The Arrhenius equation is given by:
\[
k = A e^{-\frac{E_a}{RT}}
\]
where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant, and
- \( T \) is the temperature in Kelvin.
Taking the natural logarithm of both sides, we get:
\[
\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}
\]
This equation shows a linear relationship between \(\ln k\) and \( \frac{1}{T} \). Therefore, when the graph of \(\ln k\) is plotted against \( \frac{1}{T} \), we expect a straight line with a negative slope proportional to the activation energy \( E_a \).
From the given options, the correct graph corresponding to the Arrhenius equation is the one showing a straight-line decrease with increasing \( \frac{1}{T} \), which corresponds to Option (D)