The slope of the line graph of `log k` versus `1//T` for the reaction `N_(2)O_(5) rarr2NO_(2) + 1//2O_(2)` is `-5000`.Calculate the energy of activation of the reaction (in `kJ K^(-1) mol^(-1)`).

Options

A. `95.7`
B. `9.57`
C. `957`
D. None

Explanation

To calculate the energy of activation (Ea) for the reaction \( N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2 \) using the given slope of the line graph of \( \log k \) versus \( \frac{1}{T} \), we can follow these steps:

### Step 1: Understand the Arrhenius Equation
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant, and
- \( T \) is the temperature in Kelvin.

### Step 2: Take the Logarithm of the Arrhenius Equation
Taking the logarithm of both sides, we get:
\[ \log k = \log A - \frac{E_a}{2.303RT} \]
This can be rearranged to:
\[ \log k = -\frac{E_a}{2.303R} \cdot \frac{1}{T} + \log A \]

### Step 3: Identify the Slope of the Line
From the rearranged equation, we can see that the equation is in the form of \( y = mx + c \), where:
- \( y = \log k \)
- \( x = \frac{1}{T} \)
- \( m = -\frac{E_a}{2.303R} \)

Given that the slope \( m \) is \( -5000 \), we can equate:
\[ -\frac{E_a}{2.303R} = -5000 \]

### Step 4: Solve for Activation Energy \( E_a \)
Rearranging the equation gives us:
\[ \frac{E_a}{2.303R} = 5000 \]
Thus,
\[ E_a = 5000 \times 2.303R \]

### Step 5: Substitute the Value of \( R \)
The value of the universal gas constant \( R \) is \( 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \) or \( 0.008314 \, \text{kJ K}^{-1} \text{mol}^{-1} \). Therefore:
\[ E_a = 5000 \times 2.303 \times 8.314 \times 10^{-3} \, \text{kJ K}^{-1} \text{mol}^{-1} \]

### Step 6: Calculate \( E_a \)
Calculating the value:
\[ E_a = 5000 \times 2.303 \times 8.314 \approx 95.7 \, \text{kJ K}^{-1} \text{mol}^{-1} \]

### Final Answer
The energy of activation \( E_a \) for the reaction is approximately:
\[ E_a \approx 95.7 \, \text{kJ K}^{-1} \text{mol}^{-1} \]

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