Nội dung text DPP - 5 Solutions.pdf
Class : XIIth Subject : CHEMISTRY Date : DPP No. : 5 1 (c) For a zero order reaction, R→ product Rate= ― d[R] dt = k[R] 0 = k ―d[R] = k.dt Integrating the above equation. ― ∫ d[R] = k ∫ dt ― [R] = kt + I ...(i) Where, I is integration constant At t = 0, R = [R]0 ― [R]0 = k × 0 + I I = ― [R]0 Put this value in Eq. (i) ― [R] = kt ― [R]0 or [R] = ―kt + [R]0 2 (a) For first order reaction, Half-life period (t1/2) = 0.693 k Where, k=rate constant (t1/2) = 0.693 69.3 s ―1 =0.01 s ―1 3 (b) For nth order reaction : t1/2 ∝ 1 a n―1 For second order reaction t1/2 = 1 ka = 1 0.5 × 0.2 = 100 10 = 10 min 4 (d) Topic :- Chemical Kinetics Solutions
r = K[CH3COCH3 ] a [Br2 ] b [H +] c ∴ 5.7 × 10―5 = K[0.30] a [0.05] b [0.05] c ...(1) 5.7 × 10―5 = K[0.30] a [0.10] b [0.5] c ...(2) 1.2 × 10―4 = K[0.30] a [0.10] b [0.10] c ...(3) 3.1 × 10―4 = K[0.40] a [0.05] b [0.20] c ...(4) By (1) and (2) a = 1 By (2) and (3) b = 0 By (3) and (4) c = 1 ∴ r = K[CH3COCH3 ] 1 [Br2 ] 0 [H +] 1 5 (c) Unit of rate constant = time ―1 conc (n―1) Where, n=order of reaction Given, unit of rate constant =L mol ―1 s ―1 ∴ L mol ―1 s ―1 = (s) ―1 (L mol ―1 ) n―1 = (s) ―1 (L mol ―1 ) n―1 = s ―1 (L mol ―1 ) n―1 Or 1=n-1 Or n=2 ∴ order of reaction =2 6 (c) Activation energy of a chemical reaction can be determined by evaluating rate constants at two different temperatures log k2 k1 = Ea 2.303 R ( 1 T1 ― 1 T2 ) 7 (c) Molecularity can never be fractional. 9 (d) 2SO2(g) + O2(g)⇌2SO3(g) For this reaction, rate (r1) = k[SO2 ] 2 1 [O2 ]1 ....(i) On doubling the volume of vessel, concentration would be half. Hence, Rate(r2 ) = k( [SO2 ]1 2 ) 2 ( [O2 ]1 2 ) = r1 8 r1 r2 = 8:1 10 (c) r = k[RCl]
If [RCl] = 1 2 , then rate= r 2 11 (a) A k1 B k2 C k3 D ∵ k3 > k2 > k1 As k1 is slowest hence A→B is the rate determining step of the reaction 12 (b) k = 2.303 t log10 a a ― x = 2.303 10 log10 100 80 = 2.303 10 [log 10 ― 3 log 2] = 2.303 10 [1 ― 3 × 0.3010] k = 0.0223 13 (d) Ea(A→B) = 80 kJ mol ―1 Heat of reaction (A→B) = 200 kJ mol ―1 For (B→A) backward reaction, Ea(B→A) = Ea(A→B) + heat of reaction = 80 + 200 = 280kJ mol ―1 14 (c) For endothermic reaction A→B Activation energy = 15 kcal/mol Energy of reaction = 5 kcal/mol Hence, activation energy for the reaction B→A is 15 ― 5 = 10 kcal/mol 15 (d) For zero order [A]t = [A]0 ―kt 0.5 = [A]0 ― 2 × 10―2 × 25 ∴ [A]0 = 1.0 M 16 (b) A 5 15 B 15 - 5 = 10 E nergy Progress of reaction
t = 2.303 k log a a ― x Where, k=rate constant=10―3 s ―1 a=initial amount=100 a ― x=amount left after time t=25 t=time to leave 25% reaction ∴ t = 2.303 10―3 log 100 25 = 2.303 10―3 log 4 = 2.303 × 0.6020 10―3 = 1386s 17 (d) By increasing 10 K temperature the rate of reaction becomes double. When temperature is increased from 303 K to 353 K, the rate increases in steps of 10° and has been made 5 times. Hence, the rate of reaction should increases 2 5 times i.e., 32 times. 18 (a) Temperature coefficient = rate of recation at 35°C rate of recation at 25°C = 2 Thus, increase in rate is two times, when temperature is increased 10°C. Hence, by the increase of 70°C(100-30=70°C), the increase in rate will be = (2) 7 ∵ 70° = 7 × 10° = 128 times 19 (d) log k2 k1 = Ea 2.303R ( T2 ― T1 T1T2 ) log k2 k1 = 9000 2.303 × 2 ( 5 295 × 300) log k2 k1 = 0.1103 k2 k1 = 1.288, k2 = 1.288 k ie, increase by 28.8% 20 (b) 1 2 A→2B Remember for a A→bB