Question

Enzyme activation energy \((E_{a})\) for thermal decomposition of glucose in a first-order reaction is calculated by

• A. The x-axis intercept of the Arrhenius plot
• B. The y-axis intercept of the Arrhenius plot
• C. Slope of the Arrhenius plot
• D. Rate constant at room temperature

Hint:

Arrhenius plot is a valuable tool to determine kinetic parameters such as activation energy and frequency factor from temperature-dependent rate constant data.

Answer 1

The Correct Option is C

Step 1: Start from the Arrhenius equation.
\[ k = A e^{-E_a/RT} \]
where k is the rate constant, A is the pre-exponential factor, \( E_a \) is the activation energy, R is the gas constant, and T is the absolute temperature.

Step 2: Convert it into a straight-line form.
Taking the natural log of both sides:
\[ \ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T} \]
This matches the form of a straight line, \( y = mx + c \), if we plot \( y = \ln k \) against \( x = \frac{1}{T} \).

Step 3: Identify what the slope represents.
Comparing the two equations term by term shows the slope of this line is \( m = -\frac{E_a}{R} \), while the y-intercept is \( \ln A \). Since R is a known constant (\( 8.314 \, \text{J/mol·K} \)), measuring the slope from experimental \( \ln k \) versus \( 1/T \) data and multiplying by \( -R \) gives \( E_a \) directly:
\[ E_a = -\text{slope} \times R \]

Step 4: Final Answer.
The activation energy is obtained from the slope of the Arrhenius plot.
\[ \boxed{E_a = -\text{slope} \times R} \]