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, \( k = A e^{\frac{-E_a}{RT}} \), links the rate constant \( k \) to temperature \( T \) and activation energy \( E_a \). Rearranging by taking the natural logarithm of both sides yields \( \ln k = \ln A - \frac{E_a}{RT} \). This linear form \( y = mx + b \) indicates that the slope \( m \) equals \( -\frac{E_a}{R} \), allowing determination of activation energy from the slope.

Step 2: Calculating activation energy from the slope.
Given the slope \( m = -2.55 \times 10^4 \, \text{K} \), and the universal gas constant \( R = 8.314 \, \text{J/mol·K} \), activation energy \( E_a \) can be calculated. Using the relation \( E_a = -mR \), we get \( E_a = 2.55 \times 10^4 \times 8.314 = 2.12 \times 10^5 \, \text{J/mol} \).

Step 3: Conclusion.
The calculated activation energy is \( 2.12 \times 10^5 \, \text{J/mol} \), corresponding to option (1).