Tiêu đề DPP - 10 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 10 1 (b) For first reaction, E1 = 2.303RT1T2 (T1 ― T2) log k′1 k1 ...(i) For second reaction, E2 = 2.303RT1T2 (T1 ― T2) log k′2 k2 ...(ii) Given, E1 > E2 ⇒ 2.303RT1T2 (T1 ― T2) log k′1 k1 > 2.303RT1T2 (T1 ― T2) log k′2 k2 ∴ k′1 k1 > k′2 k2 2 (d) These are the characteristics of effective collisions. 3 (b) Pseudo first order rate constant is doubled as well as rate of reaction is doubled. It may be noted that in presence of acid, hydrolysis of ethyl acetate is a pseudo-unimolecular reaction but the actual value of k depends upon the concentration reaction but the actual value of k depends upon the concentration of H + ions, otherwise rate constant of a reaction is constant at constant temperature. 4 (b) We know that, (t1/2) (t1/2) = [ a2 a1 ] n―1 Where, n=order of reaction Given, (t1/2) = 0.1s, a1 = 400 (t1/2) = 0.8s, a1 = 50 On substituting the values 0.1 08 = [ 50 400] n―1 On taking log both sides log 0.1 0.8 = (n ― 1)log 50 400 Topic :- Chemical Kinetics
log 1 8 = (n ― 1)log 1 8 0.90=(n-1)0.90 n-1=1 n=2 6 (a) For nth order reaction k = (mol L ―1 ) 1―n s ―1 For Ist order reaction Unit of k = s ―1 For zero order reaction Unit of k = mol L ―1 s ―1 = M s ―1 7 (b) For II order reaction, t1/2 = 1 Ka 8 (d) If 1 [A] 2vs times are a straight line then order of reaction is third. 9 (c) For an endothermic reaction where ∆H represents the enthalpy of the reaction, the minimum value for the energy of activation is more than ∆H, ie, Ea > ∆H 11 (c) T1 2 = T50, x = R 2 ∴ T50 = R k0 So T50 ∝ R T50 ∝ R k0 Therefore, the formula of t1/2for a zero order reaction is [R]0 2k 12 (b) The curve Y shows a gradual increase in the concentration with time. 13 (a) Acid hydrolysis of sucrose is a pseudo unimolecular or pseudo first order reaction. H Ea Ea > H
Hydrolysis of sucrose in presence of mineral acid is a biomolecular reaction. But as water is taken in large excess, so the rate of reaction only depends upon concentration of sucrose. Hence, order of the reaction is one. Therefore, it is called a pseudo first order reaction. 14 (c) For first order reaction log[A] = ― kt 2.303 + log [A]0 15 (a) r = K[A] = 4 × 10―3 × 0.02 = 8 × 10―5M sec―1 16 (b) For the first order reaction, 2N2O5→2NO2 + O2 rate dx dt = k(N2O5 ) ...(i) Given, dx dt = 2.400 × 10―5mol L ―1 s ―1 k = 3.0 × 10―5 s ―1 [N2O5 ] = ? Substituting these values in (i) 2.4 × 10―5 = 3.0 × 10―5 [N2O5 ] or [N2O5 ] = 2.4 × 10―5 3.0 × 10―5 = 0.8mol L ―1 17 (c) The definition of activation energy. 18 (d) 2.303 log K2 K1 = Ea R [T2 ― T1] T1T2 ∴ 2.303 log K2 K1 = 9 2 × 10―3 [ 10 298 × 308]; ∴ K2 K1 = 1.63;i.e., 63% increase 19 (c) Slope = - lo g [A] t 2.303 k
r = K[A] 1 ; Thus, 7.5 × 10―4 = K × 0.5; ∴ K = 15 × 10―4 sec―1 = 1.5 × 10―3 sec―1 20 (d) 2N2O5⇌4NO2 + O2 Rate =± 1 stoichiometrics coefficient d[product or reactant dt ∴ Rate = ―1 2 d[N2O5] dt = + 1 4 d[NO2] dt = d[O2] dt Or ― d[N2O5] dt = 2 d[O2] dt Or ―2d[N2O5] dt = 4 d[NO2] dt Or d[NO2] dt = 4d[O2] dt