= Solution
Using the numerical rate constants in $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$, the logarithms are
$$
\begin{array}{c|rrrr}
p/\mathrm{atm}&100&300&700&1000\\\hline
\ln k_2&-7.58306&-7.37936&-6.96963&-6.66090
\end{array}
$$
A straight-line least-squares fit is
$$
\boxed{\ln k_2=-7.68621+(1.02471\times10^{-3})(p/\mathrm{atm}).}
$$
Every logarithmic residual has magnitude below $7.2\times10^{-4}$, so the rates follow the fitted exponential to within $0.08\%$; $R^2\simeq0.999997$. Thus a constant <activation volume> describes these data well.
\Image[/past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2024/paper-2-pressure-kinetics.png]
{title=Original logarithmic pressure-rate fit and residuals for the activation-volume calculation}
Convert the slope to pascals, using $1\ \mathrm{atm}=101325$ Pa and $T=293.15$ K:
$$
\begin{aligned}
\Delta V^{\ddagger\circ}
&=-R(293.15)\frac{1.02471\times10^{-3}}{101325}\ \mathrm{m^3\,mol^{-1}}\\
&=\boxed{-24.65\ \mathrm{cm^3\,mol^{-1}}}.
\end{aligned}
$$
Multiplying all rates by a common unit-conversion factor changes the intercept but leaves this slope and volume unchanged.
Back to article page
