Step 1: Write the Arrhenius equation.
The Arrhenius equation is
\[
k = A e^{-E_a/RT}
\]
where \(k\) is rate constant, \(A\) is Arrhenius factor, \(E_a\) is activation energy, \(R\) is gas constant, and \(T\) is absolute temperature.
Step 2: Convert into logarithmic form.
Taking common logarithm on both sides, we get
\[
\log k = \log A - \frac{E_a}{2.303RT}
\]
This is in the form of a straight line equation
\[
y = c + mx
\]
where
\[
y = \log k
\qquad \text{and} \qquad
x = \frac{1}{T}
\]
Step 3: Identify the slope.
Comparing with the straight line form, the slope \(m\) is
\[
m = \frac{-E_a}{2.303R}
\]
Step 4: Conclusion.
Hence, in the graph of \( \log k \) versus \( \dfrac{1}{T} \), the slope is \( \dfrac{-E_a}{2.303R} \).
Final Answer:\( \dfrac{-E_a}{2.303R} \).