A student measures the rate constant of a reaction at five temperatures: 0.00307 s⁻¹ at 293 K…

Step 1. The values. $\dfrac{1}{T}$ is $3.413 \times 10^{-3}$, $3.300 \times 10^{-3}$, $3.195 \times 10^{-3}$, $3.096 \times 10^{-3}$ and $3.003 \times 10^{-3}\ \text{K}^{-1}$. The values of $\ln k$ are $-5.79$, $-4.60$, $-3.48$, $-2.43$ and $-1.45$.

Step 2. The fit. The best line has a slope of $-1.058 \times 10^{4}\ \text{K}$.

Step 3. The activation energy. $E_a = -R \times \text{slope} = 8.314\ \text{J/(mol K)} \times 1.058 \times 10^{4}\ \text{K} = 88\,000\ \text{J/mol} = 88.0\ \text{kJ/mol}$.

Step 4. The prediction. Put $T = 308\ \text{K}$ into the line. This gives $\ln k = -4.03$, so $k = 0.0178\ \text{s}^{-1}$.