= Solution
\b[An unweighted least-squares <Arrhenius plot> using all five data points gives $\boxed{E_a=20.7\,\mathrm{kJ\,mol^{-1}}}$.] Define the dimensionless ordinate $y=\ln(k_1/k_{\rm ref})$ with $k_{\rm ref}=10^6\,\mathrm{cm^3\,mol^{-1}\,s^{-1}}$, and set $x=1/T$. The model is
$$
y=b+mx,\qquad m=-E_a/R,\qquad b=\ln(A/k_{\rm ref}).
$$
For increasing temperatures, the five ordinates are $1.82455$, $2.56495$, $3.13549$, $3.52636$, and $4.15888$. The <linear regression> slope and intercept are
$$
m=\frac{\sum_i(x_i-\bar x)(y_i-\bar y)}{\sum_i(x_i-\bar x)^2}=-2492.15\,\mathrm K,\qquad b=11.4242.
$$
Consequently $E_a=-mR=20721\,\mathrm{J\,mol^{-1}}$. The corresponding fitted $A=e^{11.4242}10^6=9.15\times10^{10}\,\mathrm{cm^3\,mol^{-1}\,s^{-1}}$. A two-point calculation using just the endpoints gives approximately $21.5\,\mathrm{kJ\,mol^{-1}}$; the difference reflects the rounded, imperfectly collinear data rather than a unit conversion.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025-arrhenius-fit.png]
{title=Original <Arrhenius plot> and all-point least-squares fit of the supplied rate data}
Back to article page
