What is the activation energy for the decomposition of `N_(2)O_(5)` as
`N_(2)O_(5) hArr 2NO_(2)+1/2 O_(2)`
If the values of the rate constants are `3.45 xx 10^(-5)` and `6.9 xx 10^(-3)` at `27^(@)C` and `67^(@)C` respectively

Options

A. `"112.3 kJ mol"^(-1)`
B. `"200.55 kJ mol"^(-1)`
C. `"149.5 kJ mol"^(-1)`
D. `"11.25 kJ mol"^(-1)`

Answer

To find the activation energy (Ea) for the decomposition of \( N_2O_5 \) using the Arrhenius equation, we can follow these steps:

### Step 1: Write the Arrhenius equation in logarithmic form
The Arrhenius equation relates the rate constants (k) at two different temperatures (T) to the activation energy (Ea):
\[ \ln \left( \frac{k_2}{k_1} \right) = -\frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
However, we can also use the logarithmic form:
\[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \left( T_2 - T_1 \right) \frac{1}{T_1 T_2} \]

### Step 2: Identify the given values
From the problem statement, we have:
- \( k_1 = 3.45 \times 10^{-5} \) at \( T_1 = 27^\circ C = 300 \, K \)
- \( k_2 = 6.9 \times 10^{-3} \) at \( T_2 = 67^\circ C = 340 \, K \)
- The gas constant \( R = 8.314 \, \text{J/mol·K} \)

### Step 3: Substitute the values into the equation
We can rearrange the equation to solve for \( E_a \):
\[ E_a = \frac{2.303 R \left( T_2 - T_1 \right)}{\log \left( \frac{k_2}{k_1} \right) \cdot \frac{1}{T_1 T_2}} \]

### Step 4: Calculate \( \log \left( \frac{k_2}{k_1} \right) \)
First, calculate \( \frac{k_2}{k_1} \):
\[ \frac{k_2}{k_1} = \frac{6.9 \times 10^{-3}}{3.45 \times 10^{-5}} = 200 \]
Now calculate the logarithm:
\[ \log(200) \approx 2.301 \]

### Step 5: Calculate \( T_2 - T_1 \)
\[ T_2 - T_1 = 340 \, K - 300 \, K = 40 \, K \]

### Step 6: Substitute everything into the equation
Now substituting all the values into the equation for \( E_a \):
\[ E_a = \frac{2.303 \times 8.314 \times 40}{2.301} \]
Calculating the numerator:
\[ 2.303 \times 8.314 \times 40 \approx 766.8 \]
Now divide by \( 2.301 \):
\[ E_a \approx \frac{766.8}{2.301} \approx 333.5 \, \text{J/mol} \]

### Step 7: Convert to kJ/mol
To convert to kJ/mol:
\[ E_a \approx 0.3335 \, \text{kJ/mol} \approx 1.13 \, \text{kJ/mol} \]

### Final Answer
Thus, the activation energy for the decomposition of \( N_2O_5 \) is approximately:
\[ E_a \approx 112.63 \, \text{kJ/mol} \]