Title DPP - 9 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 9 1 (b) For parallel path reaction Kaverage = K1 + K2 = 1.26 × 10―4 + 3.8 × 10―5 = 1.64 × 10―4 sec―1 Also fractional yield of B = KB Kav = 1.26 × 10―4 1.64 × 10―4 = 0.7683 Fractional yield of A = KA Kav = 3.8 × 10―5 1.64 × 10―4 = 0.2317 3 (d) For first order : K = 0.693 t1/2 = 0.693 1386 = 5 × 10―4 s ―1 4 (a) Ionic reactions are instantaneous one. 5 (b) For zero order reaction, rate of reaction is independent of concentration i.e., rate of reaction does not depend upon the concentration of reactant. dx dt = k[A] 0 6 (c) t1/2 = 100s k = 0.693 t1/2 = 0.693 100 k = 6.93 × 10―3 s ―1 7 (b) The rate law for the reaction is as r= dx dt=k(A)(B) 2 (C) 0=k(A)(B) 2 on increasing the concentration of A,B and C two times. r’=dx dt=k(2A)(2B) 2 (2C) 0 =8k(A)(B) 2 Thus, the rate increases eight times. 8 (a) Activation energy is the needed by reactant molecules to gain threshold energy level. 9 (a) The rate of zero order reaction is independent of the concentration of the reactants or the Topic :- Chemical Kinetics Solutions
concentration of the reactant do not change with time. Thus, the rate of reaction remains constant. dx dt = k(a ― x) 0⇒ dx dt = k Or Rate=k 10 (b) For first order reaction, k = 2.303 t log10 a a ― x Where, a= initial concentration X= change in concentration during time‘t’. If 75% of the reaction was completed in 32 min, then k = 2.303 32 log10 100 100 ― 75 = 2.033 32 log10 4 k = 0.0433min―1 Hence, time required for the completion of 50% reaction. t = 2.303 0.0433 log10 100 100 ― 75 = 2.033 32 log10 2 = 16 min 11 (a) For the reaction : CCl3CHO + NO→CHCl3 + NO + CO Rate = dx dt = k[CCl3CHO][NO] k = dx dt × [CCl3CHO][NO] = mol/L s × mol/L × mol/L k = L mol―1 s ―1 12 (c) 2A + B→C Rate of reaction, = ― 1 2 d[A] dt = ― d[B] dt = d[C] dt ∴ ― d[A] dt = 2 d[C] dt = 2 × 2.2 × 10―3 = 4.4 × 10―3mol L ―1min―1 13 (d) For third order reaction, rate = k[A] 3
mol L ―1 s = k(mol ― L ―1 ) 3 k = 1 mol2 L ―2 s = mol―2 L 2 s ―1 14 (d) CH3COOC2H5 + H2O H + CH3COOH + C2H5OH Since, in this reaction, water is excess, it is an example of psedo first order reaction (as rate depends only on the concentration of CH3COOC2H5). 16 (d) The efficiency of an enzyme in catalyzing a reaction is due to its capacity to lower the activation energy of the reaction 17 (c) The rate of reaction is: = ― 1 2 d[A] dt = ― d[B] dt = 1 3 d[C] dt = d[D] dt 18 (a) For exothermic reaction, activation energy of reverse reaction is greater than activation energy of forward reaction, ie, Ef < Er 19 (a) 2.303 log K2 K1 = Ea R [T2 ― T1 ] T1T2 ∴ 2.303 log 3 = Ea 2 [313 ― 273] 313 × 273 ∴ Ea = 4693 cal 20 (d) lnK = lnA ― Ea RT is Arrhenius equation. Thus plots of lnK vs 1/T will give slope = Ea/R.
ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. B B D A B C B A A B Q. 11 12 13 14 15 16 17 18 19 20 A. A C D D C D C A A D