The first order reaction `2N_(2)O(g)rarr2N_(2)(g)+O_(2)(g)` has a rate constant of `1.3 xx 10^(-11)s^(-1)` at `270^(@)C` and `4.5 xx 10^(-10)s^(-1)` at `350^(@)C`. What is the activation energy for this reaction ?

Options

A. 155 kJ
B. 230 kJ
C. 68 kJ
D. 124.6 kJ

Answer

To find the activation energy (Ea) for the given first-order reaction, we can use the Arrhenius equation in the form of the logarithmic rate constant relationship:

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

### Step 1: Identify the given values
- Rate constant at \( T_1 = 270^\circ C \):
\[ k_1 = 1.3 \times 10^{-11} \, s^{-1} \]
- Rate constant at \( T_2 = 350^\circ C \):
\[ k_2 = 4.5 \times 10^{-10} \, s^{-1} \]
- Convert temperatures from Celsius to Kelvin:
\[ T_1 = 270 + 273 = 543 \, K \]
\[ T_2 = 350 + 273 = 623 \, K \]
- Universal gas constant \( R = 8.314 \, J \, mol^{-1} \, K^{-1} \)

### Step 2: Calculate the ratio of rate constants
\[ \frac{k_2}{k_1} = \frac{4.5 \times 10^{-10}}{1.3 \times 10^{-11}} = 34.6154 \]

### Step 3: Take the logarithm of the ratio
\[ \log \frac{k_2}{k_1} = \log(34.6154) \approx 1.539 \]

### Step 4: Calculate the temperature difference term
\[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{543} - \frac{1}{623} \]
Calculating each term:
\[ \frac{1}{543} \approx 0.001838 \, K^{-1} \]
\[ \frac{1}{623} \approx 0.001606 \, K^{-1} \]
Thus,
\[ \frac{1}{T_1} - \frac{1}{T_2} \approx 0.001838 - 0.001606 = 0.000232 \, K^{-1} \]

### Step 5: Substitute values into the equation
Now substituting into the equation:
\[ 1.539 = \frac{E_a}{2.303 \times 8.314} \times 0.000232 \]

### Step 6: Solve for \( E_a \)
First, calculate \( 2.303 \times 8.314 \):
\[ 2.303 \times 8.314 \approx 19.091 \]
Now rearranging the equation to solve for \( E_a \):
\[ E_a = 1.539 \times 19.091 \times \frac{1}{0.000232} \]
Calculating:
\[ E_a \approx 1.539 \times 19.091 \times 4310.344 \approx 124.6 \, kJ/mol \]

### Final Answer
The activation energy \( E_a \) for the reaction is approximately:
\[ \boxed{124.6 \, kJ/mol} \]