Arrhenius Equation — Ea, frequency factor, temperature
Chemical Kinetics
8
JEE Qs
8%
Hard
60
min
Master the two-temperature form of the Arrhenius equation and the graphical interpretation of ln k vs 1/T to solve numerical problems quickly and accurately.
🧮 Key Formulas
✅ Key Points for JEE
- 1The Arrhenius equation describes the quantitative dependence of the rate constant (k) on temperature (T) and activation energy (Ea).
- 2Activation energy (Ea) is the minimum energy reactants must possess to form products; a lower Ea leads to a faster reaction rate constant.
- 3The pre-exponential factor 'A' (frequency factor) incorporates collision frequency and the orientation factor, indicating the fraction of collisions that are effective when energy requirement is met.
- 4A plot of ln k versus 1/T yields a straight line with a slope equal to -Ea/R and a y-intercept equal to ln A, which is critical for determining Ea and A experimentally.
- 5Catalysts increase reaction rates by providing an alternative reaction pathway with a lower activation energy (Ea), thereby increasing the rate constant (k) without changing 'A' or the overall thermodynamics of the reaction.
⚠️ Common Mistakes
- ✕Using incorrect units for the gas constant R (e.g., 0.0821 L.atm/mol.K instead of 8.314 J/mol.K when Ea is in Joules).
- ✕Forgetting to convert temperature from Celsius to Kelvin, as T in the Arrhenius equation must always be in Kelvin.
- ✕Confusing the slope of ln k vs 1/T (-Ea/R) with the slope of log k vs 1/T (-Ea/(2.303R)).
📝 Practice Questions
See allQ72.For the thermal decomposition of N2O5( g) at constant volume, the following table can be formed, for the reaction mentioned below. 2 N2O5( g) →2 N2O4( g) + O2( g) x = … × 10−3 atm [nearest integer] Given : Rate constant for the reaction is 4.606 × 10−2 s−1 .
Q58.For a reaction, N2O5( g) →2NO2( g) + 12 O2( g) in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is (1) 5 times of initial pressure (2) 5/2 times of initial pressure (3) 7/2 times of initial pressure (4) 7/4 times of initial pressure
Q72. A → B The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of 1000 K . If the energy barrier with respect to reactant energy for such isomeric transformation is 191.48 kJ mol−1 and the frequency factor is 1020 , the time required for 50% molecules of A to become B is _________ picoseconds (nearest integer). [R = 8.314 J K−1 mol−1]
Q55. Consider the given figure and choose the correct option : (1) Activation energy of both forward and backward (2) Activation energy of forward reaction is E1 + E2 reaction is E1 + E2 and reactant is more stable and product is less stable than reactant. than product. (3) Activation energy of backward reaction is E1 and (4) Activation energy of forward reaction is E1 + E2 product is more stable than reactant. and product is more stable than reactant.
Q54.Which of the following graphs most appropriately represents a zero order reaction ? 2025 (23 Jan Shift 2) JEE Main Previous Year Paper (1) (2) (3) (4)
Q73.Consider a complex reaction taking place in three steps with rate constants k1, k2 and k3 respectively. The overall rate constant k is given by the expression k = . If the activation energies of the three steps are √k1k3k2 60,30 and 10 kJ mol−1 respectively, then the overall energy of activation in kJmol−1 is … … . (Nearest integer)
NCERT Chapters
- Class 12 Chemistry Ch 4: Chemical Kinetics