In the Arrhenius plot of ln k vs `(1)/(T)` , a linear plot is obtained with a slope of `-2 xx 10^(4)` K . The energy of activation of the reaction (in kJ `"mole"^(-1)`) is (R value is 8.3 J`K^(-1) mol^(-1)`)

Options

A. 83
B. 166
C. 249
D. 332

Explanation

To solve the problem, we need to use the Arrhenius equation and the information provided in the question. Here’s a step-by-step solution:

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

Taking the natural logarithm of both sides, we can rewrite the equation as:
\[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \]
This equation is in the form \( y = mx + b \), where:
- \( y = \ln k \)
- \( x = \frac{1}{T} \)
- \( m = -\frac{E_a}{R} \) (the slope)
- \( b = \ln A \)

### Step 2: Identify the Slope
From the problem, we know that the slope of the plot \( \ln k \) vs \( \frac{1}{T} \) is given as:
\[ \text{slope} = -2 \times 10^4 \, \text{K} \]
This slope can be equated to:
\[ -\frac{E_a}{R} \]

### Step 3: Rearrange to Find Activation Energy
We can rearrange the equation to solve for the activation energy \( E_a \):
\[ E_a = -\text{slope} \times R \]

### Step 4: Substitute the Values
We know:
- Slope = \(-2 \times 10^4 \, \text{K}\)
- \( R = 8.31 \, \text{J K}^{-1} \text{mol}^{-1} \)

Now, substituting these values into the equation:
\[ E_a = -(-2 \times 10^4) \times 8.31 \]
\[ E_a = 2 \times 10^4 \times 8.31 \]
\[ E_a = 166200 \, \text{J mol}^{-1} \]

### Step 5: Convert to kJ/mol
To convert from J/mol to kJ/mol, divide by 1000:
\[ E_a = \frac{166200}{1000} = 166.2 \, \text{kJ mol}^{-1} \]

### Final Answer
Thus, the energy of activation \( E_a \) of the reaction is approximately:
\[ \boxed{166.2 \, \text{kJ mol}^{-1}} \]

---