Question

For the calculation of activation energy using the Arrhenius equation, a graph of \( \ln k \) versus \( \frac{1}{T} \) graph was plotted. The slope of the straight line was found to be \( -2.55 \times 10^4 \, \text{K} \). The activation energy in \( \text{J/mol} \) is:

• A. \( 2.12 \times 10^5 \)
• B. \( 4.88 \times 10^5 \)
• C. \( 2.12 \times 10^4 \)
• D. \( 0.212 \times 10^4 \)

Hint:

The activation energy \( E_a \) can be determined from the slope of the Arrhenius plot, where the slope is equal to \( -\frac{E_a}{R} \).

Answer 1

The Correct Option is A

Step 1: The Arrhenius equation.
The Arrhenius equation relates the rate constant \( k \) to temperature \( T \) and the activation energy \( E_a \): \[ k = A e^{\frac{-E_a}{RT}} \] Taking the natural logarithm of both sides, we get: \[ \ln k = \ln A - \frac{E_a}{RT} \] This is a linear equation of the form \( y = mx + b \), where the slope \( m \) is \( -\frac{E_a}{R} \). Thus, the activation energy can be found from the slope of the line.

Step 2: Using the slope to find activation energy.
The slope \( m = -\frac{E_a}{R} \), where \( R \) is the universal gas constant (\( R = 8.314 \, \text{J/mol·K} \)). The given slope is \( -2.55 \times 10^4 \, \text{K} \). We can solve for \( E_a \) as follows: \[ E_a = -mR = 2.55 \times 10^4 \times 8.314 = 2.12 \times 10^5 \, \text{J/mol} \]

Step 3: Conclusion.
Thus, the activation energy is \( 2.12 \times 10^5 \, \text{J/mol} \), which corresponds to option (1).