DPP-33 to 34 Physical Chemistry with Answer

FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 33 Class : XIII Course : DPP No.1 Total Marks : 28 Max. Time : 28 min. Single choice Objective ('–1' negative marking) Q.1 to Q.8 (3 marks 3 min.) [24, 24] Multiple choice objective ('–1' negative marking) Q.9 (4 marks 4 min.) [4, 4] ANSWER KEY DPP No.-33 1. 4 2. 3 3. 4 4. 4 5. 3 6. 4 7. 4 8. 2 9. 2, 4 1. 2NO + 2H  N + 2H O. The experimental rate law for above reaction is, Rate = k [NO]2 [H ]. When time 2 2 2 2 is in minutes and the concentration is in moles/L, the units for k are - moles3 (1) L3  min Sol. Rate = k [NO]2 [H ] conc. (2) moles L  min moles2 (3) L2  min 1 L2 (4*) moles2  min time = k (conc.)3  k = time(conc.)2 k = 1 = moles2 L2 moles 2  min (min) L2 2. The differential rate law equation for the elementary reaction A + 2B K  3C, is : d [A] d [B] d [C] d [A] 1 d [B] 1 d [C] (1) –    dt dt dt = k [A] [B]2 (2) –   dt 2 dt  3 dt = k [A]2 [B] d [A] 1 d [B] 1 d [C] (3*) –   dt 2 dt  3 dt = k [A] [B]2 (4) None of these Sol. A + 2B K  3C (elementary reaction) d [A] 1 d [B] 1 d [C] Rate = –   dt 2 dt  3 dt = k [A] [B]2 3. For the reaction 2A  between k1, k2, and k3 is : B + 3C; if – d [A] dt = k1 [A]2; d [B] dt = k2 [A]2; d [C] dt = k3 [A] 2, the correct relation (1) k = k = k (2) 2k = k = 3 k (3) 4k = k = 3 k (4*) k1  k  k3 1 2 3 1 2 3 1 2 3 2 2 3 1 Sol. Rate = – 2 d [A] dt d [B] = dt 1 d [C] = 3 dt  k1 [A]2 = K 2 2 [A]2 = k3 [A]2 3 k1 = k = k3 2 2 3 4. Which of the following statement is incorrect? (1) Unit of rate of disappearance is Ms–1 (2) Unit of rate of reaction is Ms–1 (3) Unit of rate constant k is depend on order (4*) Unit of k for first order reaction is Ms–1 Sol. Unit of k for first order reaction is s–1. 5. The rate expression for reaction A (g) + B(g)  C(g) is rate = k[A]1/2 [B]2. What change in rate if initial concentration of A and B increase by factor 4 and 2 respectively ? (1) 4 (2) 6 (3*) 8 (4) None of these Sol. Rate = k[A]1/2 [B]2 r = k[a]1/2 [b]2 ; r = k[4a]1/2 [2b]2 r1  r2 1 1 = 2  4 = 8  r = 8 × r 6. Reaction A  B follows second order kinetics. Doubling the concentration of A will increase the rate of formation of B by a factor of : (1) 1/4 (2) 1/2 (3) 2 (4*) 4 Sol. A  B d [A] d [B] rate = – dt = dt = k [A]2  d[B]   dt  k (a)2  1  d[B]  1 = k (2a)2 = 4    dt 2 7. For the reaction 2NO  N O + O , rate expression is as follows – d[NO2 ] = K [NO ]n , where K = 3 × 10–3 mol–1 Lsec–1. If therate of formationof oxygen is 1.5 × 10–4 mol L–1 sec–1, then dt 2 the molar concentration of NO2 in mole L–1 is (1) 1.5 × 10–4 (2) 0.0151 (3) 0.214 (4*) 0.316 Sol. From the unit of K, it is evident that it is a second order reaction. – 1 d[NO2 ] = 2 dt d[O2 ] dt   – d[NO2 ] = 2 × dt d[O2 ] dt = 2 × 1.5 × 10–4 = 3 × 10–4 3 × 10–4 = K [NO ]2 = 3 × 10–3 [NO ]2   [NO ] = 0.316. 8. A first order reaction is 75% completed in 100 minutes. How long time will it take for it’s 87.5% completion? (1) 125 min (2*) 150 min (3) 175 min (4) 200 min C0 Sol.  Ct = 2 n For 75% completion, no. of half lives taken = 2 half life = 100 = 50 min 2 For 87.5% completion, no. of half lives taken = 3.  Time taken = 3 × 50 = 150 min. 9. A certain reaction A → B follows the given concentration (Molarity)–time graph. Which of the following statement(s) is/are true ? 0.5 0.4 0.3 (1) The reaction is second order with respect to A (2*) The rate for this reaction at 20 second will be 7 × 10–3 M s–1 (3) The rate for this reaction at 80 second will be 1.75 × 10–3 M s–1 (4*) The [B] will be 0.35 M at t = 60 second [A] 0.2 0.1 0 20 40 60 80 100 Time (sec) Sol. Rate for the reaction at 20 second = 0.35 50 = 7 × 10–3 Ms–1 A  B t = 0 0.4 0 t = 60 0.05 0.35 FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 34 Class : XIII Course : DPP No.2 Total Marks : 28 Max. Time : 28 min. Single choice Objective ('–1' negative marking) Q.1 to Q.8 (3 marks 3 min.) [24, 24] Multiple choice objective ('–1' negative marking) Q.9 (4 marks 4 min.) [4, 4] ANSWER KEY DPP No.-34 1. 4 2. 2 3. 3 4. 1 5. 3 6. 3 7. 4 8. 1 9. 1, 2, 3 1. Decomposition of NH4NO2 (aq) into N2 (g) and 2H2O (𝑙) is first order reaction. Which of the following graph is correct? (1) (2) (3) (4*) Sol. C = C e–kt (For Ist order) NH NO (aq)  N (g) + 2H O (𝑙) (Ist order) 2. Decomposition of HI (g) on Gold surface is zero order reaction. Initially, few moles of H2 are present in container then which of the following graph is correct ? (1) (2*) (3) (4) Sol. 2HI(g)  H (g) + I (g) (zero order) 2 2 t = 0 a b 0 b  PH2 , initial t = t a – 2x b + x PH2  (b + x)  PH2 = PH2 , initial + kt (zero order reaction) So, graph is . 3. For the zero order reaction A  B + C; initial concentration of A is 0.1 M. If A = 0.08 M after 10 minutes, then it’s half-life and completion time are respectively : (1) 10 min; 20 min (2) 2 × 10–3 min, 10–3 min (3*) 25 min, 50 min (4) 250 min, 500 min. Sol. For zero order, rate = K =  0.08  0.1 10 0.02 = 10 M min–1 half life = CO = 2K 0.1 10 2  0.02 = 25 min  completion time = CO = K 0.110 0.02 = 50 min. 4. The accompanying figure depicts the change in concentration of species A and B for the reaction A  B, as a function of time, the point of intersection of the two curves represents (1*) t1/2 (2) t3/4 (3) t2/3 (4) data insufficient to predict Sol. A  B t = 0 a 0 t = t a – x x At intersection, a – x = x x = a 2 so, this time represents half life. 5. For the reaction A  products, the graph of the fraction of A remaining as a function of time (x–axis) is a straight line with –ve slope. The order of the reaction is therefore a  x a a  x = fraction of A  (a – x) = a – kt  a = – kt So, it is zero order. 6. In the following reaction A  B + C, rate constant is 0.001 Ms–1. If we start with 1 M of A then, conc. of A and B after 10 minutes are respectively : (1) 0.5 M, 0.5 M (2) 0.6 M, 0.4 M (3*) 0.4 M, 0.6 M (4) none of these Sol. A  B + C rate constant = 0.001 Ms–1 [CO]A = 1 M From unit of rate constant, we can conclude that rate is of zero order. Decrease in concentration of A in 10 min = 0.001 × 10 × 60 = 0.6 M [CO]A t = 10 min = 1 – 0.6 = 0.4 M [CB] t = 10 min = 0.6 M. 7. The rate law for the dimerisation of NO d[NO2 ] into N O is – = k[NO ]2 . 2 2 4 dt 2 Which of the following changes will change the value of the specific rate constant k ? (1) Doubling the total pressure (2) Decreasing the pressure (3) Changing the volume of the flask (4*) Changing the temperature. 8. For an elementary reaction aA   d[A] product, the graph plotted between log vs. time gives a  dt  straight line with intercept equal to 0.6 and showing an angle of 45° with origin, then : (1*) rate constant = 3.98 time–1 and a = 1 (2) rate constant = 3.98 mol L–1 t–1 and a = 1 (3) rate constant = 1.99 time–1 and a = 1 (4) rate constant = 1.99 mol–1 L1 t–1 and a = 2 Sol. (1) Rate = – dCA = K Ca dt A  log  dCA  = log K + a log C  dt  A  log K = 0.6 K = 3.98 time–1 and (rFkk) a = tan 45º = 1. 9. For the reaction 2A + B  d [C] C with the rate law dt = k [A]1 [B]–1 and started with A and B in stoichiometric proportion. Which is/are true? (1*) unit of k is Ms–1 (2*) [A], [B] and [C] all will be linear functions of time (3*) [C] = 2kt (4) [C] = kt Sol. 2A + B  C t = 0 2a a 0 t = t 2a – 2x a – x x d[C] dt = k (2 (a – x) (a – x)–1) = 2k   d[c] =  k dt  [C] = 2 kt unit of k = Ms–1 [A] = 2 (a – x) and [C] = x [B] = (a – x).