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# Chapter 4 Chemical Kinetics

Chemical kinetics is that branch of chemistry which deals with the study of the rate of chemical reaction. In this chapter, you will be dealing with the average and instantaneous rate of reaction, the factors such as temperature, pressure, concentration and catalyst which affect the rate of reactions. Differences between elementary and complex reactions are also given. Differences between molecularity and order of a reaction are also given. Description of collision theory and derivation of integrated rate of equations for the zero and first order reactions alongwith determination of its rate constants.

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### Exercise 1

•  Q1 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. Ans: Q2 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? Ans: Q3 For a reaction, A + B → Product; the rate law is given by,  r  = k [A]½ [B]2. What is the order of the reaction? Ans: The order of the reaction = 1/2 + 2 = 2 1/2 = 2.5 Q4 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? Ans: The order of reaction is defined as the sum of the powers of concentrations In the rate law. The rate of second order reaction can be expressed as     rate = k[A]2 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: Rate = 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. Q5 A first order reaction has a rate constant 1.15 10-3s-1. How long will 5 g of this reactant take to reduce to 3 g? Ans: From the question, we can write down the following information: Initial amount = 5 g Final concentration = 3 g Rate constant = 1.15 10 - 3s - 1 We know that for a 1st order reaction, = 444.38 s = 444 s (approx) Q6 Time required to decompose SO2Cl2to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. Ans: We know that for a 1st order reaction, t½  = 0.693 / k It is given that t1/2= 60 min k = 0.693 / t½    = 0.693 / 60   = 0.01155 min-1    = 1.155 min-1 Or k = 1.925 x 10-4 s-1 Q7 What will be the effect of temperature on rate constant? Ans: 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, k = Ae - Ea / RT Where, A is the Arrhenius factor or the frequency factor T is the temperature R is the gas constant Ea is the activation energy Q8 The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea. Ans: 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: = 52897.78 J mol - 1 = 52.9 kJ mol - 1 Q9 The activation energy for the reaction 2HI(g) → H2 + I2(g) is 209.5 kJ mol-1 at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy? Ans: 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 ## Recently Viewed Questions of Class 12 Chemistry

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