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

The activation energy \( E_a \) of a reaction is a fundamental parameter in chemical kinetics that indicates the minimum energy required for reactants to undergo a successful transformation into products. It can be calculated using the \textit{Arrhenius equation}: \[ k = A e^{-E_a / RT} \] Taking the natural logarithm of both sides: \[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \] This is the equation of a straight line: \[ y = mx + c \] Where: - \( y = \ln k \)
- \( x = \frac{1}{T} \)
- Slope \( m = -\frac{E_a}{R} \)
- \( R \) = gas constant (8.314 \, \text{J/mol·K})
Hence, the \textit{activation energy} is calculated from the \textit{slope of the Arrhenius plot} (which is a plot of \( \ln k \) versus \( \frac{1}{T} \)). The slope gives \( -\frac{E_a}{R} \), and multiplying this by \( -R \) yields the value of \( E_a \).

Answer 2

The question asks how the activation energy of a first-order reaction, such as the thermal decomposition of glucose, is obtained from an Arrhenius plot. The Arrhenius equation is:

\[ k = A e^{-E_a/RT} \]

Taking the natural log of both sides gives a straight-line form, \( \ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T} \), plotted as \( \ln k \) on the y-axis against \( \frac{1}{T} \) on the x-axis. Let's check each option against this straight-line equation.

1. The x-axis intercept of the Arrhenius plot: The x-intercept is where \( \ln k = 0 \), which corresponds to a specific temperature where the rate constant equals 1. This point does not isolate \( E_a \) on its own.
2. The y-axis intercept of the Arrhenius plot: Setting \( \frac{1}{T} = 0 \) in the line equation leaves \( \ln k = \ln A \), so the y-intercept gives the pre-exponential factor A, not the activation energy.
3. Slope of the Arrhenius plot: Comparing \( \ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T} \) with the straight-line form \( y = mx + c \) shows that the slope m equals \( -\frac{E_a}{R} \). Multiplying the measured slope by \( -R \) gives \( E_a \) directly, so the slope is exactly what is needed.
4. Rate constant at room temperature: A single rate constant value at one temperature is just one point on the plot; on its own it cannot separate A from \( E_a \) without a second data point, so it cannot be used alone to find \( E_a \).

Only the slope of the line directly encodes the activation energy through its relationship to \( -E_a/R \), which is why it is the quantity used to calculate \( E_a \) from Arrhenius plot data.

Therefore, the correct answer is Slope of the Arrhenius plot.