Title DPP - 6 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 6 2 (a) For zero order reaction, t1/2 ∝ [R]0 3 (b) Effect of temperature on reaction rate is given by Arrhenius equation k = Ae ―Ea/RT 4 (b) This is Arrhenius equation. 5 (b) Let ,initial concentration=a Final concentration=a- 2 3 a= a 3 t2 3 = 2.303 k log a a/3 = 2.303 5.48 × 10―14 log 3 =2.01 × 1013s 6 (c) Let the order with respect to A and B is x and y respectively. Hence, Rate r = [A] x [B] y ...(i) On doubling the concentration of A, rate increases 4 times, 4r = [A] x [B] y ... (ii) From Eqs. (i) and (ii) 1 4 = ( 1 2 ) x ∴ X=2 ∴ order with respect to A is 2 If concentration of A and B both are doubled, 8r = [2A] x [2B] y ... (iii) From Eqs. (i) and (iii), we get 1 8 = 1 (2) x . 1 (2) y [ ∵ x = 2] 1 8 = 1 (2) 2 . 1 (2) y Topic :- Chemical Kinetics Solutions
1 8 = 1 4 × 2 y 2 y = 2 ∴ Y=1 Hence , differential rate equation is r ∝ [A] 2 [B] 1 or dC dt = kC 2 A × CB [Where, CA and CB=concentrations of A and B] 7 (d) r = k[A] n ...(i) When concentration is doubled then 4r = k(2A) n ...(ii) Divide Eq. (ii) by (i) 4 = 2 n n = 2 8 (a) t = 0.693 k log [A]0 [A] = 2.303 60 log a a 16 = 2.303 60 log 16 = 2.303 60 × 1.204 = 0.0462s = 4.6 × 10―2 s 9 (a) From the unit of rate constant (i.e.,s ―1 ), it is clear that the reaction is of first order. 2N2O5→4NO2 + O2 Hence, for first order reaction , k = 2.303 t log p0 pt ∴ 3.38 × 10―5 = 2.303 10 × 60 log 500 pt Or log 500 pt =0.00880 ∴ 500 pt = anti log 0.00880 =1.02 pt = 500 1.02 = 490 atm 11 (b)
t = 2.303 k log a a ― x ∵ x = 3 4 a ∴ t = 2.303 k log a a ― 3 4 a = 2.303 k log 4 12 (b) 2N2O5→4NO2 + O2 Rate of decomposition of N2O5 = ― 1 2 k[N2O5 ] dt Rate of formation of NO2 = 1 4 k[NO2] dt ∴ rate of decompsition of N2O5 rate of formation of NO2 = 1 2 k [N2O5] dt 1 4 k [NO2] dt or 1 2 k [N2O5] dt × 4 1 dt k[NO2] = 4 2 = 2 1 = 2:1 13 (a) k = 2.303 t log a a ― x Given, reaction is75% completed is 32 min A=100,x=75 ∴ k = 2.303 32 log 100 100 ― 75 ...(1) For 50% completion of reaction A=100, x=50 ∴ k = 2.303 t log 100 100 ― 50 ...(2) ∵ LHSof Eq.(1) = Eq.(2) ∴ RHSof Eq.(1) = Eq.(2) ∴ 2.303 32 log 100 100 ― 75 = 2.303 t log 100 100 ― 50 or 2.303 32 log 4 = 2.303 t log 2 Or t 32 = log 2 log 4 or t = 32 × log 2 2 log 2 ∴ t = 16 min ∴ reaction will be 50% completed in 16 min 14 (b) Rate ( +d[C] dt ) = k[A][B]
Thus, the order of reaction w.r.t. A=1 The order of reaction w.r.t.B=1 Total order of reaction=1+1=2 15 (a) The intersection point indicates that half of the reactant X is converted into Y. 16 (b) At T1 = 200 K, T2 = 400 K, k1 = k, k2 = 10 k ∵ log k2 k1 = Ea 2.303R ( T2 ― T1 T1 ∙ T2 ) log 10 k k = Ea 2.303R ( 400 ― 200 400 × 200) Ea = 921.2 R 17 (c) Zero order reactions occur with constant rate. 18 (a) t = 2.303 K log a (a ― x) ; Thus, K = 2.303 10 log 8 = (2.303 × 3 log 2 ) /10 19 (c) For the reaction A→B On increasing the concentration of reactant (i.e.,A) by 4 times , the rate of reaction becomes double ,hence order of reaction is1 2 . 20 (b) The rate of chemical reaction always decreases with time as reaction proceeds due to decrease in number of reactant molecules. Only for zero order reactions the rate of chemical reaction remains same.