= Solution
At fixed temperature, differentiating the <Eyring equation> with a pressure-independent prefactor and fixed concentration convention gives
$$
\left(\frac{\partial\ln k_2}{\partial p}\right)_T
=-\frac1{RT}\left(\frac{\partial\Delta G^{\ddagger\circ}}{\partial p}\right)_T.
$$
Since $\mathrm dG=V\,\mathrm dp-S\,\mathrm dT$, the derivative is the difference between the activated complex's <partial molar volume> and the sum for the reactants:
$$
\boxed{\Delta V^{\ddagger\circ}
=-RT\left(\frac{\partial\ln k_2}{\partial p}\right)_T.}
$$
If <activation volume> is approximately constant over the pressure range, integration gives
$$
\boxed{\ln k_2(p)=\ln k_2(p_0)-\frac{\Delta V^{\ddagger\circ}}{RT}(p-p_0).}
$$
Thus plot $\ln k_2$ against pressure: its slope is $-\Delta V^{\ddagger\circ}/RT$. Use pressure in pascals for a volume in $\mathrm{m^3\,mol^{-1}}$. The model neglects pressure dependence of the transmission coefficient, substantial changes in mechanism, and strong curvature from a pressure-dependent activation volume. Fixed concentration-unit conventions and compression corrections matter for non-first-order rates.<Volume of activation in the IUPAC Gold Book>[Definition and concentration-unit qualification]{ref}
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