SOME IMPORTANT GRAPH REPRESENTATION

Transcription

1 SOME IMPORTNT GRPH REPRESENTTION Use in chemical inetics : () y mx + C () y mx + C (3) y mx UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

2 (4) y mx Slope Intercept m tan zero (5) y e x (6) y e x UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

3 3 (7) y x CHEMICL KINETICS Chemical inetics involves the study of the rates and mechanism of chemical reactions. The rates of reactions : (a) The definition of rate : Consider a reaction of the form + B 3C + D () in which the molar concentration of participants are [], [B], [C] & [D]. The rate of consumption or decomposition of the one of the reactants at a given time is d[ R], where R is or B. The rate of formation of one of the products is d[ P], where P is C or D. The rate of reaction can be expressed with respect to any species in equation (). Rate d[ ] d[ B] d[ C] d[ D] 3 Thus, the rate of reaction can be defined with respect to both reactants and products. For example : 4NO (g) + O (g) N O (g) find the expression for rate of reaction. Sol. 4NO (g) + O (g) N O (g) UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

4 4 d[ NO rate ] d[ O ] d[ NO5 ] 4 (b) Rate laws and rate constant : The rate of a reaction will generally depends on temperature pressure and concentration of species involving in the reaction. The rate of reaction is proportional to the molar concentration of reacting species. i.e. + B + C + D +. Product then, rate of reaction [] a [B] b [C] c [D] d.. where [] is the concentration of reactant, [B] is the concentration of reactant B and so on. The constant a is nown as the reaction order with respect to species, b the reaction order with respect to species B and so on. The over all reaction order is equal to the sum of the individual reaction orders (a + b + c + d +..). Finally the constant is rate constant for the reaction. The rate constant dependent on concentration but also on temperature & pressure. This relationship is nown as a rate law. (c) Order of the reaction : + B + C + Product The rate law v [] a [B] b [C] c.. The order of reaction a + b + c + For example : if rate law v [] / [B] Then, it is half order in, first order in B and three half 3 order overall. Molecularity of a Reaction : The number of reacting species (atoms, ions or molecule) taing parting an elementary reaction, must collide simultaneously, in order bring about a chemical reaction is called molecularity of a reaction. Relationship between Rate law, order and the rate constant : UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

5 5 B Then, rate of reaction d[ ] [ ] n The unit of rate or reaction is mol liter sec i.e. mol L s. where M represent mol L or moles per liter & n is order of reaction. The unit of rate constant () Rate of reaction unit of rate of reaction [] n unit of [unit of concentration] n MS unit of [M] n unit of i.e., unit of [MS ] n [M S ] n [M] n n n M S mol L S Rate law Order Unit of Rate Zero MS Rate [] First order S Rate [] Second order M S Rate [][B] Second order M S Rate [][B][C] Third order M S Prob. Find the order of the reaction if unit of rate constant or the reaction is (dm 3 ) 3/ mol 3/ s. Sol. Unit of rate constant (dm 3 ) 3/ mol 3/ s (given) We now that, Unit of rate constant M n s For nth order UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

6 6 i.e. M n s 3 3/ 3/ (dm ) (mol) s mol 3 dm 3/ s mol L L dm 3 3/ s & mol L M M n s M 3/ s M n M 3/ n n i.e. it is 5 order reaction. Determing Reaction order : Using the following data for the reaction, we determine the order of the reaction with respect to and B, over all order and rate constant for the reaction [] (M) [B] (M) Initial rate (Ms ) Sol. + B Product rate of reaction [] a [B] b [.3 4 ] a [3. 5 ] b...() UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

7 7 Divide equation () by equation (3), we get 4. 3 [4.6 4 ] a [6. 5 ] b...().7 [9. 4 ] a [6. 5 ] b...(3) a 5 b [4.6 ] [6. ] 4 a 5 b [9. ] [6. ].47 (.5) a (.47) (.5) a (.5.5) (.5) a (.5) a (.5) a or taing log we can find the value of a. a Divide equation () by equation () we get a 5 b [.3 ] [3. ] 4 a 5 b [4.6 ] [6. ].5 [.5] a [.5] b [.5] [.5] b.5 [.5] b 5 [.5] b.5 [.5] b b Therefore, the reaction is second order in and first order in B and third order overall. rate [] [B] 5. 4 Ms (.3 4 M) (3. 5 ] M M s i.e. the over all rate law is rate Integrated Rate law Expression : (3.7 8 M s ) [] [B] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

8 8 Integrated rate law expression provide the predicted temporal evolution in reactant and product concentrations for reactions having an assumed order dependence. () Zero-order Reaction : Consider the following elementary reaction For zero-order reaction, the rate law is P is rate constant. rate r d[] d[p] r [] d[] d[] yields If at t, the initial concentration is [] and the concentration at t t, is [], then integration [] d[] [] t t t [] [] t This is integrated rate equation for a zero-order reaction in terms of reactant. d[p] d[p] at t, [P] and at t t, [P] [P] then integration yields [P] [P] d[p] [P] t t t [P] t UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

9 9 This is integrated rate law equation for a zero-order reaction in terms of product. i.e. [] [] t [P] Graph representation of zero-order reaction [] [] t [] t + [] y mx + c Graph of reactant vs time. [P] y t mx Graph of concentration of product vs time. [] [] t [] [] t (i) UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

10 When t then [P] and t t then [P] [P] d[p] 3 d[p] 3 t [P] d[p] 3 t [P] 3 t [P] 3t (ii) (iii) 3 t [P] [] [] below. Problem. Find the integrated rate law expression for an elementary zero order reaction given B P Sol. B P The rate law of above elementary reaction is given below d[] d[b] d[p] [] [B] d[] [] d[] [] t t t [[] [] ] t UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

11 [] [] t (i) d[b] [B] d[b] [B] t t t [B] [B] t [B] [B] t (ii) d[p] [P] d[p] t [P] t (iii) From equation (i), (ii) & (iii) we get [B] [B] [P] t [] [] () First-order reaction Consider the following elementary reaction P If the reaction is first order with respect to [], the rate law expression is Rate d[] d[p] [] is rate constant r d[] [] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

12 d[] [] yields If t, the initial concentration is [] and the concentration at t t, is [], then integrating [] d[] [] [] t [] ln t [] [] ln t [] [] [] e t.(i) or [] ln [] t (ii) Using this idea, the concentration of product with time for this first-order reaction is : [P] + [] [] [P] [P] [] [] [] [] e t Graph representation of first order reaction [P] [] ( e t ).(iii) [] [] e t UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

13 3 Plot of concentration vs time. [] ln [] t ln [] t + ln [] t / i.e. half life time of first order reaction [] ln [] t Plot of log [] vs time. [] when t t / ; then [] [] ln [] t / ln t / UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

14 4 t /.693 Problem. The half life for the first order decomposition of N O 5 is.5 5 s. How long will it tae for a sample of N O 5 to decay to 6% of its initial value? Sol. We now that, t / t 4.5 s / The time at which the sample has decayed to 6% of its initial value then t [] ln [] ( ) t.33 log 6.59 T.5 4 s Problem. Find the t 3/4 i.e. 3 4 life time of first order reaction. P Sol. P Integrated rate law expression is [] ln [] t when t t 3/4 than [] [] 3 4 [] s [] then ln [] 4 [] [] 4 t 3/4 ln 4 t 3/4 t 3/4 ln 4 ln.38 UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

15 5 (3) Second-order reaction : (Type I) Consider the following elementary reaction, P yields If the reaction is second order with respect to [], the rate law expression is is rate constant rate r d[] [] d[] [] d[] d[p] r [] d[] [] If t, the initial concentration is [] and the concentration at t t, is [], then integration [] d[] [] [] [] [] t t t t t (i) [] [] The concentration of product with time for second order reaction [] [P] [] or [] [] eff [] eff [] t then [] [] eff t [] t UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

16 [] [P] [] t / i.e. Half-life time of second order reaction (type I) eff t [] [] [] when t t / then [] [] [] eff t / [] [] [] t / Second-order reaction (Type II) [] eff Second order reactions of type II involves two different reactants and B, as follows B P 6 (ii) [B]. ssuming that the reaction is first order in both and B, the reaction rate is d[] d[b] d[p] r [][B] If t then the initial concentration are [] & [B] and the concentration at t t, are [] & The loss of reactant i.e. the formation of product is equal to [] [] [B] [B] [P] [B] [] + [] [B] then d[] [][B] d[] [][B] the integration yield [] d[] [][B] [] t [] [] d[] [B] [] [] [] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

17 7 [] d[] [] [] [] t let [B] [] The solution to the integral involving [] is given by dx x(c x) C x ln C x Using this solution to the integral, the integrated rate law expression becomes [] ln [] [] ln [] [] [] [] [] [] [] ln ln [] [] t t t [B] [] [] [B] [] [] ln ln [] [] t [B] [B] ln ln [] [] [B][] ln [B] [] [] [B] ln [B] [] [][B] t t t Graph representation of second order reaction of type I t [] [] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

18 8 [] [] t Y mx + C Plot of concentration vs time (4) nth order reaction where n : n nth order reaction may be represented as n Products the rate law is, rate d[] r [] n n where is rate constant for nth order reaction d[] n [] n d[] n [] n If at t, the initial concentration is [] and the concentration at t t, is [], then integration yields UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

19 9 [] d[] n [] [] t n Let n ( n) [] [] n [] t [] n [] t n [] n n n [] [] t () t / i.e. Half life time of nth order reaction n n n [] [] t [] [] Where t t / then [] [] n [] [] n n t / t / n (n ) [] n t / n (n ) [] n () i.e. t / [] n (3) UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

20 Thus we can say that t / of the reaction is inversely proportional to the initial concentration of reactant, except first order reaction. So, for a first order reaction (n ), t / is independent on [] for a second order reaction (n ), t / is dependent on [] t / [] for a nth order reaction t / [] n reaction. Note : For the elementary reaction, the order of reaction is equal to the molecularity of the Problem. Find the rate law for the following reaction. Sol Rate law is () () d[b] d[c] [] [] (3) d[] [] + [] ( + )[] Problem. Find the rate law for the following reaction. UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

21 Sol. () d[] [] () (3) (4) d[b] d[c] d[d] [] [B] 3 [B] [] ( + 3 ) [B] [B] 3 [B] Consecutive elementary reaction (Series reaction) : Consider the following series reaction scheme I I P In this, the reactant decays to four intermediate I, and this intermediate undergoes subsequent decay resulting in the formation of product P. The above series is elementary first order reaction. Then the rate law expression is : d[] d[i] d[p] [] () [] I [I] () I [I] (3) Let only the reactant is present at t such that [], [I], [P] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

22 then the rate law expression is d[] [] [] d[] [] [] [] [] e t (4) The expression for [] is substituted into the rate law of I resulting in d[i] [] I [I] t I [] e [I] d[i] I[I] [] e This differential equation has a standard form and after setting [I], the solution is The expression for [P] is t [I] t It I e e [] [] [P] So [P] [] + [I] + [P] [] [] [I] e It I e t I [] b Case I. Let >> I fast slow i.e. I P then I UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

23 3 and t e [P] [] [] [I] [] [] t [] t It e e e I [P] e t I I e t [] i.e. [P] t e [] e [] It t [P] [] I ( e ) The rate of formation of product can be determined by slowest step. [] [] t e () [I] [] t It (e e ) [ ] I [I] [] It ( e ) [I] It [] ( e ) () t The graph representation for case I i.e. when >> I. [P] [] I ( e ) (3) UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

24 4 Case II. I >> slow I fast I P I I t [P] [] (e ) [] [] e t The graph representation of case II i.e. when I >>. The Steady-State pproximation. UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

25 5 The steady-state approximation assume that, after an initial induction period, an interval during which the concentration of intermediate I rise from zero, and during the major part of the reaction, the rates of change of concentration of all reaction intermediate are negligibly small. d[i] Problem. Consider the following reaction I I P assuming that only reactant is present at t, what is the expected time dependence of [P] using the steady state approximation? Sol. The differential rate expression for this reaction are : d[] d[i] d[p] [] [] I [I] I [I] pplying the steady sate for I we get d[i] [] I [I] I [I] [] [I] [] I t and [] [] e [I] I [] e t UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

26 6 then d[p] I [I] I [] e t I [] e t [P] d[p] t [] e t [P] [P] a[] ( e t ) t [] ( e ) This is expression for [P]. equation Problem. Using steady state approximation find the rate law for d[p] for the following given 3 I I P Sol. d[] d[i ] d[i ] d[p] [] [I ] + [] 3 [I ] + [I ] 3 [I ] I & I are intermediate & apply steady state approximation on intermediate, we get d[i ] [I ] + [] [I ] [] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

27 7 d[i ] 3 [I ] + [I ] [I ] [I ] [] [I ] [] 3 3 and d[p] d[p] 3 [I ] 3 3 [] [] N O 5. Problem. Using steady state approximation, derive the rate law for the decomposition of On the basis of following mechanism. N O 5 a NO +NO 3 N O 5 (g) 4NO (g) + O (g) a ' 5 NO NO N O b 3 NO NO NO O NO c 5 NO N O NO NO NO Sol. The intermediate are NO & NO 3. The rate law are : d[no] b [NO [[NO 3 ] c [NO][N O 5 ] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

28 8 d[no 3] d[n O 5] a [N O 5 ] a [NO ][NO 3 ] b [NO ][NO 3 ] a [N O 5 ] + a [NO ][NO 3 ] c [NO][N O 5 ] and replacing the concentration of intermediate by using the equation above gives d[n O 5] [N O ] a b 5 a ' b' Parallel Reaction : Parallel reaction are those reaction in which the reactant can form one of two or more products. Consider the following parallel reaction in which reactant can form two products B & C. The rate law for the reactant and products are : d[] d[b] d[c] B [] c [] ( B + C )[] () B [] () C [] (3) Integration of equation () with the initial condition [] and [B] [C] yields Integration of equation (), we get [B] ( [] [] B C)t e (4) t d[b] B C [] B B [] e t ( )t UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

29 9 [B] [] B (B C)t e ( ) t B C B[] [B] e B C B C ( )t (5) Similarly [C] C[] e B C B C ( )t (6) i.e. the ratio of concentration of product is [B] [C] B C reactant. i.e. the product concentration ratio remains constant with time. The yield,, is defined as the probability that a given product will be formed by decay of the i i n n The quantum yield of product [B] is The quantum yield of product [C] is B yield of [B] B C C yield of [C] Problem. Find the quantum yield of [B] & [C] in the following reaction B B C C UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

30 3 Sol. d[b] d[b] d[c] [] [] () [] () then B C Problem. Find the quantum yield of [B], [C] & [D] in the following reaction. Sol. d[b] d[b] d[c] d[d] [] [] () [] () [] (3) The ratio of formation of product [B], [C] & [D] are d[b] d[c] d[d] : : : B 5 5 UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

31 3 C 5 5 D 5 5 Reversible reactions and Equilibrium Consider the following reaction in which the forward reaction is first order in, and the bac reaction is first order in B : B B The forward and bac rate constant are & B. Then rate law are d[] d[b] [] + B [B] [] B [B] Only reactant is present at t and the concentration of reactant and product for t > must be equal to the initial concentration of reactant. [] [] + [B] then d[] [] + B [B] [] + B ([] []) [] ( + B ) + B [] [] d[] [] ( ) [] [] B B t dx a bx ln (a bx) b Using this relationship we get UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

32 3 B [] B B [] ln [] ( ) [] t []( B) B[] ln [] ( ) [] B B []( ) [] ln [] B B []( ) [] [] B B t( + B ) ( + B )t ( B)t e ( )t [] ( + B ) B [] [] e B [] [] Then [B] B[] [] e B e B ( B)t B B e [] ( B)t ( B)t B s t, the concentration reach their equilibrium values then [] eq lim [] [] B t B & [B] eq [] [] [] It follows that the equilibrium constant of the reaction is B C [B] [] eq eq B i.e. C C is equilibrium constant in terms of concentration. B UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

33 33 t equilibrium, the forward and reverse rates must be same so, K [] eq B [B] eq Problem. Using the following equation C mechanism, (i) B (ii) B 3 C (a) Find rate of reaction? (b) Find rate of reaction when (i) is fast. Sol. (a) From the rate law d[] d[b] d[c] [] [B] [] [B] 3 [B] 3 [C] [B] is intermediate then we apply SS then we get [] [B] 3 [B] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

34 34 [B] [] 3 then (b) When (i) is fast then d[c] [] 3 3 [] [B] i.e. [] [B] & d[c] 3 3 [B] [B] Problem. Using the following equation Mechanism NO + F NO F + F (slow) F + NO NO F (fast) Find the rate of reaction. Sol. From the rate law NO + F NO F d[nof] d[f] [NO ][F ] + [F][NO ] [NO ][F ] [F][NO ] (F is intermediate) [F] [F ] d[nof] [NO ][F ] + [F][NO ] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

35 35 [F ][NO ] rrhenius Equation (Expression) The following empirical relationship between temperature (T), rate constant () and activation energy (E a ) is nown as the rrhenius expression : K e E a / RT is constant nown as the frequency factor or rrhenius pre-exponential factor. & E a is temperature independent. The unit of is always equal to the unit of rate constant () E a / RT e.() taing natural log of this equation, we get Ea ln ln RT Ea or log log.33 RT () (3) Problem. Prove that on increasing the activation energy, the rate constant will be decreasing and on increasing the temperature, the rate constant will be increasing. E Sol. e a / RT Ea ln ln RT when E a increase then the value of decreases. i.e. E a then When T increase then the value of increase. Ea RT Ea RT increases and the value of ln i.e. will be decreases and the value of ln i.e. will be UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

36 36 i.e. T then The graph of ln vs T is given below Ea ln ln RT ln y Ea ln RT T m. x + C Problem. Using the given equation find the value of & E a. ln J.3 T Sol. We now that e E a / RT Ea ln ln RT i.e. ln.3 & e i.e. Ea RT T E a R 8.34 UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

37 37 E a 83.4 J mol Problem. When temperature is increased then t / of reaction will be (a) remains constant (c) decreased (b) increased (d) first increase and then decrease Sol. We now that t /.693 E & e a / RT t / t /.693 e E a / RT E a / RT e i.e. on increasing T, t / of the reaction will decrease. i.e. T then t / i.e. The correct answer is (C). E a / RT e decrease then we can say that on increasing temperature (T), the Variation of rate constant with temperature E We now that e a / RT E ln ln a RT If and be the value of rate constant at temperature T and T, we can derive E ln a T T R T T or E log a T T.33 R T T Temperature Coefficient. UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

38 38 The ratio of rate constant of a reaction at two different temperature differing by degree is now as temperature coefficient. i.e. Temperature coefficient T T Standard Temperature coefficient to Problem. In the reaction mechanism, Ea 3, Ea3, Ea X Y Z P, 3 Find the overall rate constant ( overall ) and ctivation energy E a (overall). Sol. From the above reaction, the ate of formation of product is and d[p] d[z] 3 [Z] () [X][Y] [Z] 3 [Z] [X][Y] ( + 3 )[Z] [X][Y] [Z] [ >> 3 ] SS on intermediate. UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

39 39 then [Z] [X][Y] then we find, d[p] 3 [Z] d[p] [X][Y] 3 overall [X][Y] i.e. overall 3 () overall 3 overall. E overall RT e E E 3 RT RT 3e E RT e e i.e. overall 3 and E overall RT e E E3 E e RT RT RT i.e. E RT E E E RT overall 3 E overall E + E 3 E Problem. What is the energy of activation of the reaction if it rate doubles when temperature is raised.9 to 3 K. Sol. We now that E T T log a.33 R T T UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

40 4 E log a R 3 9 log E a Ea J E a 5.45 J Problem. plot of log versus T gave a straight line of which the slope was found to be. 4 K. What is the activation energy of the reaction. E Sol. e a / RT log log y Ea log.33 RT Ea log.33 R T m x + C where m slope of line then Ea slope.33 R E a.33 R (slope) (. 4 K). 5 J mol Fast Reaction. Fast reactions are studies by following methods () Stopped-Flow technique : For reaction that occur on timescales as short at ms ( 3 s) () Flash photolysis technique : Reaction that can be triggered by light are studied using flash photolysis. UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

41 4 (3) Perturbation-relaxation methods : chemical system initially at equilibrium is perturbed such that the system is not longer at equilibrium. By following the relaxation of the system bac toward equilibrium, the rate constant for the reaction can be determined. The temperature perturbation or T-jump are most important type of perturbation. Problem. Using the T-jump method find out the relaxation time () of the following reaction, ' ' B B Then Sol. Let a be the total concentration of ( + B) and x the concentration of B at any instant. d[b] dx rate (a x) (x) If x e is the equilibrium concentration, then x x x e or x x + x e Since d( x) dx, we have at equilibrium, d( x) dx (a x e x) (x e + x) and x x e. Hence (a x e ) x e then d( x) ( + )x r x UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

42 4 where r + relaxation rate constant then dx r x dx x t r x x e rt Then reciprocal of r i.e. It is represented by r is called relaxation time. r ( ) Problem. Find the relaxation time for the following reaction. B C Sol. Let a be the total concentration and x the concentration of B which is equal to the concentration of C. Then, the rate law is given by r dx (a x) x Now x x x e x e equilibrium concentration of x d( x) (a x e x) (x e + x) (a x e ) x x e x e x (x) UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

43 43 at equilibrium, dx, hence (a x e ) x e we get d( x) + x x e x (x) x is very small than (x) is neglected, d( x) e r ( x ) x x where r + x e is the relaxation rate constant. and d( x) dx x r x r dx r x x x e rt The relaxation time in this case is e ( x ) x e Problem. The relaxation time for the fast reaction B is µs and the equilibrium constant is. 3. Calculate the rate constant for the forward and the reverse reactions. Sol. 6 5 s s K equilibrium constant. 3 UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

44 ( ) s 5 s The Collision Theory of Bimolecular Gaseous Reaction. The reaction between two species taes place only when they are in contact i.e. the reactant species must be collide before they react. Consider the bimolecular elementary reaction. + B P rate v d[] d[b] [][B] The rate of reaction to be proportional to the rate of collision i.e. the mean sped of the molecules, their collision cross-section () and the number of densities of and B. Using inetic theory of gases, the rate of bimolecular collisions per second per cm 3 between unlie molecule is given by Z B n n (d ) B V 8T / Where n & n B are number of and B molecules, d V is the average collision diameter d db defined as m and µ is the reduced mass defined as mb m m B Z B collision frequency UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

45 45 The detailed analysis of the bimolecular collisions leads to the result that the number of collision per second per cm 3 between molecules and B is given by E rate Z / RT B e no. of collision where E Energy generated by collision then the rate of relative collision is given by dn Z B e E / RT / d(n) 8T n n B (d av ) e E / RT d[] 8RT N N [] N [B] (d V ) e n [] nb & [B] N N / / d[] 8RT N [B][] (d V) e / E / RT d[] 8RT N (d av ) T [][B]e E / RT E / RT let M N (d ) av 8R / then we now that d[] M T [][B]e d[] [][B] E / RT () E then / RT M T e () UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

46 46 The collision theory can be generalized by introducing the steric factor, P, into the equatiohn for the bimolecular rate constant. E Then PM T e / RT Relation between E a and E : E By between equation; e a / RT E By collision theory, PM T e / RT Taing natural log we get ln Ea ln () RT E ln ln P + ln M + ln T () RT Differentiate both equation () and () with respect to T we get d ln dt d ln d Ea Ea d T dt dt RT R dt d ln dt E a RT (a) and d ln dt d ln P d ln M d ln T d E dt dt dt RT d ln dt E d ln T d T dt R dt d ln dt E (a) T RT Comparing equation (a) & (a) we get E RT a T E RT UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

47 47 E a RT T RT E RT RT E a E (3) The expression for rrhenius pre-exponential factor using collision theory We now that E a / RT e [by rrhenius equation] E / RT PM T e [by collision theory] then E a / RT E / RT e PM T e e RT E / RT E / RT PM T e e RT E RT RT e E PM T e / RT e / PM T / PM T e () We now that M N d V 8R / then / 8R / V P N d T e () ctivated complex Theory of Bimolecular Reaction or Transition state Theory or Eyring Equation The activated complex forms between reactants as they collide. The difference between the energy of the activated complex and the energy of the reactants is the activation energy, E a. UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

48 48 (a) Exothermic reaction (b) Endothermic reaction ccording to Eyring the equilibrium is between reactants and the activated complex. Consider and B react to form an activated complex that undergoes decay, resulting in product formation. UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

49 49 The activated complex represents the system at the transition state. This complex is stable. # B (B) Product where (B) # is the activated complex and is the equilibrium constant between reactants and activated complex. If (B) # one of the vibrational degrees of freedom has become a translational degree of freedom. From the classical mechanics, Energy B T RT N B Boltzmann constant from the quantum mechanics, energy than hv v hv RT N RT h N The vibrational frequency v is the rate at which the activated complex move across the energy barrier i.e. the rate constant is identified by v. Then the reaction is r d[] (B) (B) # (B) [][B] # # (B) # [][B] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

50 5 then r d[] (B) # r v [][B] RT N h [][B] for conventional rate r d[] [][B] [][B] RT N h [][B] RT N h i.e. v frequency () where equilibrium constant eq Relation between and G # : equilibrium constant e # G / RT & G # H # TS # where G #, H # & S # are the standard free energy of activation, enthalpy of activation and entropy of activation. eq frequency frequency e G # / RT RT e N h RT () N h # # ( H TS )/ RT RT e N h # # H / RT S / R. e (3) UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

51 5 H # E # + n g RT RT e. e. e N h E # / RT n # g S / R (4) and n g difference is number of moles between transition state and reactant Relation between E # and E a We now that RT E # / RT n # g S / R RT e. e. e. N h N h On taing log ln # # E S R ng ln ln T RT R N h On differentiate above equation d ln dt # E RT T or d ln dt # E RT T and d ln dt E a RT (from arrhanius equation) then E RT a # E RT T E a E # + RT (5) Relation between E & E # E a E # + RT E RT E # + RT E Collision energy UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

52 5 E E # + RT Value of, using above equations From Eyring theory and rrhenius theory we have rate constant E a / RT e E # n # a / RT E / RT g S / R RT e e. e. e. N h (E RT)/ RT n # g S / R RT e. e. e. N h e # ng S / R RT. N h (6) Problem. Consider the decomposition of NOCl, NOCl(g) NO(g) + Cl (g) The rrhenius parameters for this reaction are. 3 M s and E a 4 J mol. Calculate H # and S # for this reaction with T 3 K. Sol. We now that H # E # + n g RT where n g difference in number of moles between activated complex and reactant. H # E # + ( ) RT H # E # RT & E a E # + RT then E # E a RT H # E # + RT E a RT RT E a RT H # 4 J mol (8.34 J mol K ) (3 K) 99. J mol We now that UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

53 53 e # ng S / R RT. N h # S / R e N ng h e [n g ] RT Taing log # S R ln N h e RT S # (8.34 J mol K ) S #.7 J mol K ( M s ) (6.6 J-s) (6.3 ) ln e (8.34 Jmol K ) (3 K) Note : for unimolecular n g for bimolecular n g for trimolecular n g & H # E # + n g RT [ E a E # + RT] E a RT + n g RT E a H # + RT n g RT for unimolecular E a H # + RT for bimolecular E a H # + RT for trimolecular E a H # + 3RT The Pre-equilibrium pproximation Consider the following reaction 3 B I P UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

54 54 (i) First, equilibrium between the reactants and the intermediate is maintained during the course of the reaction. (ii) The intermediate undergoes decay to form product. Then the rate law expression is d[p] 3 [I] I is in equilibrium with the reactant then [I] [][B] equilibrium constant C [I] C [][B] d[p] d[p] 3 C [][B] eff [][B] The Lindemann Mechanism eff 3 C 3 Lindemann mechanism for unimolecular reactions involves two steps. First reactants acquire sufficient energy to undergo reaction through a bimolecular collision. + * + In this, * is the activated reactant and undergoes one of two reactions. Then, rate of product formation is * + + * P d[p] [ * ] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

55 55 & rate of formation of * * d[ ] [] [][ * ] [ * ] pplying the steady-state approximation * d[ ] [] [][ * ] [ * ] [ * ] [] ( [] ) Then d[p] [] [] It state that the observed order dependence on [] depends on the relative magnitude of [] versus. t high reactant concentration, [] > and d[p] [] P n RT V [] i.e. the product formation is first order at high pressure. t low reactant concentration > [] and d[p] [] i.e. at low pressure, the rate of formation product is second order in []. d[p] Then R [][] [] uni is the apparent rate constant for the reaction defined as uni [] [] [] uni [] and uni [] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

56 56 when [] >> i.e. at high concentration uni when [] << i.e. at low concentration uni [] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

57 57 CTLYSIS catalyst is a substance that participates in chemical reaction by increasing the rate of reaction, yet the catalyst itself remains intact after the reaction is complete. The mechanism describing a catalytic process is as follows : S + C SC SC P + C where S represents the reactant; C is catalyst and P is the product. The reactant or substratecatalyst complex is represented by SC and is an intermediate. The rate expression for product formation is d[p] [SC] () Because SC is an intermediate than apply S.S.. on the formation of SC. d[sc] [S][C] [SC] [SC] [SC] [S][C] [S][C] m () m composite constant then d[p] [S][C] m (3) The relationship between initial concentration and the concentration of all species present after the reaction is : [S] [S] + [SC] + [P] [C] [C] + [SC] then [S] [S] [SC] [P] (4) & [C] [C] [SC] (5) UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

58 58 Substituting these values in equation (), we get ([S] [SC] [SC] [P]) ([C] [SC]) m K m [SC] ([S] [SC] [P]) ([C] [SC]) then rate of the reaction becomes [S] [SC] [C] [S] [C] m d[p] R [S] [C] [S] [C] m Case I. [C] << [S] i.e. much more substrate is present in comparison to catalyst. Then R [S] [C] [S] if m < [S] m [S] [C] then R [S] [C] i.e. zero order reaction with respect to substrate. & R [C] [S] [C] m UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

59 59 when concentration of substrate [S] >> m then reaction rate R [C] R max i.e. the rate of reaction will reach a limiting value where the rate becomes zero order in substrate concentration. Case II. [C] >> [S] [S] [C] R [C] m i.e. the reaction rate is first order in [S], but can be first or zero order in [C] depending on the size of [C] relative to m. Michaelis-Menten Enzyme Kinetics. Enzyme are protein molecules that serve as catalysts in a chemical reaction. mechanism. The inetic mechanism of enzyme catalyst can be described using the Michaelis-Menten E + S ES E + P E is enzyme, S substrate, ES is enzyme-substrate complex and P is product. The mechanism of above reaction is similar to catalytic mechanism. [S] [E] rate [S] [E] m But in this mechanism substrate concentration is greater than that of enzyme UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

60 6 i.e. [S] >> [E ] then rate of formation of product in enzyme catalyst is [S] [E] R [S] m () The composite constant m is referred to as the Michaelis constant in enzyme inetics and the equation is referred to as the Michaelis-Menten rate law. When [S] >> m, the Michaelis constant can be neglected, resulting new expression for the rate. R [E] R max The reciprocal equation of equation () is the Lineweaver-Bur equation i.e. [S] [E] R max[s] R [S] [S] m m R () R R [S] m max max This equation is nown as Lineweaver-Bur equation. The plot of reciprocal of rate is nown as Linewearver-Bur plot. is nown as turn over numberof the enzyme. The turn over number is the maximum number of substrate molecules per uit time that can be converted into product. UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

61 6 This is Linewearver-Bur plot. We now that d[p] [E] [S] [S] m m Case I. ` [S] >> m d[p] R [E] [S] [E] [S] i.e. rate is maximum due to all enzyme are present R R max [E] This is zero order w.r.t. substrate. Case II. If [S] m d[p] [E] [S] [E] [S] R [S] [S] m Case III. If [E] Rmax R [S] << m d[p] E [E] [S] m This is first order w.r.t. substrate. R max[s] R m UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

62 6 This is graph between initial rate and concentration of substrate. G.S. Eadie Plot We now that, R m R R [S] max max Multiplying with R, R R R R m R R [S] max max R R m R R [S] max max Multiplying with R max m, R max m R R max max m m Rmax m R max[s] R R R max m R R [S] m or R [S] R max m R m UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

63 63 R [S] R R max m m y mx + C Homogeneous and Heterogeneous Catalysis. homogeneous catalyst is a catalyst that exist in the same phase as the species involved in the reaction, and heterogeneous catalysts exist in a different phase. Enzymes surve as an example of a homogeneous catalyst, they exist in solution and catalyze reactions that occur in solution. In heterogeneous catalysis reaction, an important step in reactions involving solid catalysis is the absorption of one or more of the reactants to the solid surface. The particles absorb to the surface without changing their internal bonding. n equilibrium exists between the free and surface-absorbed species or adsorbate and surface adsorption and deadsorption can be obtained. critical parameter in evaluating surface adsorption is the fractional coverage,, defined as Number of adsorption sites occupied Total numbe rof adsorption site The variation of with pressure at fixed temperature is called adsorption isotherm. (a) The Langmuir Isotherm The simplest inetic model describing the adsorption process is nown as the Langmuir model, where adsorption is described by the following mechanism UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

64 64 R(g) + M(surface) a RM (surface) d R is reagent, M is an unoccupied absorption site of catalyst and RM is an occupied adsorption site. a and d is the rate constant for adsorption and deadsorption. Three approximations are employed in the Langmuir model : () dsorption is complete once monolayer coverage has been reached. () ll adsorption site are equivalent and the surface is uniform (3) dsorption and deadsorption are uncooperative processes. The occupancy state of the adsorption site will not affect the probability of adsorption or deadsorption for adjacent site. The rate of change in will depends on the rate constant for adsorption a, reagent pressure P and the number of vacant site which is equal to the total number of adsorption sites, N, times the fraction of sites that are open ( ) d a PN( ) The change in due to deadsorption is d d N t equilibrium, the change in with time is zero i.e. ( a PN + d N) a PN a P a P d a d a P d P d d P P where is the equilibrium constant defined as a. d This equation is the equation for the Langmuir isotherm. UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

65 65 In many instances adsorption is accompanied by dissociation of the adsorbate, a process that is described by the following mechanism : R (g) + M(surface) a RM (surface) d d a P{N( )} & d d (N) The condition for no net change leads to the isotherm. (P) / (P) / adsorption. i.e. the surface coverage now depends more wealy on pressure than for non-dissociative (b) The BET isotherm. If initial adsorbed layer can be act as a substance for further adsorption, then, instead of the isotherm leveling off the some saturated value at high pressure, it can be expected to rise indefinitely. The most widely used isotherm dealing with multiplayer adsorption was derived by Brunauer, Emmett and Teller, and is called the BET isotherm. V V mono CZ ( Z) { ( C)Z} wth Z P vapour pressure above a layer of absorbate V mono volume of monolayer coverage. P P C constant The Langmuir-Hinshelwood mechanism for adsorption and catalysis. () Unimolecular surface Reaction. is reactant and S is the vacant site on surface. If r is the rate of the reaction, then according to the Langnuir-H. hypothesis, r UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

66 66 r pply S.S.. for the formation of [S]. d[s] [][S] [S] [S] [S] [][S] If C s is the total concentration of active site on surface, then the concentration [S] of the vacant sites on the surface is equal to the product of C s and ( ). Thus [S] C s ( ) lso, the concentration of S on the surface is, [S] then C s C s [][S] [] C s( ) or [] or [] or [] [] [] thus, r [] [] [] [] ()...() The concentration is expressed in terms of partial pressure Then r P P (3) UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

67 67 or r P Two limiting cases Case I. > P + r P r P it is first order w.r.t. Case II. << P + r r P P P P eq R P eq P eq Two situations arise depending upon the pressure : () at low pressure and eq P << so that It is first order w.r.t. P or []. r eq P (a) at high pressure; and eq P >> so that r it is zero order with respect to P or []. UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

68 68 () Bimolecular surface reaction + B P rate r B P P P then it follow the rate law B r P B B B B P P B B P P B B ( P P ) B B UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

69 69 PHOTOCHEMISTRY Photochemistry process involve the initiation of a chemical reaction through the absorption of a photon by an atom or molecule. When a molecule absorbs a photon of light, the energy is photon is transferred to the molecule. The energy of a photon is given by the Planc equation : E h c hv hv Planc constant J-s speed of light in vacuum 3 8 ms v frequency of light and wave length of light The phenomena of photochemistry of photochemistry as best explained by Jablonsi diagram. So, S, S, T & T are electronic level. S singlet UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

70 7 transition. T triplet Loss of excess electronic energy through the emission of a photon is nown as rediative The process by which photons are emitted in radiative transition between S and S is nown as fluorescence. The process by which photons are emitted in radiative transition between T and S is nown as phosphorescence. The life time for phosphorescence is longer ( 6 s) than fluorescence ( 9 s) Photo physical reactions are corresponding rate expression Process Reaction Rate bsorption/excitation S + hv S a [S ] Fluorescence S S + hv f [S ] Internal conversion S S ic [S ] Intersystem crossing S T s isc [S ] Phosphoresence T S + hv p [T ] Intersystem crossing T S s isc [T ] Quantum yield Number of events number of photons absorbed rate of process r intensity of light absorbed I abs The Bear Lambert Law. When a beam of monochromatic radiation of a suitable frequency passes through a solution, it is absorbed by the solution. UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

71 7 I intensity of incident light I intensity of transmitted light and I a Intensity of the light absorbed I I bsorbance of solution, log I C I log where absorbance C I C I () concentration of solution molar extinction coefficient (unit concentration length ) or molar absorption coefficient l Transmittance; T T path length I I I I () The absorbance of a solution is additive whereas the transmittance is multiplicative. Problem. monochromatic light is incident on solution of.5 molar concentration of an absorbing substance. The intensity of the radiation is reduced to one-fourth of the initial value after UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

72 7 passing through cm length of the solution. Calculate the molar extinction coefficient of he substance. Sol. From bear Lambert law I log C I I I 4.5 5% i.e. I I.5 4 log 4 Cl cm.5 mol dm dm 3 mol cm Problem. substance when dissolved in water at 3 M concentration absorbs percent of an incident radiation in a path of cm length. What should be the concentration of the solution in order to absorb go per cent of the same solution. Sol. % absorbed then 9% transmitted then T I I 9% log I I Cl log 9% log 9 3 () 9% absorbed then % transmitted then T I I % log I I log Cl C () UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

73 73 equation () equation () log 3 9 C log log 9 log 3 C C C.8 mol dm 3. Objective Questions ased in previous years of Gate and GRF examination. Problem. In carbon-dating application of radio isotopes, 4 C emits (JRF June ) () -particle () -particle (3) --particle (4) positron Sol C N -particle ( e ) Correct answer is () Problem. With increase in temperature, the Gibb s free energy for the adsorption of a gas on a solid surface. (JRF June ) () becomes more positive from a positive value () becomes more negative from a positive (3) becomes more positive from a negative value (4) becomes more negative from a negative value Sol. From Langmuir isotherm, the fractional coverage is P P P T Then higher the pressure or temperature lower the value of fractional coverage. UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

74 74 R(g) + M(surface) RM (surface) if decreases then the formation of RM decreases. i.e. rate of formation of RM decreases. This indicate that Gibbs from energy of adsorption become positive. So, increase the temperature, the Gibbs free energy of adsorption of a gas on a solid surface become more positive from a negative value. The correct answer is (3). Problem. One of the assumption made in the conventional activated complex theory is : () equilibrium is maintained between the reactants and the activated () equilibrium is maintained between the reactants and the product (3) equilibrium is maintained between the products and the activated complex (JRF June ) (4) equilibrium is maintained between the reactants, the activated complex and the products Sol. activated complex ccording to the Eyring the equilibrium is maintained between reactants and the + B (B) # Products (B) # is the activated complex. The correct answer is () Problem. For a reaction, the rate constant at 7 C was found to be 5.4 e 5 The activation energy of the reaction is (JRF June ) () 5 J mol () 45 J mol (3) 5 J mol (4) 5 J mol Sol. From rrhenius equation E a / RT e () & 5.4 e 5 () UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

75 75 From equation () & () we get E a / RT e e 5 E a RT 5 E a RT J K mol 3 K E a 47 J mol 5 J mol The correct answer is (4) Problem. The carbon-4 activity of an old wood sample is found to be 4. disintegration min g. Calculate age of old wood sample, if for a fresh wood sample carbon-4 activity is 5.3 disintegration min g (t / carbon 4) 573 year), is (JRF June ) () 5 year () 4 year (3) 877 year (4) 67 year Sol. ctivity dn N rate constant and N no. of atom ctivity of old wood N old 4. () ctivity of new wood N new 5.3 () From equation () & () we get N N old new or N N new old (3) We now that t / (4) t 573 / UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

76 76 We now that t.33 log N N t i.e. t.33 N log N new old The correct answer is (4). T log log year Problem. Using cuvettes of.5 cm path length, a 4 M solution of a chromphone shows 5% transmittance at certain wave length. The molar extinction coefficient of the chromphre at this wave length is (log.3) (JRF June ) () 5 M cm () 3 M cm (3) 5 M cm (4) 6 M cm Sol. Transmittance T I 5 5% I bsorbance Cl log I I Cl log 5 log Cl 4 M.5 cm M cm 4 M.5 cm.5 6 M cm UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

77 77. bar is : The correct answer is (4). Problem. The rate law for one of the mechanisms of the pyrolysis of CH 3 CHO at 5 C and Rate / 3 4 [CH CHO] 3/ The overall activation energy E a in terms of the rate law is : (JRF June ) () E a () + E a () + E a (4) () E a () + E a() E a (4) (3) E a () + E a() E a(4) (4) E a () E a() + E a(4) Sol. Rate / 3 4 [CH CHO] 3/ overall [CH 3 CHO] 3/ i.e. overall overall E a / RT e i.e. overall / 4 e e E a / RT E a 4 / RT 4 / 4 / and E a / RT e e E / RT / E a a / RT e E a 4 / RT e E a / RT E a / RT E a / RT 4 e e e or E E a a E a E a4 RT RT RT RT Ea E a4 E a Ea UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

78 78 The correct answer is (3). Ea E a4 E a Ea Problem. In the Michaelis-Menten mechanism of enzyme inetics, the expression obtained as V [E] [S].4 4 V [E] The value of 3 and (Michaelis constant, mol L ) are (JRF June ) ().4, 4 ().4 8, 4 (3).4 8, 4 (4).4, 4 Sol. We now that Michaelis Menten equation is: 3[S] [E] rate V [S] m () V [S] m [S] [E] 3 V [S] m [S] [E] [S] [E] 3 3 V m [E] [S] [E] 3 3 3V Multiply this equation by m we get V V V 3 mv 3 3 m m 3[E] m 3[S] [E] 3 m V [E] V [S] [E] m or V [S] [E] 3 V () [E] m m and V [E] [S] 4 V.4 (3) [E] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

79 79 Comparing equation () & (3) we get m 4 i.e. m 4 3 and m i.e. The correct answer is (3). Problem. The Langunier adsorption isotherm is given by P, where P is the pressure P of the adsorbate gas. The Langmuir adsorption isotherm for a diatomic gas undergoing dissociative adsorption is () P () P P P (JRF Dec.) (3) (P) (4) (P) / (P) (P) / Sol. R(g) + M(surface) a RM (surface) d then P P if R (g) + M (surface) a M (surface) d (P) / (P) / i.e. the correct answer is (4). Problem. The overall rate of following complex reaction UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

80 8 (fast equilibrium) + B C (fast equilibrium) 3 + C P + The steady state approximate would be () 3 [] 3 [B] () 3 [][B] 3 (3) 3 [][B] (4) 3 [][B] (slow) (JRF Dec.) Sol. (fast equilibrium) then [ ] [] () + B C (fast equilibrium) then 3 + C P + The rate of formation of product P is [C] [][B] () (slow) d[p] 3 [][C] (3) From equation () & () we get [ ] [] & [C] [][B] d[p] then d[p] i.e. the correct answer is () 3 [] [][B] 3 [] 3 [B] UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

81 8 Problem. The species 9 Ne and 4 C emit a position and -particle respectively. The resulting species formed are respectively (JRF June ) () 9 Na and 4 B () (3) 9 Na and 4 N (4) 9 F and 4 N 9 F and 4 B Sol. and 9 9 Ne F9 e 4 4 C6 N7 e i.e. the correct answer is (b). Problem. The half life of a zero order reaction ( P) is given by ( rate constant) (JRF June ) () (3) Sol. t t / / [].33 () t/ [] (4) t/ [] P d[] [] if t t / then [] [] Thus [] [] [] t / t / i.e. the correct answer is (). t / [] Problem. The concentration of a reactant undergoing decomposition was.,.8 and.67 mol L after.,. and 3. hr respectively. The order of the reaction is (JRF Dec. ) UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568

82 8 () () (3) (4) 3 Sol. If P d[] {[] [] } t t n [] n [] [] t [] t n [] [] concentration at t and [] concentration at t [] t [] t n []..8 [.] n. [.] n. () [.8] n.3 [.8] n.3 () Equation () divide by equation () we get [.] n [.8] n n [.5] n.5385 [.5] n.5.5 [.5] n i.e. second order reaction. i.e. the correct answer is (3). UGC POINT: Institute For CSIR, NET, GTE, JM Ph: , , 6568