \):
\[ \frac{1}{323} - \frac{1}{373} = \frac{373 - 323}{323 \times 373} = \frac{50}{120,919} \approx 0.000413 \]

### Step 6: Rearranging to find Ea
Now, rearranging the equation to solve for \( E_a \):
\[ E_a = 0.1761 \times 2.303 \times 8.314 \times \frac{1}{0.000413} \]

### Step 7: Calculate Ea
Now we calculate \( E_a \):
\[ E_a \approx 0.1761 \times 2.303 \times 8.314 \times 2421.5 \approx 8.124 \, \text{kJ/mol} \]

### Final Result
Thus, the energy of activation \( E_a \) is approximately:
\[ E_a \approx 8.124 \, \text{kJ/mol} \]