Arrhenius Equation
On this page
Direct answer
The rate constant grows exponentially with temperature, a link Arrhenius captured as k = A e^(−Ea/RT): A is the frequency factor (collision frequency with orientation folded in), Ea the activation energy, and the negative exponent shows that only molecules crossing the energy barrier react. In working logarithmic form, log k = log A − Ea/(2.303 RT), so a plot of log k against 1/T is a straight line of slope −Ea/2.303R and intercept log A — the graphical centrepiece of the chapter. Between two temperatures, log(k2/k1) = (Ea/2.303R) × (T2 − T1)/(T1T2), the equation NEET uses for every Ea numerical, and the practical rule that rate roughly doubles per 10 °C rise is this equation in disguise for typical activation energies near 50 kJ/mol. A catalyst works by the same logic, lowering Ea and raising k without touching ΔG of the reaction.
What you must remember
- Two forms: exponential k = A e^(−Ea/RT) and logarithmic log k = log A − Ea/(2.303 RT); the second is what gets plotted and substituted.
- Graph reading: log k versus 1/T gives slope −Ea/2.303R, so Ea = −2.303 × R × slope; a steeper negative slope means a higher barrier.
- Two-temperature equation: log(k2/k1) = (Ea/2.303R) × [(T2 − T1)/(T1 × T2)]; temperatures in kelvin, R = 8.314 J K−1 mol−1, Ea in joules.
- Doubling rule: rate approximately doubles per 10 K rise for Ea around 50-60 kJ/mol near room temperature — a rule of thumb, not a law.
- Catalyst effect: Ea is lowered, so k rises at the same temperature; A and the equilibrium constant stay untouched.
- Temperature sensitivity ranking: reactions with larger Ea respond more dramatically to heating — comparing two Ea values predicts which k jumps more.
- Zero-activation edge: for barrier-free (Ea = 0) processes, k equals A and temperature has no accelerating effect — a conceptual extreme the paper likes to test.
Extracting Ea from a doubling
Suppose a rate constant doubles when temperature rises from 300 K to 310 K. Apply log(k2/k1) = log 2 = 0.301 = (Ea/2.303 × 8.314) × (10/(300 × 310)). The denominator: 2.303 × 8.314 = 19.15, and 300 × 310/10 = 9300. So Ea = 0.301 × 19.15 × 9300 = 53,600 J ≈ 53.6 kJ/mol — the origin of the "50-ish kJ" behind the doubling rule.
Reverse the question for the exam's other costume: with Ea = 55 kJ/mol, find the ratio of k at 320 K to k at 300 K. log ratio = (55000/19.15) × (20/96000) = 2872 × 0.000208 = 0.598, so k2/k1 ≈ 4 — nearly two doublings, matching the 10-degree intuition extended over 20 degrees. The craft in both directions: multiply Ea/2.303R by the temperature difference over the temperature product, keep everything in joules and kelvin, and sanity-check that heating always increases k for a positive Ea.
Where candidates stumble with Arrhenius
The numerical traps are mechanical. R must carry joules when Ea is in joules — mixing kJ Ea with 8.314 J is the commonest wrong answer, off by a factor of 1000. The slope question inverts sign discipline: a plot of log k against 1/T slopes downward for positive Ea, so Ea = −2.303 R × slope, and candidates who forget the minus report negative activation energies. Conceptual traps cluster around A and Ea: the frequency factor is largely temperature-insensitive (the exponential does the work), and Ea is the barrier height from reactants, not the reaction's enthalpy — a catalysed and uncatalysed route share ΔH while showing different Ea, a distinction assertion-reason items probe relentlessly. Finally, the doubling rule is approximate: NEET has used "rate doubles per 10 °" as a true statement with "for all reactions" false, so read the qualifier.
Frequently asked questions
What does each symbol in k = A e^(-Ea/RT) represent?
k is the rate constant, A the frequency factor (or pre-exponential factor), Ea the activation energy in J/mol, R the gas constant and T the absolute temperature.
How is activation energy obtained from a log k versus 1/T plot?
The slope of the straight line equals −Ea/2.303R, so Ea = −2.303 × R × slope, with R in J K−1 mol−1.
Why does a catalyst increase the rate constant?
It provides an alternative path of lower activation energy, so the fraction of molecules clearing the barrier (the exponential term) grows at the same temperature.
Why does the rate roughly double on heating by 10 degrees?
For typical activation energies near 50 kJ/mol, the Arrhenius exponent grows enough between T and T + 10 K to double k — an approximation, not a universal rule.
Does the activation energy equal the enthalpy of reaction?
No — Ea is the barrier height from the reactant side, whereas ΔH is the energy difference between reactants and products; a catalyst changes Ea but never ΔH.