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. `2.2xx10^3" J mol"^(-1)`
B. `2300 "J mol"^(-1)`
C. `2.2xx10^4" J mol"^(-1)`
D. `220 " J mol"^(-1)`

Answer

To find the activation energy (Ea) for the given reaction, we can use the Arrhenius equation in its logarithmic form. The formula we will use is:

\[ \log \left( \frac{K_2}{K_1} \right) = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

Where:
- \( K_1 = 1.5 \times 10^{-7} \, \text{s}^{-1} \) 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 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 \( \frac{K_2}{K_1} \).

\[ \frac{K_2}{K_1} = \frac{4.5 \times 10^{7}}{1.5 \times 10^{-7}} = 3 \times 10^{14} \]

### Step 3: Calculate the logarithm of the ratio
Now, we take the logarithm (base 10) of the ratio we just calculated.

\[ \log \left( \frac{K_2}{K_1} \right) = \log(3 \times 10^{14}) = \log(3) + \log(10^{14}) = 0.477 + 14 = 14.477 \]

### Step 4: Calculate the difference in the inverse of temperatures
Next, we calculate the difference in the inverse of the temperatures:

\[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{323} - \frac{1}{373} \]

Calculating each term:

\[ \frac{1}{323} \approx 0.003096 \, \text{K}^{-1} \]
\[ \frac{1}{373} \approx 0.002684 \, \text{K}^{-1} \]

Now, subtracting these values:

\[ \frac{1}{T_1} - \frac{1}{T_2} \approx 0.003096 - 0.002684 = 0.000412 \, \text{K}^{-1} \]

### Step 5: Substitute values into the Arrhenius equation
Now we substitute the values into the Arrhenius equation:

\[ 14.477 = \frac{E_a}{2.303 \times 8.314} \times 0.000412 \]

### Step 6: Solve for \( E_a \)
Rearranging the equation to solve for \( E_a \):

\[ E_a = 14.477 \times 2.303 \times 8.314 \times \frac{1}{0.000412} \]

Calculating the constants:

\[ E_a \approx 14.477 \times 2.303 \times 8.314 \div 0.000412 \]

Calculating this gives:

\[ E_a \approx 2.4 \times 10^{4} \, \text{J/mol} \]

### Final Answer
The activation energy \( E_a \) is approximately \( 2.4 \times 10^{4} \, \text{J/mol} \).

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