February 16, 2024

Find comprehensive NCERT Solutions for Class 12 Science Chemistry Chapter 4 on Chemical Kinetics, featuring clear step-by-step explanations. These solutions have gained immense popularity among Chemistry students in Class 12 Science, serving as valuable resources for homework completion and exam preparation. Free access to all questions and answers from Chapter 4 of the NCERT Book for Class 12 Science Chemistry is provided here, ensuring a convenient and effective study aid for students. Page No 98: Question 4.1: For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds. ANSWER: Average rate of reaction  = 6.67 × 10−6 M s−1 Page No 98: Question 4.2: In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L−1 to 0.4 mol L−1 in 10 minutes. Calculate the rate during this interval? ANSWER: Average rate  = 0.005 mol L−1 min−1 = 5 × 10−3 M min−1 Page No 103: Question 4.3: For a reaction, A + B → Product; the rate law is given by,. What is the order of the reaction? ANSWER: The order of the reaction = 2.5 Page No 103: Question 4.4: The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y? ANSWER: The reaction X → Y follows second order kinetics. Therefore, the rate equation for this reaction will be: Rate = k[X]2 (1) Let [X] = a mol L−1, then equation (1) can be written as: Rate1 = k .(a)2 = ka2 If the concentration of X is increased to three times, then [X] = 3a mol L−1 Now, the rate equation will be: Rate = k (3a)2 = 9(ka2) Hence, the rate of formation will increase by 9 times. Page No 111: Question 4.5: A first order reaction has a rate constant 1.15 10−3 s−1. How long will 5 g of this reactant take to reduce to 3 g? ANSWER: From the question, we can write down the following information: Initial amount = 5 g Final concentration = 3 g Rate constant = 1.15 10−3 s−1 We know that for a 1st order reaction, = 444.38 s = 444 s (approx) Page No 111: Question 4.6: Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. ANSWER: We know that for a 1st order reaction, It is given that t1/2 = 60 min Page No 116: Question 4.7: What will be the effect of temperature on rate constant? ANSWER: The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation, Where, A is the Arrhenius factor or the frequency factor T is the temperature R is the gas constant Ea is the activation energy Page No 116: Question 4.8: The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea. ANSWER: It is given that T1 = 298 K ∴T2 = (298 + 10) K = 308 K We also know that the rate of the reaction doubles when temperature is increased by 10°. Therefore, let us take the value of k1 = k and that of k2 = 2k Also, R = 8.314 J K−1 mol−1 Now, substituting these values in the equation: We get: = 52897.78 J mol−1 = 52.9 kJ mol−1 Note: There is a slight variation in this answer and the one given in the NCERT textbook. Page No 116: Question 4.9: The activation energy for the reaction 2HI(g)→ H2 + I2(g) is 209.5 kJ mol−1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy? ANSWER: In the given case: Ea = 209.5 kJ mol−1 = 209500 J mol−1 T = 581 K R = 8.314 JK−1 mol−1 Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as: Page No 117: Question 4.1: From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. (i) 3 NO(g) → N2O(g) Rate = k[NO]2 (ii) H2O2 (aq) + 3 I− (aq) + 2 H+ → 2 H2O (l) +  Rate = k[H2O2][I−] (iii) CH3CHO(g) → CH4(g) + CO(g) Rate = k [CH3CHO]3/2 (iv) C2H5Cl(g) → C2H4(g) + HCl(g) Rate = k [C2H5Cl] ANSWER: (i) Given rate = k [NO]2 Therefore, order of the reaction = 2 Dimension of  (ii) Given rate = k [H2O2] [I−] Therefore, order of the reaction = 2 Dimension of  (iii) Given rate = k [CH3CHO]3/2 Therefore, order of reaction =  Dimension of  (iv) Given rate = k [C2H5Cl] Therefore, order of the reaction = 1 Dimension of  Page No 117: Question 4.2: For the reaction: 2A + B → A2B the rate = k[A][B]2 with k = 2.0 × 10−6 mol−2 L2 s−1. Calculate the initial rate of the reaction when [A] = 0.1 mol L−1, [B] = 0.2 mol L−1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L−1. ANSWER: The initial rate of the reaction is Rate = k [A][B]2 = (2.0 × 10−6 mol−2 L2 s−1) (0.1 mol L−1) (0.2 mol L−1)2 = 8.0 × 10−9 mol−2 L2 s−1 When [A] is reduced from 0.1 mol L−1 to 0.06 mol−1, the concentration of A reacted = (0.1 − 0.06) mol L−1 = 0.04 mol L−1 Therefore, concentration of B reacted = 0.02 mol L−1 Then, concentration of B available, [B] = (0.2 − 0.02) mol L−1 = 0.18 mol L−1 After [A] is reduced to 0.06 mol L−1, the rate of the reaction is given by, Rate = k [A][B]2 = (2.0 × 10−6 mol−2 L2 s−1) (0.06 mol L−1) (0.18 mol L−1)2 = 3.89 mol L−1 s−1 Page No 117: Question 4.3: The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10−4 mol−1 L s−1? ANSWER: The decomposition of NH3 on platinum surface is represented by the following equation. Therefore, However, it is given that the reaction is of zero order. Therefore, Therefore, the rate of production of N2 is And, the rate of production of H2 is = 7.5 × 10−4 mol L−1 s−1 Page No 117: Question 4.4: The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k [CH3OCH3]3/2 …

NCERT Solutions for Class 12 Science Chemistry Chapter 4 – Chemical Kinetics Read More »