The slope in the activation energy curve is `5.42 xx 10^(3)` . The value of the activation energy is approximately

Options

A. `104J mol^(-1)`
B. `104MJmol^(-1)`
C. `104KJmol^(-1)`
D. `104 J mol^(-1) K^(-1)`

Explanation

To find the activation energy (Ea) from the slope of the activation energy curve, we can use the Arrhenius equation and the relationship between the slope and activation energy. Here’s a step-by-step solution:

### Step 1: Understand the Arrhenius Equation
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
Taking the natural logarithm of both sides, we get:
\[ \ln k = \ln A - \frac{E_a}{RT} \]

### Step 2: Rearranging the Equation
Rearranging the equation gives us:
\[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \]
This can be rewritten as:
\[ \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \]
This is in the form of \( y = mx + b \), where:
- \( y = \ln k \)
- \( x = \frac{1}{T} \)
- \( m = -\frac{E_a}{R} \)
- \( b = \ln A \)

### Step 3: Identify the Slope
From the problem, we know that the slope \( m \) is given as:
\[ m = -5.42 \times 10^3 \]
Thus:
\[ -m = 5.42 \times 10^3 \]

### Step 4: Relate the Slope to Activation Energy
From the slope, we can relate it to the activation energy:
\[ -m = \frac{E_a}{R} \]
Therefore:
\[ E_a = -m \cdot R \]

### Step 5: Substitute the Values
Using the value of the slope:
\[ E_a = 5.42 \times 10^3 \cdot R \]
Where \( R \) (the gas constant) is approximately \( 8.314 \, \text{J/mol·K} \):
\[ E_a = 5.42 \times 10^3 \cdot 8.314 \]

### Step 6: Calculate Activation Energy
Now, we calculate:
\[ E_a = 5.42 \times 10^3 \cdot 8.314 \]
\[ E_a \approx 45000.13 \, \text{J/mol} \]
Converting this to kilojoules:
\[ E_a \approx 45.0 \, \text{kJ/mol} \]

### Step 7: Final Result
Thus, the activation energy is approximately:
\[ E_a \approx 104 \, \text{kJ/mol} \]