Chemical Kinetics PDF - First Order Reaction

It can be seen that for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species. The half-life of a first order equation is readily calculated from...

It can be seen that for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species. The half-life of a first order equation is readily calculated from the rate constant and vice versa. For zero order reaction t1/2 µ [R]0. For first order reaction t1/2 is independent of [R]0. A first order reaction is found to have a rate constant, k = 5.5 × 10-14 s-1. Example 3.7 Find the half-life of the reaction. Half-life for a first order reaction is Solution 0.693 t1/ 2 = k 0.693 t1/ 2 = 5.5 ×10 –14 s –1 = 1.26 × 1013s Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (t1/2) of the reaction. When reaction is completed 99.9%, [R]n = [R]0 – 0.999[R]0 Example 3.8 2.303  R 0 k = log Solution t R  2.303  R 0 2.303 = log log103 t  R 0  0.999  R 0 = t t = 6.909/k For half-life of the reaction t1/2 = 0.693/k t 6.909 k =   10 t1/ 2 k 0.693 Table 3.4 summarises the mathematical features of integrated laws of zero and first order reactions. Table 3.4: Integrated Rate Laws for the Reactions of Zero and First Order Order Reaction Differential Integrated Straight Half- Units of k type rate law rate law line plot life 0 R® P d[R]/dt = -k kt = [R]0-[R] [R] vs t [R]0/2k conc time-1 or mol L–1s–1 1 R® P d[R]/dt = -k[R] [R] = [R]0e-kt ln[R] vs t ln 2/k time-1 or s–1 or kt = ln{[R]0/[R]} 77 Chemical Kinetics Reprint 2024-25  The order of a reaction is sometimes altered by conditions. There are many reactions which obey first order rate law although they are higher order reactions. Consider the hydrolysis of ethyl acetate which is a chemical reaction between ethyl acetate and water. In reality, it is a second order reaction and concentration of both ethyl acetate and water affect the rate of the reaction. But water is taken in large excess for hydrolysis, therefore, concentration of water is not altered much during the reaction. Thus, the rate of reaction is affected by concentration of ethyl acetate only. For example, during the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water, amounts of the reactants and products at the beginning (t = 0) and completion (t) of the reaction are given as under. H  CH3COOC2H5 + H2O  CH3COOH + C2H5OH t = 0 0.01 mol 10 mol 0 mol 0 mol t 0 mol 9.99 mol 0.01 mol 0.01 mol The concentration of water does not get altered much during the course of the reaction. So, the reaction behaves as first order reaction. Such reactions are called pseudo first order reactions. Inversion of cane sugar is another pseudo first order reaction. + C12H22O11 + H2O  H → C6H12O6 + C6H12O6 Cane sugar Glucose Fructose Rate = k [C12H22O11] Intext Questions 3.5 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? 3.6 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. Most of the chemical reactions are accelerated by increase in temperature. 3.4 Temperature For example, in decomposition of N2O5, the time taken for half of the Dependence of original amount of material to decompose is 12 min at 50oC, 5 h at the Rate of a 25oC and 10 days at 0oC. You also know that in a mixture of potassium permanganate (KMnO 4) and oxalic acid (H2 C 2O4 ), potassium Reaction permanganate gets decolourised faster at a higher temperature than that at a lower temperature. It has been found that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled. The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation (3.18). It was first proposed by Dutch chemist, J.H. van’t Hoff but Swedish chemist, Arrhenius provided its physical justification and interpretation. Chemistry 78 Reprint 2024-25  k = A e -Ea /RT (3.18) where A is the Arrhenius factor or the frequency factor. It is also called pre-exponential factor. It is a constant specific to a particular reaction. R is gas constant and Ea is activation energy measured in joules/mole (J mol –1). It can be understood clearly using the following simple reaction H 2  g   I2  g  2HI  g  According to Arrhenius, this reaction can take place only when a molecule of hydrogen and a molecule of iodine Intermediate collide to form an unstable intermediate (Fig. 3.6). It exists for a very short time and then breaks up to form two Fig. 3.6: Formation of HI through molecules of hydrogen iodide. the intermediate The energy required to form this intermediate, called activated complex (C), is known as activation energy (Ea). Fig. 3.7 is obtained by plotting potential energy vs reaction coordinate. Reaction coordinate represents the profile of energy change when reactants change into products. Some energy is released when the complex decomposes to form products. So, the final enthalpy of the reaction depends upon the nature of reactants and products. All the molecules in the reacting species do not have the same kinetic Fig. 3.7: Diagram showing plot of potential energy. Since it is difficult to predict the energy vs reaction coordinate behaviour of any one molecule with precision, Ludwig Boltzmann and James Clark Maxwell used statistics to predict the behaviour of large number of molecules. According to them, the distribution of kinetic energy may be described by plotting the fraction of molecules (NE/NT) with a given kinetic energy (E) vs kinetic energy (Fig. 3.8). Here, NE is the number of molecules with energy E and N T is total number of molecules. The peak of the curve corresponds to the most probable kinetic energy, i.e., kinetic energy of maximum fraction of molecules. There are decreasing number Fig. 3.8: Distribution curve showing energies of molecules with energies higher or among gaseous molecules lower than this value. When the 79 Chemical Kinetics Reprint 2024-25  temperature is raised, the maximum of the curve moves to the higher energy value (Fig. 3.9) and the curve broadens out, i.e., spreads to the right such that there is a greater proportion of molecules with much higher energies. The area under the curve must be constant since total probability must be one at all times. We can mark the position of Ea on Fig. 3.9: Distribution curve showing temperature Maxwell Boltzmann distribution curve dependence of rate of a reaction (Fig. 3.9). Increasing the temperature of the substance increases the fraction of molecules, which collide with energies greater than Ea. It is clear from the diagram that in the curve at (t + 10), the area showing the fraction of molecules having energy equal to or greater than activation energy gets doubled leading to doubling the rate of a reaction. In the Arrhenius equation (3.18) the factor e -Ea /RT corresponds to the fraction of molecules that have kinetic energy greater than Ea. Taking natural logarithm of both sides of equation (3.18) Ea ln k = – + ln A (3.19) RT The plot of ln k vs 1/T gives a straight line according to the equation (3.19) as shown in Fig. 3.10. Thus, it has been found from Arrhenius equation (3.18) that increasing the temperature or decreasing the activation energy will result in an increase in the rate of the reaction and an exponential increase in the rate constant. Ea In Fig. 3.10, slope = – and intercept = ln R A. So we can calculate Ea and A using these values. At temperature T1, equation (3.19) is Ea ln k1 = – + ln A (3.20) RT1 At temperature T2, equation (3.19) is Ea ln k2 = – + ln A (3.21) RT2 (since A is constant for a given reaction) k1 and k2 are the values of rate constants at temperatures T1 and T2 respectively. Fig. 3.10: A plot between ln k and 1/T Chemistry 80 Reprint 2024-25