CHEMICAL KINETICS - PowerPoint PPT Presentation

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  1. CHEMICALKINETICS The branch of chemistry which deals with the rate of chemical reactions and the factors which influence the rate of reaction is called chemical kinetics.

  2. RATE OF CHEMICAL REACTION The rate of change of concentration of any one of the reactants or products at a given time is called rate of reaction. (1) Rate of Reaction = - Decrease in concentration of Reactant Time taken(2) Rate of Reaction = Increase in concentration of product Time takenA + B → C + DRate of Reaction = -- [∆A] = -- [∆B] = [∆C] = [∆D] ∆t ∆t ∆t ∆taA + bB → cC + dDRate of Reaction = -1 [∆A] = -1[∆B] = 1[∆C] = 1 [∆D] a ∆t b ∆t c ∆t d ∆t

  3. 1.Concentration of the reactants:-the rate of reaction is directly proportional to the concentration of the reactants. 2.Temperature of the system:-The rate of chemical reaction increases with the increase in temperature. 3.Presence of catalyst:-A catalyst is a substance which do not participate in the reaction but alter the rate of reaction. 4.Nature of reactants and products:-the rate of reaction are influenced by the nature of reactants and products. A chemical reaction involves breaking of old bonds and formation of new bonds. Factors affecting the rate of reaction:-

  4. LAW OF MASS ACTION According to this law, At a given temperature the rate of a chemical reaction is directly proportional to the product of the molar concentration of the reactants. For, aA + bB → product Rate = k [A]a[B]b where, k = Rate constant or velocity constant

  5. Rate Constant Rate Constant (K) of a reaction at a given temperature may be defined as rate of reaction when the concentration of each of the reactants is unity. Characteristics of Rate constant:- • Rate constant is the measure of the rate of the reaction. Larger the value of K, faster is the reaction and vice versa. • Different reactions have different value of K. • The rate constant only depends upon temperature. • For a particular reaction, the rate constant is independent of concentration.

  6. Molecularity of a Reaction The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must colloid simultaneously in order to bring about a chemical reaction is called molecularity of a reaction. On the basis of molecularity reaction are of three type:- (a) Unimolecular Reaction. (b) Bimolecular Reaction. (c) Trimolecular Reaction.

  7. The order of reaction is defined as the sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction. Order of reaction can be 0, 1, 2, 3 and even a fraction. For a reaction , aA + bB + cC → Products Rate = [A]x+ [B]y + [C]z Then Order of Reaction , n = x + y + z ORDER OF REACTION

  8. Units of reaction Rate Constant Unit of Rate :- Rate = Concentration = mol L-1 = mol S-1L-1 Time S

  9. INTEGRATED RATE EQUATION For Zero Order Reaction:- R → P Rate = -d[R] = k[R]0 = k dt Rate = -d[R] =dt ×k d[R ] = -k dt Integrating both side [R] = -kt + c ………(1) where, c is the constant of integration At t = 0, R = R0 = Initial concentration of reactant. Substituting in equation (1) [R]0 = -k×0 + c = c Substituting the value of c in equation (1). [R] = -kt + [R]0 k = [R]0-[R] t

  10. Integrating rate equation for First Order Reaction:- R → P Rate = - d[R] = k[R] dt - d[R] = -kdt [R] Integrating this equation, we get ln [R] = - kt + c ……….(1) where, c is the constant of integration when t = 0, R = [R]0 = initial concentration of reactant. Substituting in equation (1) ln [R]0 = -k × 0 + c ln [R]0 = Substituting the value of c in equation (1) ln [R] = - kt + ln [R]0 or ln [R] = -kt [R]0 or k = 1 ln [R]0 or k = 2.303 log [R]0 t [R] t [R]

  11. Half Life Period Half Life Period of a reaction is defined as the time during which the Concentration of a reactant is reduced to half of its initial concentration. For Zero Order Reaction:- k = [R]0-[R] t At, t = t ½ , [R] = 1 [R]0 2 The Rate at t ½ becomes K = [R]0 – ½[R]0 t½ Or t½ = [R]0 2k

  12. Half Life Period For first Order Reaction:- K = 2.303 log[R]0 t [R] At t ½ , [R] = [R]0 2 The Rate at t ½ becomes k = 2.303 log[R]0 [R]/2 or t ½ =2.303 log 2 k or t ½=2.303 × 0.301 k t ½= 0.693 k

  13. For Reaction type R → P

  14. Pseudo First Order Reaction Such reactions in which the molar concentration of a reactant remains practically unchanged and the small change in concentration does not effect the rate of reaction are called Pseudo First Order Reaction. Example: Hydrolysis of Sucrose – C12H22O11 + H2O → C6H12O6 +C6H12O6 (cane sugar) (Glucose) (Fructose) Hydrolysis of Ester – CH3COOC2H5 + H2O → CH3COOH + C2H5OH

  15. Collision Theory Of Chemical Reaction According to this theory, the reactant molecules are assumed to be hard spheres and reaction is postulated to occur when molecules colloid with each other. The number of collisions that takes place per second per unit volume of the reaction mixture is known as collisions frequency Z. the value of collisions normally very high. The collisions which actually produce the products and therefore, result in the chemical reaction are called effective collisions.

  16. The main points of collisions theory:- • For a reaction to occur, there must be collisions between the reacting species. • Only a certain fraction of the total number of collisions are effective in forming the products. • For effective collisions, the molecules should possess sufficient energy as well as orientation.

  17. Temperature Dependence of the Rate of a Reaction Most of the chemical reactions are accelerated by increase in temperature. It has been found that for a chemical reaction with rise in temperature by 10ºC (or 10 K), the rate constant is nearly doubled. The temperature dependence of the Rate of a Reaction can be accurately explained by Arrhenius equation :- k = A e -Ea / RT where A = frequency factor R = Universal Gas Constant Ea = Activation Energy (J mol-1) T = Temperature in Kelvin. log10k = log10A - Ea 2.303RT

  18. Effect Of Catalyst A catalyst is the substance which alters the rate of reaction without undergoing any permanent chemical change. The action of the catalyst can be explained by intermediate complex theory. According to this theory, a catalyst participates in a chemical reaction by forming temporary bonds with the reactants resulting in an intermediate complex. This has a transitory existence and decomposes to yield products and the catalyst. For example: 2KClO3→ 2KCl + 3O2 ( MnO2) (catalyst)

  19. Concept Of Activation Energy And Transition State Theory The excess which must be supplied to the reactants to undergo chemical reaction is called Activation Energy Ea. Activation Energy = Threshold Energy – Average kinetic energy or Ea = E (threshold) + E (reactants) When the colliding molecules possess the kinetic energy equal to Ea , the atomic configuration of the species formed at this stage is different from the reactants as well as the products. This stage is called activated stage or transition state and specific configuration of this stage is called activated intermediate.

  20. Numericals:- Q. 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. Average rate = - ∆ [R] ∆ t = - (0.02 - 0.03) molL-1 25 minutes = 4 × 10-4 mol L-1 min-1 Time = 25 minutes = 25 × 60 sec Rate= - (0.02 - 0.03)molL-1 25 × 60 sec = 6.67 10-6 mol L-1 S

  21. Q. A first order reaction has a rate constant 1.15 x 10-3 S-1. How long will 5 kg of this reactant take to reduce to 3 kg. Ans. For a first order reaction:- t = 2.303 log [A]0 k [A] [A]0 = 5g/M, [A] = 3g/ M, k = 1.15 x 10-3 S-1 Where, M is molar mass t = 2.303 log 5g/M 1.15 x 10-3 3g/M t = 2.303 x 0.2218 1.15 x 10-3 t = 444 s

  22. Q. In a reaction 2A→ product , 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 time interval. Ans. Rate = - 1∆[A] 2 ∆t - 1(0.4 – 0.5) mol L-1 = 5 x 10-3 min-1 2 10 min Q. Time required to decompose SO2Cl2 to half of its initial amount in 60 minutes. If the decomposition is a first order reaction. Calculate the rate constant of the reaction. Ans. t½ = 60 min = 60 x 60 seconds k = 0.693 = 0.693 = 1.925 x 10 -4 s-1 t½3600s

  23. Q. The rate constant for a first order reaction is 60s-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value? Ans. Let initial concentration, [A]0 = a Final concentration = [A] = 1 x a 16 Rate constant = 60 s-1 t = 2.303 log [A]0 k [A] = 2.303 log a = 2.303 log 16 60 a/16 60 = 2.303 x 1.204 60 = 4.6 x10-2