Pseudo First Order Reactions
Chemical Kinetics
8
JEE Qs
8%
Hard
45
min
Always identify the reactant in significant excess and understand how its constant concentration is absorbed into the pseudo rate constant (k') for simplified kinetic analysis.
🧮 Key Formulas
✅ Key Points for JEE
- 1A reaction is classified as pseudo first order when its true overall order is higher than one, but due to one or more reactants being in large excess (or acting as a solvent), their concentrations remain practically constant throughout the reaction.
- 2The observed rate constant (k') for a pseudo first order reaction incorporates the constant concentration terms of the excess reactants, making the reaction appear first order with respect to the limiting reactant.
- 3Common examples include hydrolysis of esters in aqueous solution or inversion of cane sugar, where water (solvent) is present in large excess.
- 4The 'order' determined experimentally under such excess conditions is the pseudo order, which simplifies kinetic analysis and allows for easier determination of the true order with respect to the limiting reactant.
- 5To determine the true rate constant (k), one must divide the pseudo rate constant (k') by the constant concentration terms of the excess reactants raised to their respective orders.
⚠️ Common Mistakes
- ✕Failing to identify the reactant present in large excess, especially when it's the solvent (e.g., water in hydrolysis reactions).
- ✕Confusing the experimentally observed pseudo order with the true molecularity or the actual overall order of the reaction.
- ✕Incorrectly using the true rate constant 'k' in calculations for a pseudo-first-order scenario, instead of the pseudo rate constant 'k''.
- ✕Assuming any reaction involving a solvent is pseudo first order without verifying that the solvent is indeed present in significantly large excess.
📝 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