= Solution
Resolve the activation free energy into enthalpy and entropy in the <Eyring equation>:
$$
k_2=\frac{k_{\mathrm B}T}{hc^\circ}
\exp\left(\frac{\Delta S^{\ddagger\circ}}R\right)
\exp\left(-\frac{\Delta H^{\ddagger\circ}}{RT}\right).
$$
Taking logarithms of numerical values in fixed, consistent units gives
$$
\boxed{\ln\frac{k_2}{T}
=-\frac{\Delta H^{\ddagger\circ}}R\frac1T
+\ln\frac{k_{\mathrm B}}{hc^\circ}
+\frac{\Delta S^{\ddagger\circ}}R.}
$$
Thus for slope $s$ and intercept $b$,
$$
\boxed{\Delta H^{\ddagger\circ}=-Rs,\qquad
\Delta S^{\ddagger\circ}=R\left[b-\ln\frac{k_{\mathrm B}}{hc^\circ}\right].}
$$
The logarithms imply reference units, as made explicit in the supplied numerical plot. Approximate linearity requires the activation parameters and transmission coefficient to vary little over the temperature interval; a significant activation heat-capacity difference produces curvature.
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