First Order Kinetics + Arrhenius Equation
Chemical Kinetics
44
JEE Qs
8%
Hard
100
min
Master the integrated rate laws for first-order reactions and the Arrhenius equation, paying close attention to units and logarithmic manipulations, as these are fundamental for solving conceptual and numerical problems.
🧮 Key Formulas
✅ Key Points for JEE
- 1The half-life (t1/2) for a first-order reaction is independent of the initial concentration of the reactant, making it a unique characteristic.
- 2The unit of the rate constant (k) for a first-order reaction is time⁻¹ (e.g., s⁻¹, min⁻¹), regardless of the concentration units.
- 3A plot of ln[A] vs time (t) for a first-order reaction yields a straight line with a slope equal to -k and a y-intercept of ln[A]0.
- 4The Arrhenius equation describes the exponential relationship between the rate constant (k) and temperature (T), explaining why reaction rates generally increase with temperature.
- 5Activation energy (Ea) represents the minimum energy reactants must possess to form products; a higher Ea implies a stronger temperature dependence for the rate constant.
⚠️ Common Mistakes
- ✕Confusing natural logarithm (ln) with base-10 logarithm (log) and misapplying the 2.303 conversion factor in calculations and graphical interpretations.
- ✕Using inconsistent units for the gas constant R and activation energy Ea; ensure R is in J/mol.K when Ea is in J/mol, or kJ/mol.K when Ea is in kJ/mol.
- ✕Incorrectly assuming that half-life is independent of initial concentration for all reaction orders; this is specific to first-order reactions.
- ✕Errors in interpreting slopes and intercepts from Arrhenius plots (e.g., log k vs 1/T) to calculate Ea or A.
📝 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