Order & Molecularity — Finding order from data
Chemical Kinetics
8
JEE Qs
8%
Hard
60
min
Master the initial rate method by systematically comparing experimental data sets to deduce individual reaction orders for each reactant with meticulous calculations.
🧮 Key Formulas
✅ Key Points for JEE
- 1Order of a reaction is an experimentally determined value and is not necessarily equal to the stoichiometric coefficients in the balanced chemical equation.
- 2The initial rate method is used to determine reaction orders by systematically varying the concentration of one reactant while keeping others constant and observing the change in initial rate.
- 3Molecularity is the number of reactant species (atoms, ions, or molecules) participating in an elementary step and is always a positive integer (1, 2, or 3).
- 4Order can be zero, fractional, or negative, while molecularity is always a positive integer.
- 5The units of the rate constant (k) depend on the overall order of the reaction and can be derived from the rate law.
⚠️ Common Mistakes
- ✕Confusing reaction order (experimental) with molecularity (theoretical, for elementary steps) or stoichiometric coefficients.
- ✕Errors in applying the initial rate method, such as not isolating the effect of individual reactant concentrations.
- ✕Incorrect mathematical calculations when comparing experimental data to deduce the order with respect to each reactant.
- ✕Incorrectly determining the units of the rate constant 'k' based on the overall reaction order.
📝 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