= Solution
The slope is $-5130$ K, so the <activation enthalpy> is
$$
\boxed{\Delta H^{\ddagger\circ}=5130R=42.65\ \mathrm{kJ\,mol^{-1}}.}
$$
For $c^\circ=1\ \mathrm{mol\,dm^{-3}}$, using $k_2$ in $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ gives the numerical prefactor $k_{\mathrm B}/h=2.08366\times10^{10}\ \mathrm{K^{-1}\,s^{-1}}$. The intercept therefore yields
$$
\boxed{\Delta S^{\ddagger\circ}
=R\left[3.26-\ln(2.08366\times10^{10})\right]
=-170.45\ \mathrm{J\,K^{-1}\,mol^{-1}}.}
$$
The large logarithm must be subtracted; identifying $R\times3.26$ with the <activation entropy> would omit the transition-state attempt-frequency factor. Changing rate units without changing the standard-concentration prefactor consistently would also give a spurious entropy.
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