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FACTORS AFFECTING RATE of REACTION -.:: GEOCITIES.Ws FACTORS AFFECTING RATE OF REACTION 1)Chemical nature of the reactants a freshly exposed metallic sodium surface furnishes almost instantly when explored to air and moisture – very fast reaction . Reaction between iron and moisture (air) to form rust is slow. Reason- iron do not lose electron easily. It depends on the inherent tendencies of atoms ,molecules, or ions to undergo changes in chemical bonds. 2) Ability of the reactants to come in contact with each other. For reaction to occur the reactants must collide with each other: (a) homogeneous reaction –when reactants are in the same phase, e.g liquid, gas. (b) heterogeneous reaction –reactant only able to meet at the interface. Rate of reaction dictated by the area of contact between the phases i.e surface area plays an important role. Assoc. Prof. Dr Misni Misran SCE2230 Sesi 2005/2006 36 1 cm 0.01 cm 0.01 cm 0.01cm 1 cm 1 cm 1,000,000 cubes Total surface area = 6 cm2 total surface are = 600 cm2 3) Concentration of the Reactants The rates of both homogeneous and heterogeneous reactions are affected by the concentrations of the reactants. 4)Temperature of the system Chemical reaction will occur faster at higher temperature. e.g chemical metabolism slows down in cooler surrounding. 5) Presence of rate promoting agents or catalysts. Catalysts are substances that increase the rate of reactions without itself being used up. Best example is enzymes. ZERO –ORDER REACTION Say our reaction of interest is, aA + bB → products No need to specify the products since order of reaction is defined interms of the reactant concentration. Assoc. Prof. Dr Misni Misran SCE2230 Sesi 2005/2006 37 For zero-order , the rate of reaction is independent of the reactant concentrations. 1 d[A] Rate = − = k ' [A]D [B]D = k ' a dt a is only a coefficient, Thus d[A] − = k (mol dm-3 s-1) dt zero-order rate constant k = ak’ -d[A] = −kdt intergration from t = 0 and t = t gives [ A] d[A] = [A] − [A] = − t kdt = −kt ∫ [ A]D D ∫ t=0 or, [A] = [A]D − kt - [A] o ] slope = k [A 0 time, t Assoc. Prof. Dr Misni Misran SCE2230 Sesi 2005/2006 38 FIRST ORDER REACTION (Tertib Pertama) A → products d[A] rate = − = k[A] dt intergration : d[A] − = kdt [A] [ A] 1 t − ∫ 2 d[A] = k∫ 2 dt [ A]1 [A] t1 [A]1 ln = k(t2 − t1) [A]2 [A]D if t1 =0 s ∴ln = kt [A] −kt ln[A] = [A]D e or ln[A] = ln[A]D − kt e.g : decomposition of gaseous, N2O5, 2N2O5 4NO2 +O2 :aA products 1 d[A] v = − = k[A] 2 dt Assoc. Prof. Dr Misni Misran SCE2230 Sesi 2005/2006 39 1 d[N O ] ∴rate = − 2 5 = k[N O ] 2 dt 2 5 can also write : d[A] − = ak[A] = K [A] dt A whereK A = ak The plot of ln [A] versus t will give a slope of − K (unit of K, s-1) t = 0 at any point of t changes in [A] could be changes in physical properties. -d[A] =k decay process dt ] [A 0 time, t example : radioactive decays 238 4 234 92U →2 He+ 90Th rearrangement of methyl isonitrile to acetonitrile H CNC H C-CN 3 3 Assoc. Prof. Dr Misni Misran SCE2230 Sesi 2005/2006 40 Example : The thermal decomposition of 2,2’- azobisisobutyronitrile (AIBN) CH → 3 NC − C(CH3 )2 − NN − C(CH3 )2 CN 2NC − C •+ N2 ∆ CH 3 Has been studied in an inert organic solvent at room temperature. The progress of reaction has been monitored by the optical absorption at 350nm. The following data are obtained. t(s) 0 2000 4000 6000 8000 10000 12000 ∞ Abs 1.50 1.26 1.07 0.92 0.81 0.72 0.65 0.40 Assuming the reaction is first order calculate the rate constant. [AIBN ln = −kt [AIBN]D different in the absorbance t= 0 and t = ∞ , (AD − A∞ ) is propotional to [AIBN] initially present in solution. The instantaneous concentration [AIBN]∗ at time tα (A − A∞ ) Assoc. Prof. Dr Misni Misran SCE2230 Sesi 2005/2006 41 Thus, A − A ln ∞ = −kt A D− A∞ since AD = 1.50 and A∞ = 0. t(s) 2000 4000 6000 8000 10000 12000 ⎛ A − A ⎞ -0.246 -0.496 -0.749 -0.987 -1.240 -1.482 ln⎜ ∞ ⎟ ⎜ A ⎟ ⎝ D−A∞ ⎠ ln (A-A /A -A ) ∞ 0 ∞ slope = -1.4 - ( 0.37) / 11300 - 3000 -4 -1 = - 1.24 × 10 s = - k -4 -1 ∴ k = 1.24 × 10 s time, (s) Example 1: NH2 HNNO2 NH2 NO2 H+ + NO2 Assoc. Prof. Dr Misni Misran SCE2230 Sesi 2005/2006 42 Reaction can be monitored by removing sample at suitable interval, quenching with excess alkali and measure the intensity of yellow color of the products. The following results were obtained t (s) 0 300 780 1500 2400 3600 ∞ Absorbance 0 0.036 0.074 0.120 0.162 0.199 0.249 Show that the reaction is First Order and determine the first order rate constant. Example 2 : Say HNNO3 = a rate constant for first order reaction, K ⎛ a ⎞ a ∴kt = 2.303log10 ⎜ ⎟or ln ⎝ a − x ⎠ a − x where a – concentration of reactant at tD (a − x) is concentration at t Assoc. Prof. Dr Misni Misran SCE2230 Sesi 2005/2006 43 By assuming that the Beer-Lambert Law is obeyed, decadic absorbance. If AD is the initial absorbance and A∞ when the reaction is A α complete, ∞−At concentration at time, t (A ) ∴kt = ln ∞− AD (A ) ∞− At for the first order rate constant, ln(A∞− At ) vs t should give a straight line with slope –k ANSWER : k = 4.42×10−4 s −1 Let, [A] = [A]D − x [B] = [B]D − x -3 where x is mol dm of A and B consumed in time t d[A] d[A] − x dx − = − D = = k[A][B] dt dt dt = k([A]D − x)([B]D − x) dx = kdt ([A]D − x)([B]D − x) After integration: Assoc. Prof. Dr Misni Misran SCE2230 Sesi 2005/2006 44 1 ()[B] − x [A] ln D D = kt [B]D −[A]D ([A]D − x) [B]D OR 1 [B][A] ln D = kt [B]D −[A]D [A][B]D assuming a=b=1 and [A]D ≠ [B]D ln [B][A]0 / [A][B]0 t SECOND ORDER REACTION Consider, aA + bB → product type-1 1 d[A] rate = − = k'[A]2[B]D a dt = k'[A]2 The rate is second order w.r.t A and zero-order w.r.t B Assoc. Prof. Dr Misni Misran SCE2230 Sesi 2005/2006 45 d[A] − = k[A]2 dt Where, k = ak’ in (mol dm-3)-1 s-1 integration: [ A] 1 t ∫ 2 d[A] = −∫ kdt [ A]D [A] 0 1 1 − = kt [A] [A]D type-2 1 d[A] rate = − = k'[A][B] a dt first-order w.r.t A and B d[A] − = k[A][B] dt where K = ak' Second order is most common reactions Example : + + CH3COCH3 (aq) + H (aq) CH3COHCH3 (aq) 2NO2 (g) 2NO (g) + O2 (g) Example hydrolysis of acetyl chloride CH3COCl + H2O CH3COOH + HCl Assoc. Prof. Dr Misni Misran SCE2230 Sesi 2005/2006 46 Water in excess compared to acetyl chloride molarity of -3 H2O × 55.5 mol dm the amount of H2O consumed is almost negligible. Rate is expressed, d[CH COCl] − 3 = k[CH COCl][H O] dt 3 2 = k'[CH3COCl] where k'= k[H 2O] . The reaction therefore appears to be First-Order kinetics and is called a pseudo-First-Order reaction. HALF LIFE [A] Consider First-order reaction [A] = D ,t = t Z 1/ 2 Thus, [A] ln D/ 2 = −kt [A] 1/ 2 D ln 2 0.693 t = = 1/ 2 k k Assoc. Prof. Dr Misni Misran SCE2230 Sesi 2005/2006 47 SECOND ORDER REACTION Example : The reaction between iodoethane and tertiary amine C6H5N(CH3)2 can be followed by titrating the liberated iodide ions with silver nitrate solution. + - C2H5I + C6H5N(CH3)2 C6H5N(CH3)2C2H5 + I In an experiment at 25D C using equimolar amounts of reactants, the following values of the iodoethana concentration were obtained ; Time, s 0 89 208 375 625 803 1041 1375 1875 [C2H5I] 1.6 1.4 1.2 1.0 0.8 0.7 0.6 0.5 0.4 mol dm-3 Determine the order of reaction and the rate constant, k ANSWER : If the reaction is second-order, the rate equation, [C H I] − d 2 5 = k[C H I][C H N(CH ) ] dt 2 5 6 5 3 2 Initial concentration for both reactants = [A]D Concentration of each reactant at t = [A]t 1 1 ∴kt = − [A]t [A]t Assoc. Prof. Dr Misni Misran SCE2230 Sesi 2005/2006 48 1 A plot of vs t should give a straight line of slope, k [A]t t,s 0 89 208 375 625 803 1041 1375 1875 1 0.625 0.714 0.833 1.0 1.250 1.428 1.667 2.00 2.50 [A] A straight line plot → the reaction is second-order Slope = k = 1.00×10−3 dm3mol −1s −1 ENERGY of ACTIVATION Example : The rate constant k for the first-order decomposition of acetone-dicarboxylic acid in aqueous solution are as follows : Temp/D C 0 10 20 30 40 50 60 k,105 / s 2.46 10.8 47.5 163 576 1850 5480 Calculate the energy of activation for the reaction The Arrhenius equation for the rate constant , k ⎛ − Ea ⎞ k = Aexp⎜ ⎟ ⎝ RT ⎠ Taking logarithm, E ln k = ln A − a RT Assoc.
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