Chemical Kinetics Lecture/Lession Plan -2

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Dominick Cain
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1 Chapter Chemical Kinetics Lecture/Lession Plan - Chemical Kinetics. Rate laws of zero, first and second order reactions.. Zero order reaction Let us consider a reaction: A Product If this reaction follow a zero order rate law, then we can write a rate expression- α C 0 A where, is the concentration of the reactant A at time t. or, = kc 0 A (k is the rate constant) d = k C A t = k ( ) 0 0 [ ( ) 0 ] or, = ( ) 0 kt or, k = ( ) 0 t Hence from the above equation we can say that unit of a zero order rate constant = unit of concentration unit of time 3 = mol.lit.sec

2 4 CHAPTER. CHEMICAL KINETICS LECTURE/LESSION PLAN - If we plot the rate expression in a graph vs t, we will get a straight line with a negative slope (-k) and it will cut y axis at ( ) 0. Concentration vs time for zero order reaction As the rate for any zero order reaction does not change with change in the concentration of reactants so, a plot of reaction rate vs time for zero order reaction will be parallel to X axis Rate Vs time for zero order reaction Half life of a zero order reaction: Half life (t / ) can be defined as the time when then concentration of a specific substrate of a reaction will be half of it s initial concentration. So, when t = t /, then = ( )0

3 .. RATE LAWS OF ZERO, FIRST AND SECOND ORDER REACTIONS 5 So, the expression of half life will be ( ) 0 = ( ) 0 kt / kt / = ( ) 0 ( ) 0 t / = ( ) 0 k A plot ( ) 0 vs t / will give a straight line passing through origin with a slope k. Half life Vs initial concentration for zero order reaction.. (a)first order reaction Let us consider a reaction:a Product If this reaction follow a first order rate law, then we can write a rate expression- α where, is the concentration of the reactant A at time t.

4 6 CHAPTER. CHEMICAL KINETICS LECTURE/LESSION PLAN - = k (k is the rate constant) d = k C A dc t A = k ( ) 0 0 ln ln( ) 0 = kt If we plot this rate expression in a graph or, ln = ln( ) 0 kt ln vs t, we will get a straight line with a negative slope (-k) and it will cut y axis at ln( ) 0 or, ln ( ) 0 = kt ( ) 0 = e kt = ( ) 0 e kt If we plot this rate expression in a graph vs t, we will get an exponential decay graph and the graph will touch x axis only when the reaction will complete. From the above equation we can see that only at t =α, will be zero or the reaction will complete. Hence we can say that first order reaction will never complete or will be completed at infinite time and hence the decay graph line will never meet x-axis. Rate constant: We derived the first order rate law and from this rate expression first order rate constant can be derived. Half life of a first first reaction: ln ln( ) 0 = kt or, ln ( ) 0 = kt or, k = t ln() 0 So, unit of k will be = time = sec

5 .. RATE LAWS OF ZERO, FIRST AND SECOND ORDER REACTIONS 7 When t = t / ; then, = ( )0 So, the expression of half life will be ln ( ) 0 = ln( ) 0 kt / kt / = ln( ) 0 ln ( ) 0 t / = ln k t / = k So, from the above equation we can say that half life of a first order reaction is independent of concentration...3 (b)another expression for first order reaction Sometimes, the first order rate law can be expressed in other format where the initial concentration of reactant is a mol/litre and at any time t, the concentration is (a-x) mol/litre where x is the amount of reactant already reacted at time t. So, the first order rate law can be written by replacing = (a-x) and ( ) 0 = a ln(a-x) = ln a kt a kt = ln a-x k = ln a t a-x ln(a - x) Vs time for first order reaction

6 8 CHAPTER. CHEMICAL KINETICS LECTURE/LESSION PLAN - Figure.: ln(a - x) Vs time for first order reaction..4 (a)second order reaction (initial species is same) Let us consider a reaction: A Product If this reaction follow a second order rate law, then we can write a rate expression- α C A where, is the concentration of the reactant A at time t. or, C A = kc A (k is the rate constant) = k dc A ( ) 0 C A = k ( ) 0 = t 0 ( ) 0 + kt If we plot this rate expression in a graph vs t, we will get a straight line graph with a slope (k) and it will cut y axis at ( ) 0. Rate constant: We derived the second order rate law and from this rate expression second order rate constant can

7 .. RATE LAWS OF ZERO, FIRST AND SECOND ORDER REACTIONS 9 be derived. ( ) 0.( ) 0 k = t ( ) 0.( ) 0 Half life of a second order reaction: So, when t = t /, then = ( ) 0 So, the expression of half life will be So, unit of rate constant k will beunit = concentration concentration.concentration.time = concentration.time = mol.lit.sec / ( ) 0 ( ) 0 ( ) 0 / t / =.( ) 0 Like the first order rate law, second order rate law can be expressed in other format where the initial concentration of reactant is a mol/litre and at any time t, the concentration is (a-x) mol/litre where x is the amount of reactant already reacted at time t. So, the first order rate law can be written by replacing = (a-x) and ( ) 0 = a a-x a a a(a-x) k = t a a(a-x)..5 (b)second order reaction (initial species are different) Let us see another type of second order reaction: A + B Product The initial concentration of A and B is a and b mol/lit respectively. At any time t, x mol/lit of each reactant have converted to product. So, at time t, the remaining concentration of A and B is (a-x)

8 0 CHAPTER. CHEMICAL KINETICS LECTURE/LESSION PLAN - and (b-x) mol/lit respectively. Now with respect to product concentration (x mol/lit), the second order rate of the above reaction can be written as x a-b x o dx dx (a-x)(b-x) dx (a-x)(b-x) = k(a-x)(b-x) = k o [ (a-x) (b-x) ]dx = k t = k o a-b [ln(a-x) ln(b-x)]x o ([ln(a-x) lna] [ln(b-x) lnb]) a-b a-b [ln(a-x) a ln (b-x) b ] a-b lnb(a-x) a(b-x) k = t o t(a-b) lnb(a-x) a(b-x) ln b(a x) a(b x) Vs initial concentration for second order reaction

9 .. PSEUDO-UNIMOLECULAR REACTION OR PSEUDO-FIRST ORDER REACTION. Pseudo-Unimolecular Reaction or Pseudo-first order Reaction For a chemical reaction, aa + bb + cc products The experimentally derived rate law can be written as, ( dc ) = k.cα A Cβ B Cγ C Where, C B and C C are the concentration of substrate A, B and C respectively at time t and k is rate constant of the reaction. Now overall order of the reaction will be order(n) = α + β + γ Now if the concentration of any two reactants among the three reactants are so high that during the course of reaction their concentration change is negligible, then their concentration will not have any effect to the rate of the reaction. Hence the rate will depend only upon the concentration of one reactant whose concentration change is significant. This reaction is called pseudo unimolecular reaction. In the above reaction, let us see that the concentration of B and C is unaltered during the reaction and hence their concentration change will be considered as unity. So, the above rate expression will look like ( dc ) = k.cα A This is the unimolecular reaction rate expression. If α =, then it will be a pseudo first order reaction. Example: Acid catalysed hydrolysis of ester in aqueous medium and decomposition of sucrose in acid medium CH 3 COOEt + H O + H + = CH 3 COOH + EtOH C H O + H O + H + = C 6 H O 6 (glucose) + C 6 H O 6 (fructose) In both the reaction water is used in large excess and hence it s concentration change is negligible. So, these two are pseudo unimolecular reactio.

Lecture 12 Elementary Chemical Kinetics: Concepts

Lecture 12 Elementary Chemical Kinetics: Concepts Is it possible to understand the feasibility of processes? Yes - thermodynamics - thermodynamic state functions. Time dependence of chemical processes