Question

The temperature dependence of reaction rates is generally given by the Arrhenius equation. A plot of \( \ln k_r \) against \( 1/T \) is a straight line from which the pre-exponential factor ‘\( A \)’ and the activation energy ‘\( E_a \)’ can be determined.
The CORRECT option regarding this plot is:

• A. \( {Slope: } -E_a/R; { Intercept on the y-axis: } \ln A \)
• B. \( {Slope: } +E_a/2.303R; { Intercept on the y-axis: } A \)
• C. \( {Slope: } +E_a/R; { Intercept on the y-axis: } A \)
• D. \( {Slope: } -E_a/2.303R; { Intercept on the y-axis: } \ln A \)

Hint:

Always express the Arrhenius equation in logarithmic form \( \ln k = \ln A - \frac{E_a}{RT} \) before identifying slope and intercept from a graph.

Answer 1

The Correct Option is A

Step 1: Start with the Arrhenius equation:
\[ k = A e^{-E_a/(RT)} \] Step 2: Take the natural logarithm of both sides:
\[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \] Step 3: Compare this with the straight-line form \( y = mx + c \):
\[ {where } y = \ln k, \quad x = \frac{1}{T}, \quad m = -\frac{E_a}{R}, \quad c = \ln A \] Conclusion: The plot of \( \ln k \) vs. \( 1/T \) gives:
Slope: \( -E_a/R \)
Intercept on y-axis: \( \ln A \)