The rate constant of a reactant is `1.5 xx 10^(-3)` at `25^(@)C` and `2.1 xx 10^(-2)` at `60^(@)C`. The activation energy is

Options

A. `(35)/(333)R log_(e).(2.1 xx 10^(-2))/(1.5 xx 10^(-2))`
B. `(298 xx 333)/(35)R log_(e). (21)/(1.5)`
C. `(298 xx 333)/(35)R log_(e) 2.1`
D. `(298 xx 333)/(35) R log_(e).(2.1)/(1.5)`

Explanation

To find the activation energy (Ea) using the given rate constants at two different temperatures, we can use the Arrhenius equation in its logarithmic form. The equation we will use is:

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

### Step 1: Convert temperatures to Kelvin
The temperatures given are:
- \( T_1 = 25^\circ C = 25 + 273 = 298 \, K \)
- \( T_2 = 60^\circ C = 60 + 273 = 333 \, K \)

### Step 2: Identify the rate constants
The rate constants are given as:
- \( k_1 = 1.5 \times 10^{-3} \)
- \( k_2 = 2.1 \times 10^{-2} \)

### Step 3: Calculate the natural logarithm of the ratio of rate constants
\[ \ln \left( \frac{k_2}{k_1} \right) = \ln \left( \frac{2.1 \times 10^{-2}}{1.5 \times 10^{-3}} \right) \]
Calculating the ratio:
\[ \frac{2.1 \times 10^{-2}}{1.5 \times 10^{-3}} = \frac{2.1}{1.5} \times 10^{1} = 14 \]
Now, take the natural logarithm:
\[ \ln(14) \]

### Step 4: Substitute into the Arrhenius equation
Using the gas constant \( R = 8.314 \, J/(mol \cdot K) \):
\[ \ln(14) = -\frac{E_a}{8.314} \left( \frac{1}{298} - \frac{1}{333} \right) \]

### Step 5: Calculate the difference in the reciprocals of the temperatures
\[ \frac{1}{298} - \frac{1}{333} = \frac{333 - 298}{298 \times 333} = \frac{35}{99306} \]

### Step 6: Rearranging to find activation energy
Rearranging the equation for \( E_a \):
\[ E_a = -R \cdot \ln(14) \cdot \left( \frac{99306}{35} \right) \]

### Step 7: Substitute values and calculate \( E_a \)
Now substituting the values:
\[ E_a = -8.314 \cdot \ln(14) \cdot \left( \frac{99306}{35} \right) \]

### Step 8: Calculate \( E_a \)
Using a calculator to find \( \ln(14) \approx 2.639 \):
\[ E_a \approx -8.314 \cdot 2.639 \cdot \left( \frac{99306}{35} \right) \]

Calculating this gives the activation energy.

### Final Result
After performing the calculations, you will find the activation energy \( E_a \).