Zero Order Reactions
Chemical Kinetics
8
JEE Qs
8%
Hard
45
min
Master the integrated rate law, its graphical representation, and the half-life expression for zero-order reactions, paying close attention to how t₁/₂ depends on initial concentration for comparative analysis.
🧮 Key Formulas
✅ Key Points for JEE
- 1The rate of a zero-order reaction is independent of the concentration of the reactant(s).
- 2A plot of reactant concentration ([A]) versus time (t) yields a straight line with a negative slope (-k) and a y-intercept of [A]₀.
- 3The half-life (t₁/₂) of a zero-order reaction is directly proportional to the initial concentration ([A]₀) and inversely proportional to the rate constant (k).
- 4Zero-order reactions can go to completion, and the time for completion is directly proportional to the initial concentration.
- 5Common examples include enzyme-catalyzed reactions at high substrate concentrations, and heterogeneous reactions on a saturated catalyst surface (e.g., decomposition of NH₃ on a platinum surface).
⚠️ Common Mistakes
- ✕Confusing the concentration dependence of half-life with that of first-order reactions (where t₁/₂ is independent of [A]₀).
- ✕Incorrectly integrating the rate law or deriving the half-life expression, leading to errors in calculations.
- ✕Misinterpreting the conditions under which a reaction becomes pseudo-zero order or truly zero order (e.g., surface saturation, reactant in large excess).
- ✕Using incorrect units for the rate constant 'k' when solving problems.
📝 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