The Arrhenius Equation

\boxed{k = A\,$e^{-E_a/RT}$}

Four Forms of the Arrhenius Equation

Form 1 (Exponential): $$k = A,$e^{-E_a/RT}$$$

Form 2 (Natural log): $\ln k = \ln A - \frac{E_a}{RT}$ Graph: ln k vs 1/T → slope = $-E_a/R$; intercept = $\ln A$

Form 3 (Common log): $\log k = \log A - \frac{E_a}{2.303RT}$ Graph: log k vs 1/T → slope = $-E_a/2.303R$; intercept = $\log A$

Form 4 (Two-temperature): $\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ Used when k is known at two temperatures to find Ea.

Physical Interpretation

• $e^{-E_a/RT}$ = fraction of molecules with kinetic energy ≥ Ea (Boltzmann factor)
• As T → ∞: $$$e^{-E_a/RT}$ \to e^0 = 1$$; k → A (maximum possible rate constant)
• As T → 0: $$$e^{-E_a/RT}$ \to 0$$; k → 0 (reaction essentially stops)
• Higher Ea → steeper slope on Arrhenius plot → more temperature-sensitive reaction

NEET Worked Example

Given: k_{1} = $2.5 \times 10^{-3}$ $s^{-1}$ at $T_{1}$ = 300 K; k_{2} = $5.0 \times 10^{-3}$ $s^{-1}$ at $T_{2}$ = 310 K

$\log\frac{5.0 \times 10^{-3}}{2.5 \times 10^{-3}} = \frac{E_a}{2.303 \times 8.314}\left(\frac{1}{300} - \frac{1}{310}\right)$

$\log 2 = \frac{E_a}{19.147} \times \frac{10}{93000}$

$0.301 = E_a \times 5.616 \times 10^{-6}$

$E_a = \frac{0.301}{5.616 \times 10^{-6}} = 53{,}600 \text{ J/mol} \approx 53.6 \text{ kJ/mol}$