 Solution

The slope is $-5130$ K, so the activation enthalpy isFor $c^{\circ }=1\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$, using $k_2$ in $\mathrm{d}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$ gives the numerical prefactor $k_{\mathrm{B}}/h=2.08366\times 10^{10}{\mathrm{K}}^{-1}{\mathrm{s}}^{-1}$. The intercept therefore yieldsThe large logarithm must be subtracted; identifying $R\times 3.26$ 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.