The rate constant for a reaction is `1.5 xx 10^(-7)` at `50^(@)`C and `4.5 xx 10^(7)s^(-1)` at `100^(@)`C . What is the value of activation energy?

Options

A. `2.2 xx 10^(3) J mol^(-1)`
B. `2300 J mol^(-1)`
C. `2.2 xx 10^(4) J mol^(-1)`
D. `220 J mol^(-1)`

Explanation

To find the activation energy (Ea) for the given reaction using the Arrhenius equation, we can follow these steps:

### Step 1: Write down the Arrhenius equation
The Arrhenius equation relates the rate constants (K) at two different temperatures (T) to the activation energy (Ea):
\[ \ln \left( \frac{K_2}{K_1} \right) = -\frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
Where:
- \( K_1 = 1.5 \times 10^{-7} \, \text{s}^{-1} \) at \( T_1 = 50^\circ C \)
- \( K_2 = 4.5 \times 10^{7} \, \text{s}^{-1} \) at \( T_2 = 100^\circ C \)
- \( R = 8.314 \, \text{J/mol K} \)

### Step 2: Convert temperatures from Celsius to Kelvin
Convert the temperatures to Kelvin:
\[ T_1 = 50 + 273 = 323 \, \text{K} \]
\[ T_2 = 100 + 273 = 373 \, \text{K} \]

### Step 3: Calculate the ratio of rate constants
Calculate the ratio of the rate constants:
\[ \frac{K_2}{K_1} = \frac{4.5 \times 10^{7}}{1.5 \times 10^{-7}} = 3 \times 10^{14} \]

### Step 4: Take the natural logarithm of the ratio
Now, take the natural logarithm of the ratio:
\[ \ln(3 \times 10^{14}) = \ln(3) + \ln(10^{14}) = \ln(3) + 14 \ln(10) \]
Using \( \ln(10) \approx 2.303 \):
\[ \ln(3) \approx 1.0986 \Rightarrow \ln(3 \times 10^{14}) \approx 1.0986 + 14 \times 2.303 \approx 1.0986 + 32.242 = 33.3406 \]

### Step 5: Substitute values into the Arrhenius equation
Now substitute the values into the Arrhenius equation:
\[ 33.3406 = -\frac{E_a}{8.314} \left( \frac{1}{373} - \frac{1}{323} \right) \]

### Step 6: Calculate the difference in the reciprocals of temperatures
Calculate the difference:
\[ \frac{1}{373} - \frac{1}{323} = \frac{323 - 373}{373 \times 323} = \frac{-50}{120,139} \approx -0.000416 \]

### Step 7: Rearranging to find activation energy
Rearranging the equation to solve for \( E_a \):
\[ E_a = -33.3406 \times 8.314 \div -0.000416 \]
Calculating this gives:
\[ E_a \approx \frac{277.56}{0.000416} \approx 666,000 \, \text{J/mol} \approx 66.6 \, \text{kJ/mol} \]

### Final Answer
The activation energy \( E_a \) is approximately \( 66.6 \, \text{kJ/mol} \).

---