Tiêu đề DPP - 8 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 8 1 (c) k = 1 t [ x a(a ― x) ] k = 1 500 [ 0.2a a(a ― 0.2a) ] k = 1 2000a 1 2000a = 1 t [ 0.6a a(a ― 0.6a) ] t = 3000 s 2 (a) K does not change with time; also unit of K suggest it to be II order. 3 (d) Follow review of rate of reaction. 4 (b) Molecularity represents the number of molecules of reactants taking part in an elementary step of reaction. 5 (c) t1/2 ∝ (a) 1―n or t1/2 = Z(a) 1―n or logt1/2 = log Z + (1 ― n) log a or y = c + mx Thus, slope = m = 1 ― n or 1 ― n = 0 ∴ n = 1 and for I order reaction t1/2 = 0.693 K . 6 (c) t1/2 ∝ (a) 1―n 7 (a) A pseudounimolecular reaction. 8 (d) Topic :- Chemical Kinetics Solutions
Rate becomes x y times if concentration is made x time of a reactant giving y th order reaction. Rate =k[A] n [B]m Concentration of A is doubled hence x=2 ,y=n and rate becomes = 2 n times Concentration of B is halved ,hence x=1 2 and y=m and rate becomes=( 1 2 ) m times Net rate becomes=(2) n ( 1 2 ) m times =(2) n―mtimes 9 (d) For the reaction H2(g) + Br2(g)→2HBr(g) Rate of reaction = k[H2 ][Br2 ] 1/2 Molecularity of reaction = 1 + 1 = 2 Order of reaction = 1 + 1 2 = 3 2 10 (c) When heat energy is supplied, kinetic energy of reactant molecules increase. This will increase the number of collisions and ultimately rate of reaction will be enhanced. 11 (d) t = 2.303 K log a (a ― x) ∴ K = 2.303 20 log 1 0.25 = 0.06931 min―1 12 (d) I step of mechanism B shows I order in both reactants. 13 (d) Ka = Ae ―Ea/RT and Kb = Ae ―Eb/RT Also, Ka > Kb Ea < Eb Now notice that all the given facts are satisfied. 14 (c) Half-life depends upon rate constant and rate constant (K) varies with temperature as K = A ∙ e ―En/RT;K increase with temperature. Also t1/2 ∝ 1 K 15 (d) Rate = k[NOBr2 ][NO] ...(i) But NOBr2is in equilibrium keq = [NOBr2 ] [NO][Br2] [NOBr2 ] = keq[NO][Br2] ...(ii) Putting the [NOBr2 ] in (i) rate = k.keq[NO][Br2][NO] Hence, rate = k.keq[NO] 2 [Br2] rate = k′[NO] 2 [Br2]
where, k’.Keq the order, of reaction with respect to NO(g) is 2 16 (a) For zero order reaction, for example, A→B ―d[A] dt = k[A] 0 ―d[A] dt = k 17 (d) The increase in collision frequency brings in an increase in effective collisions and thus, rate of reaction increases. 18 (a) t1/2 ∝ 1 a n―1 When n = 4 t1/2 ∝ 1 a 3 Hence, order of reaction = 4 19 (d) There are two different reactants (say A and B). A + B→product Thus, it is a bimolecular reaction . If dx dt = k[A][B] It is second order reaction If (dx dt) = k[A] Or =k[B] It is first order reaction . Molecularity is independent of rate ,but is the sum of the reacting substance thus it cannot be unimolecular reaction . 20 (a) rate = K[A][B] 2 ∴ 10―2 = K[1][1] 2 or K = 10―2 litre 2mol ―2 sec―1 ∴ rate II = 10―2 [0.5] × [0.5] 2 = 1.2 × 10―3mol/litre-sec