Q47.For the reaction, 3A + 2B →C + D, the differential rate law can be written as : (1) + 31 d[A]dt = −d[C]dt = k[A]n[B]m (2) 13 d[A]dt = d[C]dt = k[A]n[B]m (3) −13 d[A]dt = d[C]dt = k[A]n[B]m (4) −d[A]dt = d[C]dt = k[A]n[B]m
What This Question Tests
This question tests the ability to write the correct differential rate law expression for a given reaction based on its stoichiometry, identifying reactants (negative rate) and products (positive rate) and using reciprocal stoichiometric coefficients.
Concepts Tested
Formulas Used
For a reaction aA + bB -> cC + dD, rate = -(1/a)d[A]/dt = -(1/b)d[B]/dt = (1/c)d[C]/dt = (1/d)d[D]/dt
📚 NCERT Sections This Tests
3.10 — In A Reaction Between A And B, The Initial Rate Of Reaction (R0) Was Measured
Chemistry Class 11 · Chapter 3
3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L–1 0.20 0.20 0.40 B/ mol L–1 0.30 0.10 0.05 r0/mol L–1s–1 5.07 × 10–5 5.07 × 10–5 1.43 × 10–4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ® C + D Experiment [A]/mol L–1 [B]/mol L–1 Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 × 10–3 II 0.3 0.2 7.2 × 10–2 III 0.3 0.4 2.88 × 10–1 IV 0.4 0.1 2.40 × 10–2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L–1 [B]/ mol L–1 Initial rate/ mol L–1 min–1 I 0.1 0.1 2.0 × 10–2 II – 0.2 4.0 × 10–2 III 0.4 0.4 – IV – 0.2 2.0 × 10–2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ® 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 × [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol L–1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).
3.9 — A Reaction Is First Order In A And Second Order In B.
Chemistry Class 11 · Chapter 3
3.9 A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is the rate affected on increasing the concentration of B three times? (iii) How is the rate affected when the concentrations of both A and B are doubled? 85 Chemical Kinetics Reprint 2025-26
3.16 — The Rate Constant For A First Order Reaction Is 60 S–1. How Much Time Will
Chemistry Class 11 · Chapter 3
3.16 The rate constant for a first order reaction is 60 s–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?
📋 Question Details
- Chapter
- Chemical Kinetics
- Topic
- Differential rate law
- Year
- 2014
- Shift
- 19 Apr Online
- Q Number
- Q47
- Type
- MCQ
- NCERT Ref
- Class 12 Chemistry Ch 4: Chemical Kinetics
More from this Chapter
Q93.Consider an endothermic reaction, X ⟶Y with the activation energies Eb and Ef for the backward and forward reactions, respectively. In general (1) Eb < Ef (2) Eb > Ef (3) Eb = Ef (4) There is no definite relation between Eb and Ef
Q90.Rate of a reaction can be expressed by Arrhenius equation as: k = Ae−E/RT In this equation, E represents (1) the energy above which all the colliding (2) the energy below which colliding molecules will molecules will react not react (3) the total energy of the reacting molecules at a (4) the fraction of molecules with energy greater temperature, T than the activation energy of the reaction
Q91.The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr : NO(g) + Br2( g) ⇌NOBr2( g) NOBr2( g) + NO(g) ⟶2NOBr(g) If the second step is the rate determining step, the order of the reaction with respect to NO(g) is (1) 1 (2) 0 (3) 3 (4) 2
Q68.The energies of activation for forward and reverse reactions for A2 + B2 ⇌2AB are 180 kJ mol−1 and 200 kJ mol−1 respectively. The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol−1 . The enthalpy change of the reaction (A2 + B2 ⟶2AB) in the presence of catalyst will be (in kJ mol−1 ) (1) 300 (2) 120 (3) 280 (4) 20