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Vidyamandir Classes, , Chemical Kinetics, , Chemical Kinetics, This branch of chemistry deals with the study of rates of chemical reactions and the mechanism by which, they occur., While studying reaction rates, one deals with :, (a), How fast (or slow) the reactants get converted into products, (b), The steps or paths through which the products are formed (reaction mechanism), , BASIC, , Section - 1, , Rate of a Reaction :, In general, for a reaction : A ,  B, the behaviour of the concentration of the reactant and product, as the, reaction proceeds, is shown graphically., , From the graph, it is clear that the concentration of the reactant decreases and that of the product increases, as the reaction proceeds and the rate of the change of the concentration of the reactant as well as that of the, product is also changing., Rate of a reaction can, now, be defined in two ways :, 1., , [A] [B], , t, t, where Δ[A] and Δ[B] represents the change in the concentrations of ‘A’ and ‘B’ respectively over a time, interval Δt., , Average Rate of reaction (ravg) given by : ravg , , The average rate of the reaction between a time interval (t2 – t1 = Δt) can be determined from the above, graph by locating the concentration of ‘A’ or ‘B’ on this graph at the time instants t2 and t1 as shown., If [A]2 and [A]1 are the concentrations of the reactant ‘A’ at the time instants t2 and t1, then :, ,  [A]  [A]1 , ravg    2, ,  t 2  t1 , Self Study Course for IITJEE with Online Support, , Section 1, , 1

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Chemical Kinetics, , Vidyamandir Classes, ,  [B]2  [B]1 , Similarly from the plot of ‘B’ as a function of ‘t’, we have : ravg   t  t ,  2 1 , Note : The above expressions for ravg is equivalent to the slope of the line joining the points (t2, [A]2 ) and (t1, [A]1), or (t2, [B]2) and (t1, [B]1) as shown., 2., ,  0 and is represented, Instantaneous Rate of reaction (rinst.) can be calculated from ravg in the limit Δt , as :, rinst  , , Note : , , d[A] d[B], , dt, dt, , The above expression for rinst. is equivalent to the slope of the tangent from the plot of the concentration, of A’ or ‘B’ at any time instant ‘t’., , , , The rate of the reaction (rinst.or ravg) is always calculated as a positive quantity., , , , The rate of the change of the concentration of the reactant will be a negative quantity since its, concentration is decreasing with time., , , , The rate of the change of the concentration of the product will be a positive quantity since its concentration, is increasing with time., , , , The magnitude of the rates of the change of the concentration of reactants and products will be equal, in this case, as one mole of ‘A’ gives one mole of ‘B’ in the above reaction., , , , The rate of a reaction at any temperature depends on the concentration of the reactants and sometimes, on the concentration of some foreign substances (e.g a catalyst) being used in the reaction as well. The, representation of this dependence of the rate of the reaction on the concentrations is known as rate, law and this rate law is determined experimentally., , , , The above expression for rinst. is called as differential rate law., , In general for a reaction :, mA + nB ,  pC + qD, The rate of reaction can be expressed as follows :, Rate  , , 1 d[A], 1 d[B], 1 d[C], 1 d[D], , , , m dt, n dt, p dt, q dt, , Illustrating the concept :,  2NO(g) + Br2(g) is found to be, The rate of formation of NO(g) in the reaction 2NOBr(g) , 4, 1.6  10 M/s. Find the rate of overall reaction and rate of consumption of NOBr., , 2, , Section 1, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , We have :, , Chemical Kinetics, , d[NO], = 1.6 × 10–4 M/s., dt, , Now, Rate of overall reaction = –, Rate of consumption of NOBr =, , 1 d[NOBr], 1 d[NO] 1 d[Br2 ], , , = 0.8 × 10–4 M/s, 2, dt, 2 dt, 1 dt, , d[NOBr], = –1.6 × 10–4 M/s, dt, , Order of a Reaction :, By performing a reaction in actual in laboratory and carefully examining it, it is possible to express the rate, law as the product of concentrations of reactants each raised to some power., For example consider the reaction :, P + 2Q ,  R, The differential rate law is written as :, Rate  , , d[P], 1 d[Q], , dt, 2 dt, , Also, Rate can be expressed as Rate = k [P]m [Q]n, where k is called as rate constant or velocity constant or specific reaction rate., k is a characteristic of a reaction at a given temperature. It changes only when the temperature changes., The powers m and n are integers or fractions. m is called as order of reaction with respect to P and n is, called as order of reaction with respect to Q., The overall order of reaction = m + n, Note : The values of m and n are calculated from the experimental data obtained for a reaction and the powers m, and n are not related to the stoichiometric coefficients of the reactants., , Units of k :, In general, the rate law for a nth order reaction can be taken as :, dc,   kc n, dt, , dc, , n,  Note: rinst.   dt  kc , , , , where k : rate constant ; c : concentration and, , n : order of reaction, , dc / dt, , , , k=, , , , Units of k  (mol/L)1 – n (time)–1, , cn, , Self Study Course for IITJEE with Online Support, , Section 1, , 3

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Chemical Kinetics, , Vidyamandir Classes, , For a ‘zero’ order reaction (n = 0) :, Units of k  (mol/ L)1 (time)–1, , , Units are : mol/L/sec, mol/L/min, ... etc., , For a first order reaction (n = 1) :, Units of k  (time)–1, , , Units are : sec–1, min–1, hrs–1 etc., , For a second order reaction (n = 2) :, Units of k  (mol/ L)–1 (time)–1, , , Units are : L/mol/sec, L/mol/min, .... etc., , From the rate laws for the reactions given below, determine the order with respect to, Illustration - 1, each species and the overall order:, –, , +, , 2HCrO4 + 6I– + 14H ,  2Cr3+ + 3I2 + 8H2O, –, –, Rate = k [HCrO4 ][I ]2[H+]2, , (i), , (ii) H2O2 + 2I– + 2H+ ,  I2 + 2H2O, Rate = k [H2O2][I–], , SOLUTION :, (i), , The order of the reaction with respect to [HCrO4–] is 1; with respect to [I–] is 2 and with respect to [H+], is 2. The overall order of the reaction is 1 + 2 + 2 = 5, , (ii), , The order of the reaction with respect to [H2O2] is 1 and with respect to [I–] is 1. The overall order of, the reaction is 1 + 1 = 2, , Note : , , In (i) the stoichiometric coefficient of I– is 6 whereas the power coefficient (n) in the rate law is 2., , , , Reaction (i) may not take place in a single step. It may not be possible for all the 22 molecules to be, in a state to collide with each other simultaneously. Such a reaction is called a complex reaction., , , , A complex reaction takes place in a series of a number of elementary reactions., , Illustration - 2, obtained., , For a reaction 2 NO(g) + 2 H2(g) ,  N2(g) + 2 H2O(g) ; the following data were, e, , 1., 2., 3., , [NO](mol/L), 5 × 10–3, 15 × 10–3, 15 × 10–3, , (a), , Calculate the order of reaction., , (b), , Find the rate constant., , (c), , Find the initial rate if [NO] = [H2] = 8.0 × 10–3 M, 4, , [H2](mol/L), 2.5 × 10–3, 2.5 × 10–3, 10 × 10–3, , Section 1, , Rate(mol/L/s), 3 × 10–5, 9 × 10–5, 3.6 × 10–4, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , Chemical Kinetics, , SOLUTION :, Assuming rate law can be expressed as follows :, rate = k [NO]x [H2]y, By analysing the data :, From observations 1 and 2, we see that [H2] is constant and when [NO] is tripled, the rate is also tripled., , rate (r)  [NO], , x=1, From observations 2 and 3, we see that [NO] is constant ; when [H2] is increased four times, the rate also, increases four times., rate  [H2], , y=1, , , r = k [NO] [H2O], , , , The order of reaction w.r.t NO and H2 is 1 and the overall order of reaction is 1 + 1 = 2., Initial rate = k [NO] [H2] = 2.4 × (8 × 10–3)2 = 1.536 × 10–4 mol/L/s., , NOW ATTEMPT IN-CHAPTER EXERCISE-A BEFORE PROCEEDING AHEAD IN THIS EBOOK, , nth ORDER REACTIONS, , Section - 2, , Zero Order Reactions :, The rate law for zero order reactions (n = 0) is written as :, dc, d t = –k, , . . . . .(i), , Clearly, zero order reactions are those, whose rates are not affected by change in concentrations of reactants, (i.e., independent of concentration). The rates of such reactions only depend upon temperature. Most of, photochemical reactions are zero order reactions. Other examples are : decomposition of HI over the, surface of gold and NH3 over tungsten., Equation (i) can be rearranged and integrated to get the variation of the concentration of the reactants as a, function of time., Ct, , t, ,  dc    k dt, C0, , Thus,, , k =–, , 0, ,  C t  C0 , t, , , , [Ct – C0] = –k [t – 0], , . . . . .(ii), , where C0 = Initial concentration of the reactant, and Ct = Concentration of the reactant at any time instant ‘t’ after the reaction started, , Self Study Course for IITJEE with Online Support, , Section 2, , 5

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Chemical Kinetics, , Vidyamandir Classes, , , , From the above expression, it is clear that if we plot Ct as a function of time (t) then it will be a straight, line with a negative slope = –k and Y– Intercept of C0., , , , One can also define, Half Life (t1/2) which is equal to the time taken for the reactant’s concentration, to drop to 50% of its initial value., , C0, 2k, Thus, for a Zero order reaction, half life is directly proportional to initial concentration of the reactant., , Putting Ct = 0.5 C0 in equation (ii), we get t = t1/2 =, , Note : The equation obtained on integrating the differential rate law is known as Integrated Rate Law., For example : equation (ii) is the Integrated Rate Law for the Zero Order Reaction., , First Order Reactions :, The differential rate law for a first order reaction A ,  B; is given as :, dc, = –k c, dt, , ., , . . . . .(i), , This equation can be rearranged and integrated as follows to get the integrated rate law for a first order, reaction :, Ct, , , C0, , t, , dc,    k dt, c, , , , C, ln 0  kt, Ct, , or, , C, 2.303 log10 0  kt, Ct, , 0, , . . . . .(ii), , [ln is the Natural Log (base ‘e’)  loge], Sometimes, it is important to remember the log10 for the following values :, log10 2 = 0.301, log10 3 = 0.477, log10 4 = 0.602, log10 5 = 0.699, log10 6 = 0.778, log10 7 = 0.845, Also, remember that lnx  loge x = 2.303 log10 x, Putting Ct = 0.5 C0 in equation (ii), we get t = t1/2, , 2.303log10 2 0.693, =, k, k, , Thus, for a first order reaction half life is independent of the initial concentration of the reactant., For the reaction, , A, , Initial conc., , C0, , After time ‘t’, , 6, , Section 2, , Ct  C0 – x, , , , , B, , (following first order), , 0, x, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , Chemical Kinetics, , The integrated rate law can also be written as :, 2.303 log10, , C0,  kt, C0  x, , or, , 2.303 log10, , 1,  kt, 1 , , x, where x is the concentration of the reactant consumed and  is the degree of dissociation = C, 0, , Features of a First Order Reaction :, 1., , A first order reaction must follow above form of rate law for all time instants., This means if we are given value of C0 and values of x at different time instants [i.e (C0 – x) as, value of reactants after t], the values of k can be calculated for different time instants by using the, above first order law. If the reaction for which the data were given is a first order reaction, then all, values of k will approximately equal to each other., , 2., , The time for half reaction for a first order reaction is independent of initial concentration of reactants., , 3., , The concentration of reactants in a first order reaction decreases exponentially with time (see figure), [Ct = C0e–kt from (ii)], , Note that plot of log10 Ct vs t is linear. It is important to note that equation of this straight line is of the, form :, y = m x + b. Comparing it with Ist order rate law as follows :, k t = 2.303 log10, , , , C0, Ct, ,  k , log10C t  ,  t  log10C0 is the equation of line.,  2.303 , ,  k , Note that slope of the line = ,  and Y-intercept (OA) = log10 C0,  2.303 , , , Rate constant of a first order reaction can also be calculated by measuring the concentration of the, reactant at two time instants (if the initial concentration is not known)., If C1 and C2 are the reactant’s concentrations at two time instants ‘t1’ and ‘t2’ respectively, then we, have :, , Self Study Course for IITJEE with Online Support, , Section 2, , 7

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Chemical Kinetics, , and, , Vidyamandir Classes, , C, 2.303 log10 0  kt1, C1, , . . . . .(iii), , C, 2.303 log10 0  kt 2, C2, , . . . . .(iv), , Subtracting (iv) from (iii), we get :, C, 2.303 log10 1  k  t 2  t1 , C2, , , , After Ist half life : Ct = C0/2, nd, , After 2 half life : Ct = C0/2, , , , After, , nth, , Ct, half life :, C0, , 2, , Thus, k can be evaluated., , , Ct 1, , C0 2, , , , Ct, C0, , 1,  , 2, , 2, , n, , 1,    (where n = t/t = No. of half lives), 1/2, 2, , Note : The differential rate law can be integrated for a nth order reaction as follows :, Using :, Ct, , , , , C0, , dc, = –k cn, dt, , dc, cn, , [Provided that all the reactants are present in the same molar concentrations], , t, ,    k dt, 0, , , , 1  1, 1 ,  n 1  n 1   kt, (n  1)  C t, C0 , , . . . . .(v), , The above equation is the integrated rate law for a nth order reaction., Also, linear plot can be obtained for 1 / c nt  1 vs t. (n  1), Putting Ct = 0.5C0 in equation (v), we get t = t1/2=, From the above expression, it is clear that t1/2 , , 8, , Section 2, , 1, (n  1) C0n 1 k, ,  2n 1  1, , , , . . . . .(vi), , 1, Cn0 1, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , Chemical Kinetics, , Order of a reaction (n) can also be obtained by measuring the half life (t1/2) of a reaction at different values, of initial concentration (C0) of the reactant molecules.,  If C01, C02 are the initial concentrations of the reactants and (t1/2)1, (t1/2)2 are the corresponding half, lives, then using (vi), we get :, ,  t1/2 1  C02 n 1, =,  t1/2 2  C01 , log10, , Taking log on both sides, we get :, , Illustration - 3, , n=1+, ,  t1/ 2 1,  t1/ 2 2, , C , log10  02 ,  C01 , , The half life of first order decomposition of nitramide is 2.1 hour at 15C., NH2NO2 (aq) ,  N2O(g) + H2O(l), , If 6.2 gm of NH2NO2 is allowed to decompose, find :, (a), , time taken for nitramide to decompose 99%, , (b), , volume of dry N2O gas produced at this point at STP., , SOLUTION :, (a) Using first order kinetics, we have :, C0, kt = 2.303 log10 C, t, , , (b), , 0.693,  100 ,  t  2.303 log10 , , 2.1,  00  99 , , , 0.693 , k , , t1/2 , , , , , t = 13.96 hours, , 6.2 gm of NH2NO2  0.1 mol, NH2NO2(aq.) ,  N2O(g) + H2O(), Moles at t = 0, , 0.1 mol, , Moles at t = 13.96 hrs, , 0.1– 0.99 × 0.1, , 0.99 × 0.1, , , , Moles N2O = 0.099, , , , VN 2 O at STP = 0.099 × 22.4 L = 2.217 L at STP, , Self Study Course for IITJEE with Online Support, , Section 2, , 9

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Chemical Kinetics, , Vidyamandir Classes, , Molecularity, As already discussed, the order of a reaction is an experimental concept. The theoretical aspect of chemical, kinetics is molecularity., A complex chemical reaction is understood in terms of various indirect steps called elementary processes., The study of a reaction in terms of elementary processes is called as reaction mechanism. Now various, elementary steps occur at different rates. The slowest elementary process in the reaction mechanism is, called as rate determining step., Molecularity is defined as the number of ions or molecules or atoms taking part in an elementary process of, the reaction mechanism., In the rate determining step, when one molecule takes part, it is said to be a unimolecular reaction ; two, molecules take part, it is said to be a bimolecular reaction ; three molecules take part, it is said to be a, termolecular reaction., , Unimolecular :, 1., , Cyclopropane ,  propene, , 2., , O3 (g) ,  O2 (g) + O (g), , 3., , N2O5 (g) , , , N2O4 (g) + 1/2 O2 (g), , Bimolecular :, 1., , NO (g) + O3 (g) ,  NO2 (g) + O2 (g), , 2., , 2HI (g) ,  H2 (g) + I2 (g), , Termolecular :, 1., Note : , , 2 NO (g) + O2 (g) ,  2 NO2 (g), For a reaction : A ,  B, , in the rate law :, , rate = k [A]m [B]n, , Neither the order of reaction (m + n) nor the molecularity of a reaction can be predicted from, stoichiometric coefficient of a balanced reaction. The order of reaction is always to be determined, experimentally and molecularity is determined theoretically after studying the reaction mechanism., However as a theoretical idea sometime, we can have an approximate order of reaction equal to, molecularity (i.e., the number of molecules taking part in slowest elementary step for complex, reactions)., , , Order of a reaction can be fraction also. For example consider the following reaction:, (i), , H2 (g) + Br2 (g) ,  2 HBr (g), rate = k [H2] [Br2]1/2, , (determined experimentally), , 1 3, order of reaction = 1  , 2 2, , 10, , Section 2, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , (ii), , Chemical Kinetics, , CH3CHO (g)  CH4 (g) + CO (g), rate = k [CH3CHO]3/2, order of reaction =, , (determined experimentally), , 3, 2, , Also note that sum of stoichiometric coefficient (1 + 1 = 2) is not equal to the order of, reaction., , Pseudo First Order Reactions :, The molecularity of acidic hydrolysis of sucrose and esters is 2, whereas their order is 1. In both the, reactions water is in excess so that its concentration remains constant throughout the reaction. The rate of, reaction therefore depends only on the concentration of sucrose and ester in two reactions respectively., So the reactions in which the molecularity is 2 or 3 but they conform to the first order kinetics are known, as pseudo first order reactions OR pseudo unimolecular reactions., C12H22O11 + H2O + H+ ,  C6H12O6 (glucose) + C6H12O6 (fructose), CH3COOC2H5 (ester) + H2O + H+ ,  CH3COOH + C2H5OH, (In both the reactions, H+ ion acts as a catalyst), , Illustration - 4, , The rate law for the decomposition of gaseous N2O5,, , d [ N 2O5 ], 1,  k[ N 2O5 ], N2O5(g) ,  2NO2(g) + 2 O2 (g) is observed to be : r  , dt, A reaction mechanism which has been suggested to be consistent with this rate law is, K, , eq, , ,  NO2(g) + NO3(g), N2O5(g) , k, , 1  NO (g) + NO(g) + O (g), NO2(g) + NO3(g) , 2, 2, , (fast equilibrium), (slow), , k, , 2  2NO (g), NO(g) + NO3(g) , (fast), 2, Show that the mechanism is consistent with the observed rate law., , SOLUTION :, r = k1 [NO2] [NO3], , . . . . .(i), , and from the fast equilibrium step,, K eq , , [NO 2 ][NO3 ], [N 2O5 ], , Thus, [NO2] [NO3] = K [N2O5], , . . . . .(ii), , Self Study Course for IITJEE with Online Support, , Section 2, , 11

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Vidyamandir Classes, , Chemical Kinetics, , Methods of determining the order of a reaction, The following methods are commonly used for determining the order of a reaction :, , 1. Use of differential rate expressions :, We know that for an nth order reaction, r = k cn, Taking log on both sides, we have : log10 r = log10k + n log10c, Thus, if we plot log10 r as a function of log10c, it should be a straight line with a slope ‘n’., , , 2., , The order of the reaction can be evaluated., , Use of integrated rate laws :, This method can be used both analytically and graphically., In the analytical method, a certain order of the reaction is assumed and rate constant of the reaction is, calculated from the given date. The constancy of the ‘k’ values obtained suggests that the assumed order of, the reaction is correct. If not, we assume a different order for the reaction and again calculate the values of, ‘k’ using the new rate expression. This is essentially a trial and error method., In the graphical method, if the plot of log c versus t is a straight line then the reaction is of first order. Similarly,, the integrated expression for second order can be utilized to ascertain if the reaction order is 2 and so on., , 3., , Half life method :, Provided that all the reactants are present in the same molar concentrations, we have already derived the, expression for calculating the order of a reaction given by :, log10, , n=1+, ,  t1/ 2 1,  t1/ 2 2, , C , log10  02 ,  C01 , , where C01, C02 are the initial concentrations of the reactants and (t1/2)1, (t1/2)2 are the corresponding half, lives, , 4., , Isolation method :, Sometimes, the kinetics of a reaction are studied in successive experiments by keeping the concentrations of, all but one reactant in excess so that the result gives the order with respect to the reactant whose concentration, is changing significantly. The synthesis of HI (g) from H2 (g) and I2 (g) is pseudo first order with respect to, H2 (g) in the presence of large excess of I2 (g) and also pseudo first order with respect to I2 (g) in the, presence of large excess of H2 (g). Hence, overall the reaction is of second order., , Self Study Course for IITJEE with Online Support, , Section 2, , 13

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Vidyamandir Classes, , Chemical Kinetics, , 4., , Kinetics of some reactions can be studied by measuring optical rotation of reaction mixture at different, interval of time., , (i), , Inversion of Cane Sugar (C12H22O11), C12H22O11 + H2O + H+ ,  C6H12O6 (glucose) + C6H12O6 (fructose), The rate is measured by measuring the change in the angle of rotation (optical activity) by a polarimeter., Sucrose is dextro-rotatory, glucose is dextro-rotatory and fructose is leavo-rotatory. The change, produced in rotatory power in time t gives a measure of x, the quantity of sucrose decomposed in that, time. The total change in the rotatory power produced at the end of the reaction gives the measure of, C0, the initial concentration of sucrose., If r0, rt and r represent rotations at the start of reaction, after time t and at the end of reaction, respectively, then, , (ii), , , , C0  r0 – r, , , , C0 – x  rt – r , , , , r0  r, kt = 2.303 log10 r  r, t , , and, , x  r0 – rt, , Consider the following first order reaction on which A, B and C are optically active compounds which, rotate the place polarized light in the clockwise or anticlockwise direction., , A(soln.)  B(soln .) + C(soln.), Time, , 0, , t, , , , Total rotation in degrees, , r0, , rt, , r, , Let the specific optical rotations of A, B and C per unit concentration be, ra , rb and rc (including +ve, or –ve sign)., Let the initial concentration of A be C0 and the decrease in concentration till time t be x., A(soln .)  B(soln .) + C(soln .), , At time = 0, , C0, , 0, , 0, , At time = t, , C0  x, , x, , x, , At time = , , 0, , C0, , C0, , Such that, , 1, Co, k = In, t Co x, , Optical rotation at time = 0,, 16, , Section 3, , ra C0 = r0, , .…(i), Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , Chemical Kinetics, ra (C0  x )  x rb  x rc = rt ., , Optical rotation at time = t,, , ra C0  x(rb + rc  ra ) = rt, Optical rotation at time  ,, , r0  x (rb + rc  ra ) = rt, , .…(ii), , C0 (rb + rc ) = r, , .…(iii), , From equation (ii), x, , rt  r0, (rb + rc  ra ), , .…(iv), , 1, Since constant involved in x is (r + r  r ) , the same constant must appear in expression of C0 ., b c a, Thus, subtracting equation (i) from equation (iv), we get :, C0 =, , r  r0, (rb + rc  ra ), , .…(v), , Subtracting equation (iv) from equation (v), , C0  x , , , 5., , r  r1, (rb  rc  ra ), , 1 r r, k  ln  0, t r  rt, , In first-order reactions involving gases, sometime measuring the pressure of the reaction mixture is very, good method for measuring reaction rates., For example consider decomposition of arsine gas (AsH3)., AsH3 (g) ,  As (s) +, , 3, H (g), 2 2, , The rate of reaction is measured as the increase in pressure of the reaction mixture. Note that there is an, increase in number of moles of the gaseous products to the right, so as the reaction proceeds, there will be, an increase in pressure of contents (P  n)., Let the initial pressure of AsH3(g) is P0, if x is the decrease in pressure of AsH3(g) after time t., AsH3(g), ,  , , , As (s), , C0  initial pressure, , P0, , 0, , Ct  partial pressure, , P0 – x, , 0, , +, , 3, H 2 (g), 2, 0, 3, x, 2, , Arsenic is solid, so P(As) = 0, Self Study Course for IITJEE with Online Support, , Section 3, , 17

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Chemical Kinetics, , Vidyamandir Classes, , After time t, let Pt be the total pressure, then, Pt = P(AsH3) + P(H2) = (P0 – x) +, , , Pt = P0 +, , x, 2, , , , 3x, 2, , x = 2 (Pt – P0), , Now C0  P0, and Ct  P0 – x  P0 – 2 (Pt – P0)  3P0 – 2Pt, , , , kt = 2.303 log10, , P0, 3 P0  2 Pt, , On similar pattern, please try to write the expression for Ist order rate law for following first-order, reactions. (in terms of Po and Pt ), 1, O (g), 2 2, , 1., , N2O(g) ,  N2(g) +, , 2., , (CH3)3C – O – O – C(CH3)3 (g) ,  2 (CH3)2C = O(g) + C2H6(g), , ACTIVATION ENERGY (Ea), , Section - 4, , A mixture of magnesium and oxygen does not react at room temperature. But if a burning splinter is introduced, to the mixture, it burns vigorously. Similarly a mixture of methane and oxygen does not react at room, temperature, but if a burning match-stick is put in the mixture, it burns rapidly. Why it happen like this, that, some external agents has to be introduced in order to initiate the reaction ?, According to the theory of reaction rates “ for a chemical reaction to take place, reactant molecules, must make collisions among themselves ”. Now in actual, only a fraction of collisions are responsible, for the formation of products, i.e., not all collisions are effective enough to give products. So the collisions, among reactant molecules are divided into two categories :, Effective collisions, , and, , In-effective collisions, , Effective collisions are collisions between the molecules which have energies equal to or above a certain, minimum value. This minimum energy which must be possessed by the molecules in order to make an, effective collision (i.e., to give a molecule of products) is called as threshold energy. So it is the effective, collisions which bring about the occurrence of a chemical reaction., Ineffective collisions are the collisions between the molecules which does not posses the threshold energy., These can not result in a chemical reaction., , 18, , Section 4, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , Chemical Kinetics, , Now most of the times, the molecules of reactants do not possess the threshold energy. So in order to make, effective collisions (i.e., to bring about the chemical reaction), an additional energy is needed to be absorbed, by the reactant molecules. This additional energy which is absorbed by the molecules so that they achieve, the threshold energy is called as energy of activation or simply activation energy. It is represented as Ea., Note : , , The progress of the reaction can studied in a graph with energy of the reacting system. You can find, H and Ea from graph below., , , , All the colliding molecules must collide at a proper orientation for a collision to be effective, other than, “sufficient energy” factor. A reaction having more number of reactants (high molecularity) is expected, to be slower because it is less probable that all the reactant molecules, with energy greater than, activation energy collide simultaneously at proper orientation., , A reaction which needs higher activation energy is slow at a given temperature., 1, O (g) ,  NO2(g) is faster at ordinary temperature whereas the following, 2 2, , For example : NO(g) +, reaction :, , 1, O (g) ,  CO2(g) is slower at the same temperature as the value of Ea for the second, 2 2, reaction is much higher., , CO(g) +, , Arrhenius Equation :, Arrhenius proposed the following equation for the determination of activation energy (Ea) at a given, temperature T., k  e E a / RT, , , , k  A e E a / RT, , where Ea = activation energy, A = frequency factor and k = rate constant, Simplifying the equation :, log10 k , , Ea,  log10 A, 2.303 RT, , Self Study Course for IITJEE with Online Support, , Section 4, , 19

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Vidyamandir Classes, , Chemical Kinetics, , Now a plot of log10 k or n k vs 1/T will be a straight line whose equation is given above. If k1 is the, value of the rate constant at T1 K and k2 is the value of rate constant at T2 K for a reaction whose energy, of activation is Ea , then, ,  T2  T1 , k2, Ea, k1  A e Ea /RT1 , Divide, and, take, log, on, both, sides, to, get, :, log, , , , 10, e, , k1 2.303RT  T1T2 , and k 2  A e Ea / RT2 , This equation is called as Integrated Arrhenius equation., This equation is very useful in determining the value of Ea of, a reaction by knowing the values of k1 and k2 at different, temperatures T1 and T2., , Note : , , For most of the reactions, a 10° C rise near the room temperature, the rate constant is almost, k35 C, , doubled i.e. k, , 25 C, , 2, , ., , This can be explained by the fact that the number of molecules possessing energy greater than the, threshold energy increases tremendously as shown by the shaded area (a–b–f–e–a) in the figure., , , 20, , The fraction of molecules having internal energy greater than or equal to Ea is equal to e E a /RT ., This shows that as temperature increases, the number of molecules crossing this energy barrier, increases tremendously, accounting for the increase in the rate constant., , Section 4, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , , Chemical Kinetics, , Rate constant ‘k’ can also be increased by reducing ‘Ea’ to ‘Eac’ through the addition of a catalyst in, the system., k1  Ae E a / RT and k 2  Ae  E ac / RT, , , Note :, , log10, , E  E ac, k, log10 2  a, k1 2.303RT, , (k 2 ) f (E a ) f  (E ac )f, , (k1) f, 2.303 RT, , log10, , and, , (k 2 )b (E a ) b  (E ac ) b, , (k1 )b, 2.303RT, , . . . .(i), , . . . .(ii), , log10, , (E a )f  1 1 , (k 2 )f, ,   , (k1 )f 2.303R  T1 T2 , , . . . .(iii), , log10, , (k 2 )b, (Ea )b  1 1 , ,   , (k1 )b 2.303R  T1 T2 , , . . . .(iv), , where (Ea)f and (Ea)b are the activation energies of the reaction in forward and backward reaction, and, (Eac)f and (Eac)b are the activation energies of catalysed reaction in forward and backword reaction., , Illustration - 6, , The activation energy of the reaction : A + B ,  products is 105.73 kJ/mol. At 40C,, the products are formed at the rate of 0.133 mol/L/min. What will be rate of formation of products at 80C ?, SOLUTION :, Let the rate law be defined as, At T1 : r1 = k1 [A]x [B]y, At T2 : r2 = k2 [A]x [B]y, , ,  k2 , r2 = r1  k ,  1, , Self Study Course for IITJEE with Online Support, , Using Arrhenius equation, find k at 40C., E a  T2  T1 , k2, log10 k  2.303 R  T T , 1,  1 2 , , Section 4, , 21

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Vidyamandir Classes, , Chemical Kinetics, , , , , , log10, , k 2 105.73 103 , 40, , , , k1 2.303  8.31  313  353 , , k2,  2.0, k1, , , , k2,  100, k1, , , , r2 = 0.133  100 = 13.3 mol/L/min, , Illustration - 7 The activation energy of a non–catalysed reaction at 37C is 200 kcal/mol and the, activation energy of the same reaction when catalysed decreases to only 60.0 kcal/mol. Calculate the ratio of, rate constants of the two reactions., SOLUTION :, , kc, 1,  log10 k  2.303 R T  E a  E ac , , , , Illustration - 7, , k, 1, log10 c , (200 103  60 103 ), k 2.303  2  310, , log10 k c  98.0, k, , , kc,  1098, k, , Note : Decrease in the activation energy in forward and backward direction by the addition of catalyst is same. i.e., (Ea)f – (Eac)f = (Ea)b – (Eac)b .Thus rate constant in forward and backward directions increases by same, kf, kf, , , factor. Since, for a reaction A , , adding catalyst will not alter Keq of the reaction.,  B : K eq , kb, kb, , IN-CHAPTER EXERCISE - B, 1., , Show that for a first order reaction, the time required to complete 99.9 % of the reaction is 10 times, that required for the half of the reaction., NOW ATTEMPT IN-CHAPTER EXERCISE-B FOR REMAINING QUESTIONS, , 22, , Section 4, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , FACTORS AFFECTING RATE OF REACTIONS, , Chemical Kinetics, , Section - 5, , Factors affecting Rate of Reactions, The rate of reaction depends upon following factors :, , 1., , Concentration :, In general the rate of a reaction is directly proportional to the concentration of reactants, i.e., the rate, increases as the concentration of reactant(s) increases. For gaseous reaction, rate is proportional to the, partial pressures of reactant(s)., , 2., , 3., , Nature of Reactants :, , , The rate of reactions in which complex molecules are taking part is slower than those in which simple, molecules take part. A chemical reaction involves the rearrangement of atoms (i.e., breaking and, reforming of bonds), hence the rearrangement of molecules involving many bonds is rather slow, process and consequently the rate of a reaction is slower., , , , Physical state of reactants also play key role in determining reaction rates. The greater is the surface, area of a solid surface, the faster is the rate of reaction involving solid molecules. For example, the, burning of wood is slower than the burning of a pulverized wood (due to increased surface area)., , Effect of Catalyst :, The catalyst in general enhances the rate of reactions with out actually taking part in the reaction. The, catalyst is used up during the reaction but at the end of reaction it is recovered as such. The phenomenon of, increase in the rate of a reaction with the help of a catalyst is known as catalysis. Catalysts generally lower, the activation energy which enables more reacting molecules to cross the energy barrier and hence increased, rate of reaction., , 4., , Effect of Temperature :, The rate of a reaction increases by increasing the temperature. It is quite clear from Arrhenius equation, ,  k  A eE / RT  , that for small rise in temperature rate of reaction increases tremendously (increase in, a, , exponential). In fact it is one of most significant factors that affects the rate most strongly. A 10C rise, in temperature, for most of the reactions, doubles the rate of reaction. On increasing the temperature,, the number of molecules possessing activation energy increases (i.e, effective collisions) by a large quantity,, as compared to the total increase in molecular collisions., , Self Study Course for IITJEE with Online Support, , Section 5, , 23

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Chemical Kinetics, , Vidyamandir Classes, , RADIOACTIVE DECAY, , Section - 6, , It is a process in which an unstable nucleus spontaneously looses energy by emitting ionizing particles and, radiation. This decay results in an atom of one type (called the parent nuclide) transforming to an atom of, a different type (called the daughter nuclide). This is a stochastic (random) process i.e. it is impossible to, predict when a given atom will decay, but if a large number of similar atoms are given, the decay rate is, predictable. The substances which undego radioactivity are called as radioactive substance. It was discovered, by Henry Becquerel for atoms of Uranium. Later it was discovered that many naturally occurring compounds, of heavy elements like radium, thorium etc also emit radiations. At present, it is known that all the naturally, occurring elements having atomic number greater than 82 are radioactive. e.g. radium, polonium, thorium,, actinium, uranium, radon etc., The S.I. unit of activity is becqueral (B.q). One Bq is defined as one decay per second. Other unit of, acticvity is Curic (Ci) and is related to Bq as : 1 Ci  3.7  1010 Bq, , Type of decay :, Following are the decay generally occurring in nature :, , (i), ,   decay : In this decay, a helium nucleus (42 He) is emitted.,   particles are He nuclei. they carry a charge of 2e and mass equal to (roughly) 4m p . They, have less penetrating power but high ionizing power and they can be deflected by electric and, magnetic fields., An   decay is represented as :, A, A 4, 4, Z X  Z  2Y  2 He, Parent, Daughter, nucleus, nucleus, , (  particle), , 234,  90, Th  24 He, If the above reaction is spontaneous i.e. Q-value (or disintegration energy) is positive, it can also be, represented as :, , e.g., , 238, 92 U, , where, , 238, 92, , 234, U  90, Th  42 He  Q, , Q, , , , m 238, 92, , , , m 238, m, U  90, Th  24 He  931.5 MeV, , and m A X represents mass of the nuclide ZA X in amu (or  ), Z, (ii), ,   decay : In this decay, a nucleus spontaneously by emitting an electron (e  ) or a positron, ( e  mass equal to me and charge +e). They can be deflected by electric and magnetic fields., They have low ionising power but high penetrating power., , 24, , Section 6, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , Chemical Kinetics, , Type of  decay :, (a), ,  , ,  decay : An e 01e is emitted by the nucleus as :, A, ZX, ,  ZA 1 Y  10 e  v (anti  neutrino), , In this decay, a proton transforms into a neutron within the nucleus emitting a e  :, 1, on, , e.g., (b), , 32, 15, ,  11 p  01 e  v, 32, p  16, S  01e  v, , , , , , , 0,   decay : A positron e  1e is emitted by the nucleus as :, A, Z, , X  ZA 1 Y 10 e  v(neutrino), , In this decay, a proton transforms into a neutron within the nucleus emitting a e :, 1, 1, 1 p  0n, ,  10 e  v, , The symbols v and v represents neutrino and anti-neutrino, these are neutral particles with zero rest, mass they travel with speed of light, and they very weakly interact with matter., , Application of Radioactivity and Radioisotopes :, Radioisotopes find numerons uses in different areas. Some important applications of radioisotopes are as, follows :, (i), , Tracers : By incorporating a small amount of a radioisotope in a reaction system, one can trace, the course of the reaction. Such a sample of radioisotope is called tracer. Since all the isotopes of an, element are chemically equivalent, the monitored path of the isotope will indicate the path of the, reaction. Consider the following esterification reaction :, , By labelling the oxygen atom of methanol with 18O and then using it in the esterification, it can be, proved that the starred oxygen comes from the alcohol and not from the acid as the ester is found, enriched with 18 O isotope. The use of 14 C as a radioactive tracer using labelled compound is well, known. The dynamic nature of chemical equilibria has been established by the use of labelled, compounds., , Self Study Course for IITJEE with Online Support, , Section 6, , 25

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Chemical Kinetics, (ii), , Vidyamandir Classes, , Age of Minerals Rocks : It involve determination of either a spaces formed during a radioactive, decay or of the residual activity of an isotope which is undergoing decay. The former may be, illustrated by helium dating. Helium present in uranium mineral has almost certainly been formed, from   particles., A gram of uranium in equilibrium with its decay products produces approximately 107 g of helium, per year. So if the helium and uranium contests of a mineral are known, the age of the mineral can be, estimated., 238 which has a half life of, The lather can be illustrated by considering a rock containing 92, U, 238, 4.5  109 years. We have seen that in the uranium decay series, 92 U after a series of decay give, 206, the stable isotope 82, Pb as the end product. Asuming that initially the rock did not contain any lead,, 238, 206, we can determine the age of the rock by measuring the ratio of 92, U and 82, Pb and using the, , equation :, N t = N 0e kt, , where N 0 and N t are the amounts of uranium present initially (t = 0) and after the lapse of time t,, respectively and k is the decay constant., (iii) Radiocarbon Dating : Radiocarbon (14, 6 C ) dating of historical wooden derived objects is based, on knowledge that the cosmic ray intensity has been practically constant for thousands of years., 14, 6 C, , is formed in the upper atmosphere by the action of cosmic radiation on, 14, 7, , 1, N  10 N  14, 6 C  1H, , The 14C so produced is eventually converted into carbon dioxide, which is turn is incorporated, into plants and trees by the process of photosynthesis and then finds way into animals which eat, plants. Because of the natural plant-animal cycle, an equilibrium is set up and all living matter, contains the same small proportion of as it contains the same small proportion of as it occurs in, the atmosphere. Once the plant or animal dies, the uptake of carbon dioxide by it ceases and the, level of in the dead begins to fall due to the decay which undergoes., 14, 14, 6C7n, ,  01 , , The half life (t1/ 2 ) period of 14C is 5770 years. A comparison of the  - activity of the dead, matter with that of the carbon still in circulation enables measurement of the period of isolation of the, material from the living cycle. The method, however, ceases to be accurate over period longer than, two or three half life periods of 14C . The proportion of 14C to 12 C in living matter 1:1012., 26, , Section 6, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , Chemical Kinetics, , Illustration - 8 The beta activity of 1g of carbon made from green wood is 15.3 counts per minute. If the, activity of 1 g of carbon derived from the wood of an Egyption mummy case is 9.4 counts per minute the, same conditions, how old is the wood of the mummy case ?, SOLUTION :, , k=, , 0.693 0.693, =, = 1.20×104 yr 1, t1/2, 5770, , log, , N0, kt, =, N t 2.303, , N, 1.20×104 × t, 15.3, = log 0 = log, 2.303, Nt, 9.4, Hence t =, , 2.303, 1.20×10, , 4, , log, , 15.3, 9.4, , t = 3920 years., , Illustration - 9, , One of the hazards of nuclear explosion is the generation of Sr 90 and its subsequent, incorporation in bones. This nuclide has a half life of 28.1 year. Suppose one microgram was absorbed by a, new born child, how much Sr 90 will remain in his bones after 20 years., SOLUTION :, Given t1/ 2 = 28.1year, m = 106 g, t = 20 year, m = ?, °, , 2.303, m, log °, , m, , , , t=, , , , 20 =, , , , m = 6.1×107 g, , 2.303× 28.1 106, log, 0.693, m, , Illustration - 10 0.1 g atom of radioactive isotope X A ( half-life 5 days ) is taken. How many number of, Z, atoms will decay during eleventh day?, SOLUTION :, N = 0.1 g atom, °, , and, , t = 10 days, , and, , Self Study Course for IITJEE with Online Support, , t1/ 2 = 5 days, Section 6, , 27

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Chemical Kinetics, , Vidyamandir Classes, , 2.303, N, 0.693 2.303, 0.1, log °, , =, log, t, N, 5, 10, N, Amount left after 10 days = 0.0250 g atom, =, , , , Similarly if t = 11 days, , , , , 0.693 2.303, 0.1, =, log, 5, 11, N, Amount left after 11 days = 0.0218 g atom, Amount decayed in 11 day  0.0250  0.0218,  3.2×103 g atoms, = 3.2× 6.023×1023 ×103 atoms = 1.93×10 21 atoms, , Nuclear Stability :, , , It has been found out that for a nuclei to be stable ratio of, neutrons to protons (n/p ratio) should lie in the range 1 to, 1.5., , , , On plotting the graph of no. of neutrons to no. of protons,, the stable nuclei lie in a well defined belt, called stability, belt., , , , The nuclei with atomic number upto 20 (Z = 20, n = 20 i.e. upto 40, 20 Ca ) have n/p ratio close to 1 i.e., plot is linear with slope equal to 45 . With increase in atomic number, the graph becomes curved, whose slope increases gradually. Which means that as no. of protons increase, more no. of neutrons, are required to held them together., , , , The nuclei with n/p ratio lying above or below the stability belt are unstable. They undergo positron, emission or K-capture or lose  or   particles so that their n/p ratio falls within the stability belt., , , , With increase in atomic number beyond 20, a nuclide (below the stability belt) decays by  , emission or K-electron capture so that n/p ratio increase to, , ( N  1), so as to go up into the stability, (Z  1), , belt., , , Above the stability belt, the nuclide being neutron-rich decays by  , emission so that n/p ratio, decreases to, , 28, , Section 6, , ( N  1), so as to come down into the stability belt., ( Z  1), , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , Chemical Kinetics, , , , When the atomic number increases beyond 82, the stability is attained by   emission. The greater, importance of this emission lies in the fact that number of protons decreases and so the proton-proton, repulsion decrease., Hence the stability increases., , Note : , , It has been observed that the maximum number of stable nuclei are those which contain even number, of protons and even number of neutrons. The remaining stable nuclei contain either odd number of, protons or odd number of neutrons. The number of stable nuclei containing odd number of protons, and odd number of neutrons is the least., e.g. there are only 4 nuclei (H, Li, B and N ) having odd number of both protons and neutrons., No. of n, even, even, odd, odd, , , , No. of p, even, odd, even, odd, , No. of such nuclei, 160, 56, 52, 4, , A free neutron is unstable. However, a neutron is stable inside a nucleus due to the presence of, protons. Protons exert a repulsive force on each other and they need a “glue” to stay together in a, nucleus which is provided by neutrons through nuclear force., , Nuclear Fission & Fusion :, (a), , Nuclear Fission, The breaking of a heavy nucleus into two or more fragments of comparable masses, with the release of, tremendous energy is called as nuclear fission. The most typical fission reaction occurs when slow moving, neutrons strike 92U 235 . The following nuclear reaction takes place :, 92 U, , , , 235, ,  0 n1  56 Ba141 36 Kr 92  30 n1  200MeV, , If more than one of the neutrons produced in the above fission reaction are capable of inducing a, fission reaction (provided U 235 is available), then the number of fissions taking place at successive, stages goes increasing at a very brisk rate and this generates a series of fissions. This is known as, chain reaction. The chain reaction takes place only if the size of the fissionable material ( U 235 ) is, greater than a certain size called the critical size., , , , It the number of fission is a given interval of time goes on increasing continuously, then a condition, of explosion is created. In such cases, the chain reaction is known as uncontrolled chain reaction., The forms the basis of atomic bomb., , Self Study Course for IITJEE with Online Support, , Section 6, , 29

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Chemical Kinetics, , , (b), , Vidyamandir Classes, , In a chain reaction, the fast moving neutrons are absorbed by certain substances known as, moderators (like heavy water), then the number of fissions can be controlled and the chain reaction is, such cases is known as controlled chain reaction. This forms the basis of a nuclear reactor., , Nuclear Fusion, The process in which two or more light nuclei are combined into a single nucleus with the release of, tremendous amount of energy is called as nuclear fusion. Like a fission reaction, the sum of masses before, the fusion (i.e. of light nuclei) is more than the sum of masses after the fusion (i.e. of bigger nucleus) and, this difference appears as the fusion energy. The most typical fusion reaction is the fusion of two deuterium, nuclei into helium., 1 H 2  2 He4  21.6 MeV, For the fusion reaction to occur, the light nuclei are brought closer to each other (with a distance of, 1H, , 2, , 1014 m ). This is possible only at very high temperature to counter the repulsive force between nuclei., Due to the is reason, the fusion reaction is very difficult to perform. The inner core of sun is at very high, temperature, and is suitable for fusion, In fact the source of sun’s and other start’s energy is the nuclear, fusion reaction., , Laws of Radioactive Decay :, The laws of Radioactive decay are as follows :, , , , , The disintegration of a radioactive substance is random and spontaneous., Radioactive decay is purely a nuclear phenomenon and is independent of any physical and chemical, conditions., The radioactive decay follows first order kinetics, i.e., the rate of decay is proportional to the number of, undecayed nuclei in a radioactive substance at any time t. If d/N be the number of nuclei disintegrating in, time dt, the rate of decay is given as, , dN, ., dt, , dN,  N, dt, where  is called as decay or disintegration constant., , From first order kinetic rate law :, , Rate of decay (R) i.e. Number of nuclei decaying per unit time is defined as , , , R (t )  , , dN, dt, , dN, dt, , Let N 0 be the number of nuclei at time t = 0 and N t be the number of nuclei after time t, then according, to integrated first order rate law, we have:, , 30, , Section 6, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , Chemical Kinetics, , Nt  N 0 e t, , , , N, N, t  n 0  2.303 og 0, Nt, Nt, , and, , R(t)   N 0 e t  N where R0 is the decay rate at, , t 0, , , , The half life (t1/2 ) period of a radioactive substance is defined as the time in which one-half of the radioactive, substance is disintegrated. If N 0 be the number of nuclei at t  0 , then in a half life t1/ 2 , the number of, nuclei decayed will be N 0 ., Nt  N 0 e t, , …(i), , , , N0,  N 0 et1/2, 2, , …(ii), , n, , Nt  1 ,    n : number of half lives  t, From (i) and (ii), N0  2 , t1/ 2, , The half life (t1/2 ) and decay constant () are related as : t1/ 2 , , , 0.693, , , The mean life () of a radioactive substance is equal to the sum of life times of all atoms divided by the, number of all nuclei and can be calculated as follows :, No. of nuclei decaying between time t and t  dt are equal to dN = R(t) dt, , , dN  N 0 e t dt, , The life of these nuclei is thus  N 0 e t dt  t, , , Total life of all nuclei, Average life () =, Total nuclei, , , , , , ,  0, ,  tN 0et dt, N0, , , , 1, , , Illustration - 11 At a given instant there are 25% undecayed radio-active nuclei in a sample. After 10, seconds the number of undecayed nuclei reduces to 12.5% Calculate (i) mean-life of the nuclei, and (ii) the, time in which the number of undecayed nuclei will further reduce to 6.25%., SOLUTION :, , 25.0%  12.5%,  time taken is one half life, , , , t1/ 2  10 sec, , , , t  10sec, , or, , , , 0.693, t1/ 2, , or, , , , 1,  0.1sec, , , 12.5%  6.25%,  time taken is one half life, , Self Study Course for IITJEE with Online Support, , Section 6, , 31

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Chemical Kinetics, , Vidyamandir Classes, , Illustration - 12, , A radioactive sample of 238U decays to Pb through a process for which the half-life is, 9, 238, U after a time of 1.5 10 years., 4.5  109 years. Find the ratio of number of nuclei of the Pb to, Given (2)1/3  1.26., SOLUTION :, n, , Nt  1 , t, 1.5  109 1,    where n , , , N0  2 , t1/ 2 4.5  109 3, 1/3, , Nt  1 ,  , N0  2 , , , , , , 1, 1.26, , 238, , t 0, t t, , x,  1.26, x y, , , , Illustration - 13, 84 Po, 3, , 210 is, , U  Pb, x, , x y, y, , 84 Po, , (N ), y,  t Pb  0.26, x  y ( Nt )U, , , , 210 decays, , with a-particle to 82 Pb 206 with a half life of 138.4 days. If 1.0 g of, , placed in a sealed tube, how much helium will accumulate in 69.2 days ? Express the answer in, , cm at STP.., SOLUTION :, , t1/ 2  138.4 days, t  69.2 days, , 32, , t, , No. of half-lives n , , , , Amount of Po left after 69.2 days , , t1/ 2, , , , 69.2 1, , 138.4 2, , , , 1, , , , g  0.707 g, (2)1/2, Amount of Po used in 69.2 days  1  0.707  0.293g, , , , Now 84 Po, , , , 210 g Po on decay will produce = 4 g He, , , , 0.293 g Po decay will produce , , , , 5.581103  22400,  31.25 cm3, Volume of He at STP , 4, Section 6, , 210, ,  82Pb 206  2 He 4, , 4  0.293,  5.581 103 g He, 210, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , Illustration - 14, , Chemical Kinetics, , In nature a decay chain series starts with 90 Th 232 and finally terminates at 82 Pb, , 208, , .A, , thorium ore sample was found to contain 8  105 ml of He at STP and 5 107 g of Th 232 . Find the age of, ore sample assuming that source of He to be only due to decay of Th232 . Also assume complete retention of, 232,  1.39  1010 years ]., He with in the ore. [t1/ 2 of Th, , SOLUTION :, 90 Th, , 232, ,  82 Pb 208  6 2 He 4  41 e0, , , , 6  22400 ml He is formed by 232 g thorium decay, , , , 8  108 ml He is formed by , , 232  8 105, g thorium decay  1.38 107 g, 6  22400, , After time t, amount of thorium  5  107 g, At t = 0, amount of thorium in the sample  5  107  1.38  107  6.38 10 7 g, Now, t , , 2.303  1.39  1010, 6.38 10 7, m, 2.303, log, log  ,  4.89  109 years, 0.693, , m, 5  107, , Illustration - 15 An experiment requires minimum  activity produced at the rate of 346   particles per, 99, , minute. The half life period of 42 Mo99 which is a emitter is 66.6 hrs. Find the minimum amount of 42 Mo, required to carry out the experiment in 6.909 hours., SOLUTION :, To carry out experiment, Activity at the last minute of 6.909th hour = 346 dpm., Now, t , , 2.303, log, , , 6.909 , , A, At, , 2.303  66.6,  A , log   , 0.693,  346 , ,  A  6.909  0.693, log    ,  0.03121,  346  2.303  66.6, , , A  346  1.0745  371.77, , , 371.77,  66.6  60  35728.5  60  2143710, 0.693, 99, Amount of Mo required to carry out experiment in 6.909 hour ,  2143710  3.52  1016 g, 23, 6.023 10, , Now, A   N, , , , , N , , Self Study Course for IITJEE with Online Support, , Section 6, , 33

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NOW ATTEMPT IN-CHAPTER EXERCISE-C BEFORE PROCEEDING AHEAD IN THIS EBOOK, Chemical Kinetics, , Vidyamandir Classes, , ADVANCED KINETICS, , 1., , Section - 7, , Parallel / Side Reactions (Elementary Reactions) :, , Rate of consumption of ‘A’ in A , B :, , d[A],   k1[A], dt 1, , Similarly, Rate of consumption of ‘A’ in A , C :, , , d[A] ,  Rate of Re action  k1 [A]  , , dt 1 , , , d[A],   k 2 [A], dt 2, , d[A] d[A] d[A], , , dt, dt 1 dt 2 = –k1[A] – k2[A], , , , Overall (net) rate of consumption of ‘A’ =, , , , d[A], = – (k1 + k2)[A] = –koverall [A], dt, , , , The Overall reaction is also of the same order (here, Ist order) with koverall = k1 + k2, , . . . . (i), , Check yourself that solution of (i) is :, , [A] [A]0 e(k1  k2 ) t, Also, from the rate equation :, , d[B], d[C],   k1[A] and,   k 2 [A], dt, dt, , Substituting the value of [A] (t) in the differential equation for [B] and [C] and integrating, we get :, t, , [B]  [B]o   k1[A]0 e (k1  k 2 )t dt  [B]0 , 0, t, , and [C]  [C]o   k 2 [A]0 e (k1  k 2 )t dt  [C]0 , 0, , k1[A], 1  e (k1  k 2 )t, (k1  k 2 ), , , , k 2 [A], 1  e (k1  k 2 )t, (k1  k 2 ), , , , , , , , [B] k1, Also, if [B]0 and [C]0 = 0, we get : [C]  k, 2, , 34, , Section 7, , Self Study Course for IITJEE with Online Support

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Chemical Kinetics, , Vidyamandir Classes, , Illustrating the concept :, , k1 = 10–3s–1 and k2 = 2 × 10–3s–1. Find the overall half life of the reaction., SOLUTION :, koverall = k1 + k2 = 3 × 10–3s–1, t1/ 2 , , 2., , 0.693, 0.693, , s  231 sec . [Note : Reactions are of first order]., koverall 3  103, , Consecutive Reactions (Elementary Reactions) :, k, , k, , 1  B , 2, A , C, , d[A],   k1[A], dt, , . . . .(i), , d[B],   k1[A]  k 2 [B], dt, , . . . .(ii), , d[C],   k 2 [B], dt, , . . . .(iii), , On solving the above differential equations, we get :, [A]  [A]0 e  k1t, , . . . .(iv), , k1, e k1t  e  k 2 t, k 2  k1, , , , . . . .(v), ,  k e k 2 t  k e k1t, 2, [C]  [A]0  1  1, , k 2  k1, , , , , , , , . . . .(vi), , [B]  [A]0, , , , Note : [A] + [B] + [C] = [A]0 [Assuming [B]0 = [C]0 = 0], , Self Study Course for IITJEE with Online Support, , Section 7, , 35

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Vidyamandir Classes, , Chemical Kinetics, d[B], 0, dt, k1[A] = k2 [B]max., , [B] is maximum when, , , . . . . (vii), , d[B],   k1e k1t  k 2e  k 2 t  0  k e k1t  k e  k 2 t, 1, 2, dt, k, 1, t, n 2, k 2  k1 k1, , Using (v),, , , 1, k2, Thus, at t = k  k  n k , [B] is maximum., 2, 1, 1, , Illustration - 16, , k  0.1 s 1, , k 1 / 30 s 1, , 1, 2, A ,  B , C, , Find the time at which concentration of B is maximum. Also, find the concentration of A, B, and C at this, instant. Take [A]0 = 1M, SOLUTION :, 1, k2, 1, 0.1, [B] is maximum at t = k  k  n k  0.1  1/ 30  n 1 / 30  15  n 3 sec., 2, 1, 1, , [A] t = 15 n3 = [A]0 e  k1t, 1.5, , = 1 e 0.115 n 3  e n (3, , ), ,  31.5 , , 1, 3 3, , M, ,  d[B] ,  0, When [B] is max : k1[A] = k2 [B]max. ,  dt, , , , , 1, 1, 1, ,  [B]max ., 10 3 3 30, , , , [B]max. , , 1, M, 3, , Also, [A] + [B] + [C] = [A]0, [C] = 1 , , 36, , Section 7, , 1, 3 3, , , , 1, 4, 1 , M, 3, 3 3, , Self Study Course for IITJEE with Online Support

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Chemical Kinetics, , Vidyamandir Classes, , SUBJECTIVE SOLVED EXAMPLES, Example - 1, , The vapour pressure of two miscible liquids ‘A’ and ‘B’ are 300 and 500 mm of Hg respectively., In a flask, 10 mol of ‘A’ is mixed with 12 mol of ‘B’. However, as soon as ‘B’ is added, ‘A’ starts polymerising, into a completely insoluble solid.This polymerisation follows first-order kinetics. After 100 minutes, 0.525, mol of a solute is dissolved which arrests the polymerisation completely. The final vapour pressure of the, solution is 400 mm of Hg. Estimate the rate constant of the polymerisation reaction. Assume negligible, volume change on mixing and polymerisation, and ideal behaviour for the final solution., [Given, log1011 = 1.04], SOLUTION :, We have :, PA0  300 mm Hg, , and, , PB0  500 mm Hg, , nA0 = 10 mol, , and, , nB0 = 12 mol, , Let the amount of A after 100 min is reduced to nA., At this stage :, nTotal = nA + nB + nSolute = nA + 12 + 0.525, Mole fractions of A and B in the solution will be, A , , Since, , nA, 12, B , and, n A  12.525, n A  12.525, , PTotal = PA0  A = PB0  B , , we get, 400  300 , , nA, 12,  500 , n A  12.525, n A  12.525, , Solving for nA, we get :, nA = 9.90 mol, nA ,  An, For the first-order kinetics, we have :, kt  2.303log10, , or, , , 38, , [A]0, [A]t, , 10, 9.9, –4, 1, k =1.00  10 min = 1.67  10–6 s1, k(100 min)  2.303log10, , Subjective Solved Examples, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , Chemical Kinetics, , Example - 2 A first order reaction : A ,  B requires activation energy of 89 kJ/mol. When 20% solution, of A was kept at 27C for 40 minutes, 25% decomposition took place. What will be the percent decomposition, in the same time in a 30% solution maintained at 37C ? Assume that the activation energy remains constant, in this range of temperature., SOLUTION :, Note : It does not atter whether you take 20%, 30%, 40% or 70% of ‘A’ because first order reaction is independent, of the initial concentration of reactant., At 27C, 20% of A decomposes 25%, C0, 1, 1,  kt = 2.303 log10 C = 2.303 log10 1  =  n 1  [ = 0.25], t, , , k(40) =  n, , 100, 75, , 1, 4, n min 1, 40 3, Using Arrhenius equation find k at 310K., , , , k (at 300 K) =, , E a  T2  T1 , k 310, , , , log10 k, 2.303 R  T1T2 , 300, 3, log10 k 310  89 10  310  300   0.5, k 300 2.303  8.31  300  310 , , , , k(at 310 K) = k(at 300K)  10, , Now calculate % decomposition at 310K using first order kinetics., C0, kt = 2.303 log10 C, t, , , k  40 = n, , , , n, , , , log10, , 1,  1, 4, ,  , n   10   40  0.91, 1    40, 3, , 1, 0.91, ,  0.40, 1   2.303, , 1,  2.5, 1 , , , , 1, 1 , , , ,  = 0.6, ,  = 0.6  60.0 % decomposition of A at 310 K., , Self Study Course for IITJEE with Online Support, , Subjective Solved Examples, , 39

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Chemical Kinetics, , Vidyamandir Classes, , Example - 3, , A certain reaction A + B ,  products ; is first order w.r.t. each reactant with, k = 5.0 × 10–3 M–1s–1. Calculate the concentration of A remaining after 100s if the initial concentration of A, was 0.1 M and that of B was 6.0 M. State any approximation made in obtaining your result., SOLUTION :, C, Using, 2.303 log10 0A  k 0 t, C tA, , A + B ,  products, Given : Rate = k [A][B], , (2nd Order reaction), , Now, since [B] >> [A], [B] can be assumed to, remain constant throughout the reaction. Thus,, the rate law for the reaction, becomes :, Rate  k0 [A], , , , 2.303 log10, , , , log e, , where k0 = k [B] = 5.0 × 10–3 × 6.0 s–1, , , Thus, the reaction is now of first order [Pseudo, first order]., +, , 0.1M, 0.1 – x, , 0.1, 3, C tA, , [ loge x = 2.303 log10 x], , = 3.0 × 10–2 s–1, , Note : A, , 0.1,  k0 t = 3, C tA, , C tA  10e 3  5  10 3 M, , Check your approximation :, [B]change = x = 0.1 – 5 × 10–3, , k, , B ,  products, , = 0.095 M, , 6.0M, 6.0 – x, , % [B]change =, , 0.095, × 100% = 1.58%, 6, , Example - 4, , The gas phase decomposition of N2O5 to NO2 and O2 is monitored by measurement of, total pressure. The following data are obtained., , Ptotal (atm), , 0.154, , 0.215, , 0.260, , 0.315, , 0.346, , Time (sec), , 1, , 52, , 103, , 205, , 309, , Find the average rate of disappearance of N2O5 for the time interval between each interval and for the total, time interval. [Hint : Integrated rate law is NOT to be used]., SOLUTION :, 2N2O5 (g), Initial Pressure (at t = 0), At t = t, , 40, , P0, P0 – 2x, , Subjective Solved Examples, , , , , 4NO2 (g), 0, 4x, , +, , O2 (g), 0, x, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , Now :, , Chemical Kinetics, , , , PTotal = (P0 – 2x) + 4x + x, , x=, , 1, (P, – P 0), 3 Total, , PN 2O5 = P0 – 2x = (5P0 – 2PTotal) = Pt, 2, Thus, PN 2 O5  (Pt1 – Pt2) where Pt2 and Pt1 are the total pressures at time instants t2 and t1 (t2 > t1), 3, respectively, , Ptotal (atm), , Time (sec), , 0.154, , PN 2O5, t, , = Avg. Rate of disappearance of N2O5 (atm/sec.), , 1, , 2 (0.154  0.215), = –7.97 × 10–4, 3, (52  1), 0.215, , 52, , 2 (0.215  0.260), = – 5.88 × 10–4, 3 (103  52), 0.260, , 103, , 2 (0.260  0.315), = – 3.59 × 10–4, 3 (205  103), 0.315, , 205, , 2 (0.315  0.346), = – 1.99 × 10–4, 3 (309  205), 0.346, , 309, , Example - 5, , In an ore containing Uranium, the ratio of U238 to Pb206 nuclei is 3. Calculate the age of, the ore, assuming that all the lead present in the ore is the final stable product of U238. The half life of U238 is, 4.5  109 years., SOLUTION :, Note : The radio active decay follows first order kinetics. Here, we take N0  C0 and Nt  Ct and   k, The first order rate equation for radioactive decay is :,  t  2.303 log, , N0, Nt, , 0.693, where  = t, 1, 2, , N0 = initial nuclei (at t = 0), Nt = final nuclei (at t),  = decay constant or disintegration constant, Self Study Course for IITJEE with Online Support, , Subjective Solved Examples, , 41

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Chemical Kinetics, , Vidyamandir Classes, , U238 ,  Pb206, N0  x, 0, Nt  x – y, y, , , N0, x, , Nt x  y, , Using t = 2.303 log10, Given :, , xy, 3, y, , x, xy, , , , x, 4, , xy 3, , 2.303, 4,  4.5  109 log10, 0.693, 3, , , , t=, , , , t = 1.85  109 years., , Example - 6, , The nucleidic ratio of 1H3 to 1H1 in a sample of water is 8.0 × 10–18 : 1. Tritium undergoes, decay with a half life period of 12.0 years. How many tritium atoms would a 10.0 gm of such sample contain, 36 years after the original sample is collected ?, SOLUTION :, The ratio of tritium atoms to that of H–atoms, will be same as the ratio of moles of T-atoms, to that of H–atoms, since 1 mole of T2O  2, mole of T atoms and 1 mole of H2O  2 mole, of H atoms., , , , , ,   N T O  NT2O  2,  40, , =  1018  6 1023   2,  9, , , = 5.33 × 106 atoms, , Calculate the initial number of tritium atoms., 10 gm = mass of T2O + mass of H2O, = n T 2 O  22  n H 2 O  18, , , , , ,  8 1018 n H 2O  22  n H 2 O  18, ,  n H 2O  18, 10 5,  n H 2 O  , 18 9, , 36, No. of half lives   3, 12, ( x), x, , 1, 1, Use : N t  N 0    N 0  , 2, 2, , 3, , 1,   5.33  106 atoms, 8, = 6.66 × 105 atoms, , 5, 18 40, , 1018,  n T2O   8 10, 9, 9, , 42, , Subjective Solved Examples, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , Example - 7, , Chemical Kinetics, , Dimethyl ether decomposes according to the following reaction :, , CH3 – O – CH3 (g) ,  CH4 (g) + CO(g) + H2 (g), At a certain temperature, when ether was heated in a closed vessel, the increase in pressure with time was, noted down., Time (min), Pressure (mm Hg), (i), , 0, , 10, , 20, , 30, , 420, , 522, , 602, , 678, , Show that the reaction is first order., , (ii), , Compute the pressure of CO (g) after 109 minutes., , SOLUTION :, CH3 – O – CH3 (g) ,  CH4 (g) + CO (g) + H2 (g), time CH3 – O – CH3 CH4 CO H2, t=0, , C0  P0, , 0, , 0, , 0, , t=t, , Ct  P0 – x, , x, , x, , x, , , , (all are gases), , Pt = P0 + 2x, 1, (P – P0), 2 t, , , , x=, , , , C0, P, 2 P0,  0 , C t P0  x 3P0  Pt, , Now find k1, k2 and k3 using the first order kinetics, kt = 2.303 log10, , 2 P0, 3P0  2Pt, , k1 =, , 2.303, 2 (420), log10, = 0.0129 min–1, 10, 3(420)  522, , k2 =, , 2.303, 2 (420), log10, = 0.0122 min–1, 20, 3(420)  602, , k3 =, , 2.303, 2 (420), log10, = 0.0123 min–1, 30, 3(420)  678, , As k1  k2  k3 , the reaction is first order., 1, kaverage = (k1 + k2 + k3) = 0.0127 min–1, 3, 0.693, t1/ 2 ,  54.56 min, , k, 109, 2, No. of half lives =, 54.56, Self Study Course for IITJEE with Online Support, , Subjective Solved Examples, , 43

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Vidyamandir Classes, , Chemical Kinetics, , 2, , 420, 1, (Pt )CH3OCH3  P0   , 4, 2, = 105 mm Hg = P0 – x, x = PCO = 420 – 105 = 315 mm Hg, , , , , Example - 8, , The decomposition of N2O5 according to following reaction is first order reaction :, 2N2O5(g) ,  4 NO2(g) + O2(g), , After 30 min. from start of the decomposition in a closed vessel, the total pressure developed is found to be, 250 mm of Hg and on complete decomposition, the total pressure is 500 mm of Hg. Calculate the rate, constant of the reaction., SOLUTION :, 2N2O5(g) ,  4NO2(g) + O2(g), t=0, , P0, , 0, , 0, , t=t, , P0 – 2x, , 4x, , x, , 2P0, , P0/2, , t=, , –, , P0 : initial presure ; Let Pt : pressure at 30 min and, P : pressure at the end of decomposition, , , Pt = P0 + 3x, , , , P0 , , , , x=, , 1, (P – P0), 3 t, , and, , P = 2P0 +, , 1, 5, P0 = P0, 2, 2, , 2, 2, P , × 500 = 200 mm Hg, 5, 5, , For the first order kinetics, C0, kt = 2.303 log10 C, t, , C0 : initial concentration ; Ct : final concentration, C0, P0, 1, 50, x  (250  200) , Now C  P  2x, and, 2, 3, t, 0, , , , , 44, , C0, 200, 6, , , Ct 200  2  50 5, 3, k, , 1, 6,  2.303 log10  6.08  10 3 min 1, 30, 5, , Subjective Solved Examples, , Self Study Course for IITJEE with Online Support, , NOW ATTEMPT OBJECTIVE WORKSHEET BEFORE PROCEEDING AHEAD IN THIS EBOOK

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Vidyamandir Classes, , Chemical Kinetics, , THINGS TO REMEMBER, 1., , k, , mA  nB ,  pC  qD, m, ,  Rate of reaction = k  A , ,  Bn, , , , 1 d  A, 1 d  B, 1 d C, 1 d  D, , , , m dt, n dt, p dt, q dt, where k depends only on T., ,  Units of k = (mol/L)1 – n (time)–1, 2., , k, , A , B, , nth order reaction : Half life  t1/ 2  , ,  n  1 kC0n  1, ,  t1/ 2 , , 1, C0n  1, ,  1, 1, 1 , ,   kt, ,  n  1  C nt  1 Cn0  1 , , and, 3., , 2n  1  1, , k, First order reaction : A , B, C, C0, 1, kt  2.303 log10 0  2.303 log10,  2.303 log10, Ct, C0  x, 1 , 0.693, Half life  t1/2  , (independent of initial concentration of the reactant), k, , x, , t, 1, C t  C0   where x , = No. of half lives, t1/ 2, 2, , 4.  Arrhenius Equation : k  Ae Ea / RT, where A = Arrhenius factor OR Pre-exponential factorand, , Ea = Activation energy, ,  k can be increased by increasing T OR reducing Ea (addition of catalyst), (i), , k, Ea  1 1 , Temperature effect : log10 2 ,   , k1 2.303R  T1 T2 , , (ii), , Catalyst effect : log10 k 2  E a  E ac, k1 2.303 RT, , SOLUTIONS TO IN-CHAPTER EXERCISE-B, 1., , t 99.9 % , t1/ 2 , , 2.303, k, , log10, , 2.303log10 2, k, , C0, C0  0.999 C0, , , t 99.9%, t1/ 2, , , , , 2.303, k, , log10 1000 , , 3, log10 2, , Self Study Course for IITJEE with Online Support, , 3  2.303, k, ,  10, , Things to Remember & Solutions, , 45

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Vidyamandir Classes, , My Chapter Notes, , Self Study Course for IITJEE with Online Support

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Vidyamandir Classes, , Illustration - 1, , Self Study Course for IITJEE with Online Support