Tiêu đề DPP - 4 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 4 1 (a) K = 2.303 k log a (a ― x) = 2.303 40 log 0.1 0.025 = 0.0347 ∴ r = K × [A] = 0.0347 × 0.01 = 3.47 × 10―4 M/min 3 (b) For zero order reaction x = kt = 0.2 mol dm―3 h ―1 × 30 60 h = 0.1 mol dm―3 Now, concentration = 0.05 mol dm―3 Hence, initial concentration = 0.1 + 0.05 = 0.15 mol dm―3 4 (c) For the reaction, 2X + Y→Z Rate = ― 1 2 d[X] dt = d[Z] dt = 0.05 mol L ―1 min―1 ― 1 2 d[X] dt = 0.05 ― d[X] dt = 2 × 0.05 = 0.1 mol L ―1 min―1 5 (b) X(g)→Y(g) + Z(g) The reaction is a first order reaction. Topic :- Chemical Kinetics Solutions
Hence , k = 0.693 t1/2 = 2.303 t log a (a ― x) 02234 0.693 10 = 2.303 t log a a/10 Or t = 2.303 × 10 0.693 × log 10 =33min 6 (a) To be solved with the help of formula, log k2 k1 = Ea 2.303 R [ T2 ― T1 T1T2 ] T1 = 273 + 27 = 300 K T2 = 273 + 67 = 340 K log 6.9 × 10―3 3.45 × 10―5 = Ea 2.303 × 8.31[ 340 ― 300 340 × 300] log 200 = Ea 19.1379 × 40 102000 2.3010 = Ea 19.14 × 4 10200 , Ea = 19.14 × 10200 × 2.3010 4 = 112304.907 J = 112.3 kJ 7 (c) dx dt = K(a ― x) 2 is differential form of II order. Integrate it to get (c). 8 (d) Am + B n→ABx In this case, Overall order of reaction = m + n Hence, code 3 is wrong 9 (a) For the first order reaction, t1/2 = 1n2 k Or t1/2 = 0.693 k 10 (d) Order of reaction is an experimentally determined quantity and thus, cannot be predicted from the given equation. 11 (b) The rate for first order reaction is expressed as A→products Rate= ― d[A] dt
Rate=k[A] And the rate constant (k) is given as k = 2.303 t log [A0 ] [At ] or ― k = t 2.303 log [At ] [A0 ] 12 (d) t1/2 = 1 (a) n―1 Where, n=order of reaction a= initial concentration For first order reaction, n=1 t1/2 = 1 a n―1 a = 1 a 0 = 0 Thus for a first order reaction,t1/2 is independent of initial concentration. 13 (b) Relation between (t1/2)and initial concentration of reactant for (n ― 1)order reaction t1/2 ∝ [R] 2―n 0 14 (a) K = 2.303 40 log 0.1 0.025 ∴ K = 0.03466 min―1 rate = K × 0.01 = 0.03466 × 0.01 = 3.47 × 10―4M min―1 15 (c) Rate of endothermic reactions increase with increase in temperature while that of exothermic reactions decrease with increasing temperature. 16 (d) k = 2.303 t log10 a a ― x For half-life period, x = a 2 t = 2.303 k log10 a a ― a 2 t = 2.303 k log10 2 ∵ t = t1/2
t1/2 = 2.303 × 0.3010 k t1/2 = 0.693 k 17 (d) ∵ Rate constant (k′) > rate constant (k′ ′) Greater the rate constant lesser will be the activation energy ∴ E′a < Ea′ ′ 18 (d) The reactant concentration drop from 0.8 to 0.4 M, i.e.,50% takes place in 15 minute. K = 2.303 15 log 0.8 0.4 = 0.693 15 = 0.0462 min―1 Also, t = 2.303 K log 0.1 0.025 = 2.303 0.0462 log 0.1 0.025 = 30 min 19 (c) For II order, t = 1 Ka x (a ― x) ∴ t = 1 8 × 10―5 × 1 ( 0.5 0.5) = 1.25 × 10―4 minute 20 (a) The Arrhenius equation can be written as log k = logA ― Ea 2.303 RT On comparing this equation with standard equation of straight line y = mx + c, we get, y = log k x = 1 T m = ― Ea 2.303R c = logA Hence, on plotting graph between log k(y ― axis)andl T (x-axis), we get a line with slope equal to m = ― Ea 2.303 R