Tiêu đề DPP - 1 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 1 1 (a) It is the definition of molecularity. 2 (c) ( dx dt ) = k[NO] 2 [o2] = k( nNO V ) 2 ( nO2 V ) ( dx dt) = k V 3 (nNo) 2 (nO2 ) ( dx dt) = k(nNo) 2 (nO2 ) ( V 2 ) 3 = 8( dx dt) 3 (c) No doubt order cannot be predicted by merely looking chemical reaction but this can be treated as standard example of II order reaction. 4 (a) Rate = ― [ dc dt] = [ ― dn dt] 1 V [ ∵ c = n V ] ∴ ― [ dc dt] = ― 1 RT[ dP dt]. [c = P RT] 5 (a) For zero order reaction Rate =[A] 0 = k mol L ―1 s = k K=mol L ―1 s ―1 6 (b) A k1 B, A k2 C, By Arrhenius equation, R1 = A′e ―Ea1/RTand k2 = A′e ―Ea2/RT Topic :- Chemical Kinetics Solutions
(A′ is Arrhenius constant) (Since, Ea2 = 2Ea1) ∴ k2 = A′e ―2Ea′ |RT k1 k2 = A′e ―Ea1|RT A′e ―2Ea|RT = e Ea1|RT ∴ k1 = k2e Ea1/RT 7 (d) For the reaction, 2A + B→A2B According to rate laws, Rate ∝ concentration of reactants rate = k[A] 2 [B] Where, k=rate constant 8 (d) This is activation state and orientation concept for mechanism of reactions. 9 (b) Rate depends upon the slowest step. Hence, from equation O + O3→2O2 r = k[O3 ][O] And from equation O3⇌O2 +O Keq = [O2 ][O] [O3 ] [O] = Keq[O3 ] [O2 ] ∴ r = k[O3 ] Keq[O3 ] [O2 ] = k′[O3 ] 2 [O2 ]―1 10 (a) Amount of A left in n1 halves = [A0] 2 n1 Amount of B left in n2 halves = [B0] 2 n2 Also if [A0] 2 n1 = [B0] 2 n2 when A decays to n1 halves and B decays to n2 halves. ∵ [A0 ] = 4[B0] ∴ 4 = 2 n1 2 n2 = (2) n1―n2 or (n1 ― n2 ) = 2 ∴ n2 = n1 ― 2 ...(i) Now, T = n1 × t1/2A and T = n2 × t1/2B ∴ n1 × t1/2A n2 × t1/2B = 1 or n1 × 5 n2 × 15 = 1
or n1 n2 = 3 ...(ii) ∴ By Eqs. (i) and (ii) n1 = 3,n2 = 1 Thus, T = 3 × 5 = 15 minute 11 (c) ∵ On doubling the concentration of A, the rate of reaction becomes two times. ∴ The order of reaction w.r.t. A is 1 ∵ On doubling the concentration of B, the rate of reaction does not change. ∴ the order of reaction respect to B is 0 ∵ on doubling the concentration of C, the rate of reaction becomes four times ∴ the order of reaction w.r.t. C is 2 ∴ the overall order of reaction=1+0+2=3 12 (c) For nth order; unit of rate constant may be derived by K = rate [reactant] n 13 (c) r = K[N2O5 ] = 6.2 × 10―4 × 1.25 = 7.75 × 10―4M/s 14 (c) A → product Initially a 0 After time t (a-x) x After t1/4 (a ― a 4 ) a 4 For the first order kinetics , k = 2.303 t log ( a a ― x ) k = 2.303 t1/4 log a 3a 4 t1/4 = 2.303log 4 3 k = 0.29 k 15 (a) The order of reaction is zero. Suppose the following reaction take place . A + B→product ∴ rate = [A][B] ―1 ∴ order =1+(-1)=0 16 (d) Pseudo first order reactions are those reactions which are not truly first order but show first order kinetics under specific conditions. For examples, acidic hydrolysis of an ester and hydrolysis of cane sugar. 17 (d)
The differential rate law for the reaction, 4NH3(g) +5O2(g)→4NO(g) +6H2O(g) is Rate = ― 1 4 d[NH3 ] dt = ― 1 5 d[O2 ] dt = + 1 4 d[NO] dt = + 1 6 d[H2O] dt 18 (a) 79Au198 ― B 80Hg198 k = 0.693 t1/2 = 0.693 65 After 260 hr, k = 2.303 260 log a a ― x 0.693 65 = 2.303 260 log a a ― x a a ― x = 16 1 1 ― x = 16 x = 15 16 g = 0.9375 g 19 (d) Rate =k[NO2Cl] Hence ,rate determining step is NO2CL→NO2 + CL 20 (b) RCl + NaOH→ROH + NaCl Rate = k[RCl] For this reaction rate of reaction is depends upon the concentration of RCl It means, the rate of reaction is halved by reducing the concentration of RCl by one half