First Order Reactions — t½, integrated rate law
Chemical Kinetics
8
JEE Qs
8%
Hard
75
min
Master the derivations of integrated rate laws and half-life for first-order reactions, focusing on their application in various problem types, especially those involving gas-phase reactions and multiple half-lives.
🧮 Key Formulas
✅ Key Points for JEE
- 1The half-life (t½) of a first-order reaction is independent of the initial concentration of the reactant, a key distinguishing feature.
- 2The integrated rate law allows calculation of the rate constant, concentration of reactant at any time, or time required for a specific extent of reaction.
- 3A plot of ln[A] vs time (t) for a first-order reaction yields a straight line with a slope of -k and a y-intercept of ln[A]0.
- 4The units of the rate constant 'k' for a first-order reaction are always reciprocal time units (e.g., s^-1, min^-1, hr^-1).
- 5Many natural processes like radioactive decay follow first-order kinetics, making the half-life concept broadly applicable.
⚠️ Common Mistakes
- ✕Confusing natural logarithm (ln) with base-10 logarithm (log) and forgetting the conversion factor (2.303 for log to ln).
- ✕Incorrectly identifying initial ([A]0) and final ([A]t) concentrations or pressures in problem statements.
- ✕Applying first-order rate equations to reactions that are zero-order or second-order without verifying the order.
- ✕Errors in algebraic manipulation of the integrated rate law or half-life formulas, leading to incorrect calculations.
📝 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