IIT JEE Main Advanced Physical Chemistry 12th Chemical Kinetics

Chemical Kinetics

Contents Topic

Page No.

Theory

01 - 06

Exercise - 1

07 - 24

Exercise - 2

25 - 33

Exercise - 3

34 - 38

Exercise - 4

39 - 41

Answer Key

42 - 43

Syllabus Chemical Kinetics : Rates of chemical reactions; Order of reactions; Rate constant; First order reactions; Temperature dependence of rate constant (Arrhenius equation).

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CHEMICAL KINETICS Rate/Velocity of chemical reaction : The rate of change of concentration with time of different chemical species taking part in a chemical reaction is known as rate of reaction of that species. c mol / lit. = = mol lit–1 time–1 = mol dm–3 time–1 Rate = t sec Rate is always defined in such a manner so that it is always a positive quantity.

Types of Rates of chemical reaction : For a reaction R  P Average rate =

Total change in concentrat ion c  [R]  [P] = = – = Total time taken t t t

Instantaneous rate : rate of reaction at a particular instant.  c  dc d [R] d [P] Rinstantaneous = tlim =– =  = 0  dt dt dt  t 

Relation between reaction rates of different species involved in a reaction : For the reaction : N2 + 3H2  2NH3 1d [H2 ] d [N2 ] 1 d [NH3 ] =  3 dt = Rate of reaction =  dt 2 dt

Order of reaction : Let there be a reaction m1A + m2B  products. Now, if on the basis of experiment, we find that R  [A]P [B]q Where p may or may not be equal to m1 and similarly q may or may not be equal to m2. p is order of reaction with respect to reactant A and q is order of reaction with respect to reactant B and (p + q) is overall order of the reaction.

Integrated rate laws : Zero order reactions : For a zero order reaction General rate law is, Rate = k [conc.]º = constant If C0 is the initial concentration of a reactant and Ct is the concentration at time ‘t’ then Rate = k =

C0  Ct ' t'

or

kt = C0 – Ct

or Ct = C0 – kt

Unit of K is same as that of Rate = mol lit–1 sec–1.

First Order Reactions : k=

C0 2.303 log C t t

Half life time (t1/2) t1/2 =

ln2 0.693 2.303 log 2 = = k k k

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CHEMICAL KINETICS # 1

Second order reaction : 1 1 – Ct C 0 = kt

Pseudo first order reaction : A second order (or of higher order) reactions can be converted into a first order reaction if the other reactant is taken in large excess. Such first order reactions are known as pseudo first order reactions.  For A + B  Products [Rate = K [A]1 [B]1 2.303

b(a  x )

k = t (a  b) log a (b  x ) Table : Characteristics of Zero, First, Second and nth Order Reactions of the Type A  Products Zero Order

Differential Rate law

First-Order

 Ä[ A ] = k[A]° Ät



(Integrated Rate law) [A]t = [A]0 – kt

[ A ] = k[A] t

In [A]t = –kt + In [A]0

nth order

Second-Order



[ A ] = k[A]2 t

1 1 = kt + [A]t [ A ]0



Ä[ A ]  k [ A ]n Ät 1

(A t )

n1



1 ( A 0 )n 1

 (n  1) kt

Linear graph

[A]t v/s t

Half-life

t1/2 =

In [A] v/s t

[ A ]0 2k

t1/2 =

(depends on [A]0)

0.693 k

(independent of [A]0)

1

1 v/s t [A]

( A t )n1

1 t1/2 = k[ A ] 0

t 1/ 2 

v/s

t

1 ( A 0 )n1

(depends on [A]0)

Methods to determine order of a reaction : By comparison of different initial rates of a reaction by varying the concentration of one of the reactants while others are kept constant r = k [A]a [B]b [C]c

if

[B] = constant [C] = constant

then for two different initial concentrations of A we have

r01 = k [A0]1a

r02 = k [A0]2a r01



r02

 [A ]    0 1   [ A 0 ]2 

a

Method of half lives : The half lives of each order is unique so by comparing half lives we can determine order for nth order reaction

t1/2  t1 / 2 t1' / 2



1 [R0 ]n1

(R '0 )n1 (R 0 )n1

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CHEMICAL KINETICS # 2

Methods to monitor the progress of the reaction : Pressure measurement : Progress of gaseous reaction can be monitored by measuring total pressure at a fixed volume & temperature. Let there is a 1st order reaction A(g) 

k=



nB(g)

P0 (n  1) 2.303 log nP  P t 0 t

Final total pressure after infinite time = Pf = nP0 Formula is not applicable when n = 1, the value of n can be fractional also. Do not remember the formula but derive it for each question.

By titration method : By measuring the volume of titrating agent we can monitor amount of reactant remaining or amount of product formed at any time. It is the titre value . Here the milliequivalent or millimoles are calculated using valence factors. V0 = vol. of titrant used at t = 0 Vt = vol. of titrant used at ‘t’ V = vol. of titrant used at t =  k=

[this is exclusively for HCl.]

V  V0 2.303 log V  V t  t

Optical rotation measurement : It is used for optically active sample. It is applicable if there is at least one optically active species involved in chemical reaction. where are r0, rt, r are angle of optical rotation at time K=

t = 0, t = t and t = 

Y  Y0 2.303 log  t r  rt

Effect of temperature on rate of reaction : In early days the effect of temperature on reaction rate was expressed in terms of temperature coefficient which was defined as the ratio of rate of reaction at two different temperature differing by 10ºC (usually these temperatures were taken as 25ºC and 35ºC) T.C. =

K t  10  2 to 3 ( for most of the reactions) Kt

Arrhenius theory of reaction rate :

Enthalpy (H) HR

Ea1

Ea2

Threshold enthalpy or energy

Reactants H =  Hp – HR = Ea1 – Ea2

HP

 HR = Summation of enthalpies of reactants  HP = Summation of enthalpies of reactants  H = Enthalpy change during the reaction Ea1 = Energy of activation of the forward reaction Ea2 = Energy of activation of the backward reaction

Products

Progress of reaction (or reaction coordinate)

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CHEMICAL KINETICS # 3

T2

Fraction of molecule

T2 > T1

T1 Ea

e E a / RT  represents fraction of molecules K.E. having energy greater Ea

rate  e E a / RT dependence of rate on temperature is due to dependence of k on temperature.

k  e Ea / RT k  Ae Ea / RT

[Arrhenius equation]

General characteristics of catalyst : P.E.

A catalyst drives the reaction through a low energy path and hence Ea is less. That is, the function of the catalyst is to lower down the activation energy. Ea = Energy of activation in absence of catalyst. E’a = Energy of activation in presence of catalyst. Ea – E’a = lowering of activation energy by catalyst.

Ea E'a

HR HP

Products

Reaction Coordinate

Molecularity and Order : The number of molecules that react in an elementary step is the molecularity of the elementary reaction. Molecularity is defined only for the elementary reactions and not for complex reactions. No elementary reactions involving more than three molecules are known, because of very low probability of near-simultaneous collision of more than three molecules. The rate law for the elementary reaction aA + bB  products

rate = k[A]a[B]b, where a + b = 1, 2 or 3.

For an elementary reaction, the orders in the rate law equal the coefficients of the reactants.

Mechanism of a reaction : Reactions can be divided into Elementary / simple / single step Complex / multi-step ELEMENTARY REACTION : These reaction take place in single step without formation of any intermediate T.S.

Ep

P.E. Er Reaction coordinates

For elementary reaction we can define molecularity of the reaction which is equal to no of molecules which make transition state or activated complex because of collisions in proper orientation and with sufficient energy

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CHEMICAL KINETICS # 4

COMPLEX REACTION : Reaction which proceed in more than two steps. or having some mechanism. ( sequence of elementary reaction in which any complex reaction procceds) T.S.

intermediate

E Reaction coordinates

For complex reaction each step of mechanism will be having its own molecularity but molecularity of net complex reaction will not be defined. CALCULATION OF RATE LAW/ ORDER

MECHANISM IN WHICH R.D.S. GIVEN If R.D.S. involves only reactant, product or catalyst on reactant side rate law of R.D.S. = rate law of reaction RDS is having intermediate on reactant side To calculate order, we have to specify [ intermediate] in expression of rate law in terms of conc. of [R], [P] or catalyst with the help of same equilibrium step given in mechanism. MECHANISMS IN WHICH RDS NOT SPECIFIED : STEADY STATE APPROXIMATION : At steady state

d [intermediate ] =0 dt

COMPLICATIONS IN 1st ORDER REACTION PARALLEL 1st ORDER REACTION OR COMPETING FIRST-ORDER REACTIONS

1 1 1 Teff = T1 + T2

k1 [B] = k [C] 2

Ea 

(remember)

(remember)

E a1 k 1  E a 2 k 2 k1  k 2

REVERSIBLE 1ST ORDER REACTION ( both forward and backward ) A t=0 t=t t = teq.

a a –x a – xeq.

B 0 x xeq.

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CHEMICAL KINETICS # 5

kfa x = k k f b

kf + kb =

1  e

( k f  k b ) t



(remember)

 x eq.  1   n  x  x  t  eq. 

(remember)

SEQUENTIAL 1ST ORDER REACTION OR CONSECUTIVE FIRST-ORDER REACTIONS k

t=0 t

k

1 2  A  B  C a 0 0 a–x y z

k 1a y = k  k { e  k1t – e  k 2 t } 2 1

all first order equation

(remember)

k1 1 tmax. = k  k  n k 1 2 2

CASE-

(remember)

k1 >> k2 k

1 

A

CASE  :

B

k

2  

C

k2 >> k1

Conc [C]

a

[B]

t

[A]

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CHEMICAL KINETICS # 6

PART - I : OBJECTIVE QUESTIONS *Marked Questions are having more than one correct option. Section (A) : Rate of Reaction A-1.

For a hypothetical reaction A  L, the rate expression is dC A rate = – dt (A) Negative sign represent that rate is negative (B) Negative sign indicates to the decrease in the concentrations of reactant (C) Negative sign indicates the attractive forces between reactants (D) None of the above is correct

A-2.

xA + yB  zC. If  (A) 1,1, 1

A-3.

d[C] d[B] d[ A ] = then x,y and z are : = 1.5 =  dt dt dt (B) 3, 2, 3 (C) 3, 3, 2

In the following reaction :  d[ A ] 

(D) 2, 2, 3

xA  yB  d[B] 

log  dt  = log  dt  + 0.3     where –ve sign indicates rate of disappearance of the reactant. Thus, x : y is : (A) 1 : 2 (B) 2 : 1 (C) 3 : 1 (D) 3 : 10 A-4.

In the formation of sulphur trioxide by the contact process, 2SO2 (g) + O2 (g)  2SO3 (g) d(O 2 ) The rate of reaction is expressed as – = 2.5 x 104 mol L-1 sec-1 dt The rate of disappearance of (SO2) will be (A) 5.0 x 10—4 mol L–1 s–1 (B) 2.25 x 10—4 mol L–1 s–1 —4 –1 –1 (C) 3.75 x 10 mol L s (D) 50.0 x 10—4 mol L–1 s–1

A-5.

Rate of formation of SO3 in the following reaction 2SO2 + O2  2SO3 is 100 g min–1. Hence rate of disappearance of O2 is : (A) 50 g min-1 (B) 40 g min–1 (C) 200 g min–1 (D) 20 g min–1

A-6.

A reaction follows the given concentration (M)–time graph. The rate for this reaction at 20 seconds will be: 0.5 0.4 0.3 0.2 0.1

0

20 40 60 80 100 Time/second

(A) 4 × 10–3 M s–1 A-7.

(B) 8 × 10–2 M s–1

(C) 2 × 10–2 M s–1

(D) 7 × 10–3 M s–1

For the reaction : N2 + 3H2  2NH3. If the rate of disappearance of hydrogen is 1.8 × 103. What is the rate of formation of ammonia. (mole per litre per sec.) (A) 1.8 × 103 (B) 1.2 × 103 (C) 2.7 × 103 (D) 0.9 × 103

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CHEMICAL KINETICS # 7

A-8.

The rate of a reaction is expressed in different ways as follows ; + 1/2(d[C]/dt) = – 1/3 (d[D]/dt) = + 1/4 (d[A]/dt) = – (d[B]/dt) The reaction is : (A) 4 A + B  2C + 3D

(B) B + 3D  4A + 2C

(C) 4A + 2B  2C + 3D

(D) B + (1/2) D  4A + 3

A-9.

ln the reaction; A + 2B  3C + D, which of the following expression does not describe changes in the concentration of various species as a function of time : (A) {d [C] / dt} = – {3d [A] / dt} (B) {3d [D] / dt} = {d [C] / dt} (C) {3d [B] / dt} = – {2d [C] / dt} (D) {2d [B] / dt} = {d [A] / dt}

A-10.

For a chemical reaction 2X + Y  Z, the rate of appearance of Z is 0.05 mol L–1 per min. The rate of disappearance of X will be (A) 0.05 mol L—1 per hour (B) 0.05 mol L—1 per min —1 (C) 0.1 mol L per min (D) 0.25 mol L—1 per min

Section (B) : Rate Law B-1.

B-2.

B-3.

What is the order of a reaction whose rate = KCA3/2 CB–1/2 ? (A) 2 (B)1

(C)– 1/2

(D) 3/2

The rate of certain hypothetical reaction A + B + C  products, is given, by r = – [C]1/4. The order of reaction is given by : (A) 1 (B) 1/2

(C) 2

The rate constant of nth order has units : (A) Litre1–n mol1-n sec–1

(B) Mol1–n litre1-n sec

2

2

(C) Mol1 n litre n sec–1

dA = K [A]½ [B]1/3 dt

(D) 13/12

(D) Mole1–n litren-1 sec–1

B-4.

The rate constant of reaction changes when : (A) Volume is changed (B) Concentrations of the reactants are changed (C) Temperature is changed (D) Pressure is changed

B-5.

For which of the following, the units of rate constant and rate of the reaction are same (A) First order reaction (B) Second order reaction (C) Third order reaction (D) Zero order reaction

B-6.

For a chemical reaction, 2A + 2B  C + D, the order of reaction is one with respect to A and one with respect to B. The initial rate of the reaction is 4 × 10–2 mol L–1 s–1. When 50% of the reactants are converted into products, the rate of the reaction would become(A) 2 × 10–2 mol L–1s–1 (B) 1 × 10–2 mol L–1s–1 –2 –1 –1 (C) 4 × 10 mol L s (D) 2 × 10–1 mol L–1 s–1

B-7.

For a gaseous reaction the rate equation is v = k[A][B]. If the volume of the gaseous system is suddenly reduced to 1/3 of initial volume. The rate would become – (A) 1/9 times (B) 9 times (C) 1/6 times (D) 6 times

B-8.

aA + bB  Product, dx/dt = k [A]a [B]b . If concentration of A is doubled, rate is four times. If concentration of B is made four times, rate is doubled. What is relation between rate of disappearance of A and that of B ? (A) – {d [A] / dt} = – {d [B] / dt} (B) – {d [A] / dt} = – {4 d [B] / dt} (C) – {4 d [A] / dt} = – {d [B]/ dt} (D) None of these

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CHEMICAL KINETICS # 8

B-9.

For the reaction, 2NO(g) + 2H2(g)  N2(g) + 2H2O(g) the rate expression can be written in the following ways : {d [N2] / dt} = k1 [NO][H2] ; {d[H2O] / dt} = k[NO][H2] ; {– d[NO] / dt} = k1 [NO] [H2] ; {–d[H2] / dt} = k1 [NO][H2] The relationship between k, k1 , k1 and k1. is : (A) k = k1 = k1 = k1 (B) k = 2k1 = k1 = k1 (C) k = 2k1 = k1 = k1 (D) k = k1 = k1 = 2 k1

B-10.

For rate constant is numerically the same for three reactions of first, second and third order respectively. Which of the following is correct : (A) if [A] = 1 then r1 = r2 = r3 (B) if [A] < 1 then r1 > r2 > r3 (C) if [A] > 1 then r3 > r2 > r1 (D) All

B-11.

In acidic medium the rate of reaction between (BrO3)¯ and Br¯ ions is given by the expression. – [d (BrO3–) /dt] = K [BrO3– ] [Br¯] [H+]2 It means : (A) Rate constant of overall reaction is 4 sec–1 (B) Rate of reaction is independent of the concentration of acid (C) The changes in pH of the solution will not affect the rate (D) Doubling the concentration of H+ ions will increase the reaction rate by 4 times.

B-12.

For the irreversible process, A + B  products, the rate is first–order w.r.t. A and second–order w.r.t. B. If 1.0 mol each of A and B introduced into a 1.0 L vessel, and the initial rate was 1.0 × 10–2 mol L–1 s–1 , rate when half reactants have been turned into products is : (A) 1.25 × 10–3 mol L–1 s–1 (B) 1.0 × 10–2 mol L–1 s–1 (C) 2.50 × 10–3 mol L–1 s–1 (D) 2.0 × 10–2 mol L–1 s–1

Section (C) : The integrated rate laws C-1.

A first order reaction follows the expressions (A) Cte

k 1t

= C0

(B) Ct = C0 e k 1t

(C) ln

C0 = –k 1t Ct

(D) ln

Ct = k 1t C0

C-2.

Which of following statement is false (A) A fast reaction has a larger rate constant and short half-life (B) For a first order reaction, successive half lives are equal (C) For a first order reaction, the half-life is independent of concentration (D) The half life of a reaction is half the time required for the reaction to go to completion

C-3.

Half life period of a zero order reaction is – (A) Independent of concentration (B) Directly proportional to initial concentration (C) Inversely proportional to concentration (D) Directly proportional to the square of the concentration

C-4.

Wrong data for the first order reaction is – (A) t0.5 =100s, t0.75 = 200s (C) Both the above

(B) t0.75 = 32 min t0.5 = 16min (D) t0.5 = 100s, t0.75 = 150s

C-5.

What is the half life of a radioactive substance if 75% of any given amount of the substance disintegrates in 60 minutes – (A) 2 Hours (B) 30 Minutes (C) 45 Minutes (D) 20 Minutes

C-6.

For an elementary reaction X (g)  Y (g) + Z (g) the half life periods is 10 min. In what period of time would the concentration of X be reduced to 10% of original concentration (A) 20 Min. (B) 33 Min (C) 15 Min (D) 25 Min

C-7.

99% of a first order reaction was completed in 32 min. When will 99.9% of the reaction complete – (A) 50 Min (B) 46 Min (C) 49 Min (D) 48 Min

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CHEMICAL KINETICS # 9

C-8.

In the presence of an acid, the initial concentration of cane sugar was reduced from 0.20 to 0.10 molar in 5 hours and from 0.2 to 0.05 molar in 10 hours. The reaction is of – (A) Zero order (B) First order (C) Second order (D) Third order

C-9.

For a second order reaction, if the conc. of a reactant decreases from 0.08M to 0.04M in ten minutes, what would be the time taken for the conc. to decreases to 0.01M – (A) 20 minutes (B) 30 minutes (C) 50 minutes (D) 70 minutes

C-10.

The rate constant for a second order reaction 8.0 × 10–4 litre mol–1min–1. How long will it take a 0.5M solution to be reduced to 0.25M in reactant – (A) 8.665 × 102 min (B) 8.0 × 10–4 min (C) 2.50 × 103 min (D) 4.0 × 10–4 min

C-11.

A radioactive isotope decomposes according to the first order with half life period of 15 hrs. 80% of the sample will decompose in (A) 15 × 0.8 hr. (B) 15 × (log 8) hr. (C) 15 × (log5 / log2) hr. (or 34.83) hr. (D) 15 × 10/ 8 hr.

C-12.

Radioactive decay follows (A) Zero order kinetics (C) First order kinetics

C-13.

(B) Second order kinetics (D) Pseudo first order kinetics

The t0.5 for the first order reaction, PCI5(g)  PCI3(g) + Cl2(g) is 20 min. The time in which the conc. of PCI5 reduces to 25% of the initial conc. is close to(A) 22 min (B) 40 min (C) 90 min (D) 50 min

C-14.

The rate constant of reaction 2 A + B  C is 2.57 × 10–5 It mole–1 sec–1 after 10 sec. 2.65 × 10–5 It. mole–1 sec–1 after 20 sec. and 2.55 × 10–5 It. mole–1 sec–1 after 30 sec. The order of the reaction is: (A) 0 (B) 1 (C) 2 (D) 3

C-15.

For a first order reaction, the plot of log C against ‘t’ (logC vs 't') gives a straight tine with slope equal to : (A) (k / 2.303) (B) (– k / 2.303) (C) (ln k / 2.303) (D) – k.

C-16.

In a certain reaction, 10% of the reactant decomposes in one hour, 20 % in two hours, 30% in three hours and so on the dimensions of the rate constant is : (A) hour–1 (B) mole litre–1 sec–1 (C) litre mole–1 sec–1 (D) mole sec–1

C-17.

In presence of HCl, sucrose gets hydrolysed into glucose and fructose. The concentration of sucrose was found to reduce from 0.4 M to 0.2 M in 1 hour and to 0.1 M in total of 2 hours. The order of the reaction is : (A) zero (B) one (C) two (D) None of these

C-18.

In a first order reaction the reacting substance has half-life period of ten minutes. What fraction of the substance will be left after an hour the reaction has occurred ? (A) 1/6 of initial concentration (B) 1/64 of initial concentration (C) 1/12 of initial concentration (D) 1/32 of initial concentration

C-19.

K1 K2 In the following first order reactions (A)   Product, (B)   Product, the ratio k1 /k2 if 90% of (A) has

been reacted in time 't' while 99% of(B) has been reacted in time 2t is : (A) 1 (B) 2 (C) 1/2 C-20.

(D) None of these

Two substances A (t1/2 = 5 min) and B (t1/2 = 15 min) are taken in such a way that initially [A] = 4[B]. The time after which both the concentration will be equal is : (Assume that reaction is first order) (A) 5 min (B) 15 min (C) 20 min (D) concentration can never be equal

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CHEMICAL KINETICS # 10

C-21.

If a I-order reaction is completed to the extent of 60% and 20% in time intervals, t1 and t2, what is the ratio, t1 : t2 ? (A) 6.32 (B) 5.58 (C) 4.11 (D) 8.33

C-22.

For a reaction A  Products, the concentration of reactant are C0 , aC0 , a2C0, a3C0 ........... after time interval 0, t, 2t ........where 'a' is constant. Then : (A) reaction is of 1st order and K = (1/1) ln a (B) reaction is of 2nd order and K = (1/tC0)(1–a)/a (C) reaction is of 1st order and K =

C-23.

(D) reaction is of zero order and K =

1  1 ln   t a

The rate constant for the reaction A  B is 2 × 10–4 It. mol–1 min–1. The concentration of A at which rate of the reaction is (1/12) × 10–5 M sec–1 is : (B) (1/20) 5 / 3 M

(A) 0.25 M

C-24.

 1 1 ln  a  t  

(C) 0.5 M

(D) None of these

Graph between concentration of the product and time of the reaction A  B is of the type

Hence graph

between – d[A]/dt and time will be of the type : (–d[A]/dt)

(B)

(A)

(C)

(D)

Time

C-25.

A graph plotted between log t50% vs. log concentration is a straight line. What conclusion can you draw from this graph.

(A) n = 1 ; t1/2  a (C) n = 1 ; t1/2 = (0.693 / k)

(B) n = 2, t1/2  1/a (D) None of these

C-26.

What will be the order of reaction and rate constant for a chemical change having log t50% vs log concentration of (A) curves as : (A) 0, 1/2 (B) 1, 1 (C) 2, 2 (D) 3, 1

C-27.

A drop of solution (volume 0.05 mL) contains 3.0 × 10–6 moles of H+. If the rate constant of disappearance of H+ is 1.0 × 107 mole litre–1 sec–1. How long would it take for H+ in drop to disappear : (A) 6 × 10–8 sec (B) 6 × 10–7 sec (C) 6 × 10–9 sec (D) 6 × 10–10 sec

Section (D) : Methods to determine The rate law D-1.

When concentration of reactant in reaction A  B is increased by 8 times, the rate increase only 2 times. The order of the reaction would be (A) 2 (B) 1/3 (C) 4 (D) 1/2

D-2.

The data for the reaction A + B  C is Exp. [A]0 [B]0

initial rate

1 2 3 4

0.10 0.80 0.10 0.80

0.012 0.024 0.012 0.024

(A) r = k [B]3

0.035 0.035 0.070 0.070 (B) r = k [A]3

(C) r = k [A] [B]4

(D) r = k [A]2 [B]2 .

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CHEMICAL KINETICS # 11

D-3.

A + B  Product,

dx =k [A]a [B]b dt

 dx   = k, then order is :  dt 

If 

(A) 4

(B) 2

(C) 1

(D) 0

D-4.

One litre of 2M acetic acid and one litre of 3M ethyl alcohol were mixed. The esterification takes place according to the reaction: CH3COOH + C2H5OH  CH3COOC2H5 + H2O If each solution is diluted by one litre water the rate would become – (A) 4 times (B) 2 times (C) 0.5 times (D) 0.25 times

D-5.

In the reaction : A + 2B + C  D + 2E Ther rate of reaction remains unchanged if the conc. of B is doubled and that of A and C is kept constant. What is the order with respect to B. (A) 0 (B) 1/2 (C) 1 (D) 3

D-6.

The following data pertain to reaction between A and B : S.No. [A] [B] Rate mol.l–1 mol.l–1 mol.l–1sec–1 I 1 × 10–2 2 × 10–2 2 × 10–4 II 2 × 10–2 2 × 10–2 4 × 10–4 III 2 × 10–2 4 × 10–2 8 × 10–4 Which of the following inference(s) can be drawn from the above data – (a) Rate constant of the reaction 10–4 (b) Rate law of the reaction is k[A][B] (c) Rate of reaction increases four times on doubling the concentration of both the reactant, Select the correct answer – codes :– (A) a,b and c (B) a and b (C) b and c (D) c alone

D-7.

Using the data given below the order and rate constant for the reaction : CH3CHO(g)  CH4(g) + CO(g) would be – Experiment no. Initial conc. Initial rate [mol/] [mol.lit–1sec 1] I 0.10 0.020 II 0.20 0.080 III 0.30 0.180 IV 0.40 0.320 Answer is – (A) 2,[k = 2.0 /mol sec] (B) 0,[k = 2.0 mol/ sec] (C) 2,[k = 1.5 /mol sec] (D) 1,[k = 1.5 sec–1]

D-8.

For the reaction 2A + 3B  products, A is in excess and on changing the concentration of B from 0.1 M to 0.4 M, rate becomes doubled, Thus, rate law is : (A)

D-9.

dx = k[A]2 [B]2 dt

(B)

dx = k[A] [B] dt

For the reaction A  Products, – Time [A]

0 20 mol

5 min 18 mol

(C)

dx = k[A]0 [B]2 dt

(D)

dx = k[B]1/2 dt

d [A] = k and at different time interval, [A] values are : dt

10 min 16 mol

At 20 minute, rate will be : (A) 12 mol /min (B) 10 mol/min

15 min 14 mol

(C) 8 mol/min

(D) 0.4 mol/min

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D-10.

 dx  dx   = k[A]2 . If log   is plotted against log [A], then graph is of the type :  dt   dt 

A  Product and 

(A)

(B)

(C)

(D)

D-11.

If a is the initial concentration of reaction, then the half life period of a reaction of nth order is directly propotional to (A) an (B) an–1 (C) a1–n (D) an+1

D-12.

The half-life period for a reaction at initial concentrations of 0.5 and 1.0 moles litre-1 are 200 sec and 100 sec respectively. The order of the reaction is (A) 0 (B) 1 (C) 2 (D) 3

D-13.

Consider a reaction A   B + C. If the initial concentraton of A was reduced from 4 M to 2 M in 1 hour and from 2 M to 1 M in 0.5 hour, the order of the reaction is(A) One (B) Zero (C) Two (D) Three

Section (E) : Methods to monitor the progress of reaction E-1.

Consider the reaction 2A(g)  3B(g) + C(g). Starting with pure A initially, the total pressure doubled in 3 hrs. The order of the reaction might possibly be (A) zero (B) first (C) second (D) unpredictable from this data

E-2.

Formation of NO2F from NO2 and F2 as per the reaction 2NO2(g) + F2(g)  2NO2F(g) is a second order reaction, first order with respect to NO2 and first order with respect to F2. If NO2 and F2 are present in a closed vessel in ratio 2 :1 maintained at a constant temperature with an initial total pressure of 3 atm, what will be the total pressure in the vessel after the reaction is complete? (A) 1atm

(B) 2 atm

(C) 2.5 atm

(D) 3 atm

E-3.

In a gaseous state reaction, A2 (g)  B(g) + (1/2)C (g), The increase in pressure from 100 mm to 120 mm is notices in 5 minutes. The rate of disapearance of A2 in mm min–1 is : (A) 4 (B) 8 (C) 16 (D) 2

E-4.

The decomposition of a gaseous substance (A) to yield gaseous products (B), (C) follows First order kinetics. If initially only (A) is present and 10 minutes after the start of the reaction the pressure of (A) is 200 mm Hg and that of over all mixture is 300 mm Hg, then the rate constant for 2A  B + 3 C is : (A) (1/1200) ln 1.25 sec–1 (B) (2.303 /10) log 1.5 min–1 –1 (C) (1/10) ln 1.25 sec (D) None of these

E-5.

In the reaction NH4NO2 (aq.)  N2 (g) + 2 H2O (l) the volume of N2 after 20 min and after a long time is 40 ml and 70 ml respectively. The value of rate constant is : (A) (1/20) In (7/4) min–1 (B) (2.303 /1200) log (7/3) sec–1 –1 (C) (1/20) log (7/3) min (D) (2.303 / 20) log (11/7) min–1

E-6.

 / Cu –N2Cl   

–Cl + N2 Half-life is independent of concentration of reactant. After 10 minutes

volume of N2 gas is 10 L and after complete reaction it is 50 L. Hence rate constant is: (A) (2.303 /10) log 5 min–1 (B) (2.303 /10) log 1.25 min–1 –1 (C) (2.303 /10) log 2 min (D) (2.303 /10) log 4 min–1

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CHEMICAL KINETICS # 13

Section(F) : Effect of Temperature F-1.

The activation energies of the forward and backward reactions in the case of a chemical reaction are 30.5 and 45.4 KJ/mol respectively. The reaction is – (A) Exothermic (B) Endothermic (C) Neither exothermic nor endothermic (D) Independent of temperature

F-2.

Chemical reaction occurs as a result of collisions between reacting molecules. Therefore, the reaction rate is given by (A) Total number of collisions occurring in a unit volume per second (B) Fraction of molecules which possess energy less than the threshold energy (C) Total number of effective collisions (D) None of the above

F-3.

The chemical reactions in which reactants require high amount of activation energy are generally (A) Slow (B) Fast (C) Instantaneous (D) Spontaneous

F-4.

For a reaction for whi ch the activation energies of forward and reverse reactions are equal (A) H = 0 (B) S = 0 (C) The order is zero (D) There is no catalyst

F-5.

The activation energy of reaction is equal to(A) Threshold energy for the reaction (B) Threshold energy + Energy of the reactants (C) Threshold energy – Energy of the reactants (D) Threshold energy + Energy of the products

F-6.

Which of the following explains the increase of the reaction rate by catalyst (A) Catalyst decreases the rate of backward reaction so that the rate of forward reaction increases (B) Catalyst provides extra energy to reacting molecules so that they may reduce effective collisions (C) Catalyst provides an alternative path of lower activation energy to the reactants (D) Catalyst increases the number of collisions between the reacting molecules.

F-7.

The activation energy of the reaction A + B  C + D + 38 k.cal is 20 k.cal, What would be the activation energy of the reaction, C + D A + B (A) 20 k.cal (B) –20 k.cal (C) 18 k.cal (D) 58 k.cal

F-8.

Rate of which reactions increases with temperature : (A) of any (B) of exothermic reactions (C) of endothermic (D) of none.

F-9.

For a zero order reaction. Which of the following statement is false : (A) the rate is independent of the temperature of the reaction. (B) the rate is independent of the concentration of the reactants. (C) the half life depends upon the concentration of the reactants. (D) the rate constant has the unit mole It–1 sec–1.

F-10.

A large increase in the rate of a reaction for a rise in temperature is due to (A) increase in the number of collisions (B) the increase in the number of activated molecules (C) The shortening of mean free path (D) the lowering of activation energy

F-11.

The first order rate constant k is related to temp. as log k = 15.0 – (106 /T) Which of the following pair of value is correct ? (A) A = 1015 and E = 1.9 × 104 KJ (B) A = 10–15 and E = 40 KJ (C) A = 1015 and E = 40 KJ (D) A = 10–15 and E = 1.9 × 104 KJ.

F-12.

The decomposition of N2O into N2 & O2 in presence of gaseous argon follow second order kinetics with k = (5.0 × 1011 L mol1 s1) e (A) 5.0 x 1011 J



41570 K T

(B) 41570 J

(K stands for Kelvin units). The energy of activation of the reaction is : (C) 5000 J

(D) 345612.98 J

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CHEMICAL KINETICS # 14

F-13.

How much faster would a reaction proceed at 25°C than at 0°C if the activation energy is 65 kJ? (A) 2 times (B) 5 times (C) 11 times (D) 16 times

F-14.

The rate constant, the activation energy and the frequency factor of a chemical reaction at 25°C are 3.0 × 10–4 s–1, 104.4 KJ mol–1 and 6.0 × 1014 s–1 respectively. The value of the rate constant as T   is : (A) 2.0 × 1018 s–1 (B) 6.0 × 1014 s–1 (C) infinite (D) 3.6 × 1030 s–1

F-15.

For a given reaction, energy of activation for forward reaction (Eaf) is 80kJ.mol1. H = 40kJ.mol1 for the reaction. A catalyst lowers Eaf to 20 kJ.mol1. The ratio of energy of activation for reverse reaction before and after addition of catalyst is : (A) 1.0 (B) 0.5 (C) 1.2 (D)2.0

Section (G) : Mechanism of reactions G-1.

For the reaction H2 (g) + Br2 (g)  2HBr (g) the experiment data suggested that r = k[H2][Br2]1/2 The molecularity and order of the reaction are respectively : (A) 2, 3/ 2 (B) 3/2 , 3/2 (C) Not defined, 3/2 (D) 1,1/2

G-2.

For the reaction NO2 + CO  CO2 + NO the experimental rate expression is – molecules of CO involves in the slowest step will be (A) 1 (B) 2 (C) 3

G-3.

dc = k[NO2]2 the number of dt

(D) depends on mechanism

The reaction of hydrogen, and iodine monochloride is represented by the equation : 2HCl(g) + 2(g) H2(g) + 2Cl(g) This reaction is first–order in H2(g) and also first–order in Cl(g). Which of these proposed mechanism can be consistent with the given information about this reaction ? Mechanism  : H2(g) + 2Cl(g) 2HCl(g) + 2(g) Slow

Mechanism  : H2(g) + Cl(g)   HCl(g) + H(g) (A)  only G-4.

G-5.

fast HI(g) + Cl(g)   HCl(g) + I2(g) (B)  only (C) both  and 

The reaction, X + 2Y + Z  N occurs by the following mechanism (i) X+Y M very rapid equilibrium (ii) M+ZP slow (iii) O+YN very fast What is the rate law for this reaction (A) Rate = k[Z] (B) Rate = k[X] [Y]2 [Z] (C) Rate = [N]

(D) neither  nor 

(D) Rate = k[X] [Y] [Z]

In the Lindemann theory of unimolecular reactions, it is shown that the apparent rate constant for such a reaction is kapp =

k1C where C is the concentration of the reactant k1 and  are constants. Calculate the 1  C

value of C for which kapp has 90% of its limiting value at C tending to infinitely large values, given  = 9 x 105. (A) 106 mole/litre (B) 104 mol/litre (C) 105 mole/litre (D) 5 x 105 mol/litre G-6.

G-7.

Trimolecular reactions are uncommon because (A) the probability of three molecules colliding at an instant is very low. (B) the probability of three molecules colliding at an instant is high. (C) the probability of three molecules colliding at an instant is zero. (D) the probability of many molecules colliding at an instant is high. Select the correct statements : (A) the molecularity of an elementary reaction indicates how many reactant molecules take part in the step. (B) the rate law of an elementary reaction can be predicted by simply seeing the stoichiometry of reaction. (C) the slowest elementary step in sequence of the reactions governs the overall rate of formation of product. (D) a rate law is often derived from a proposed mechanism by imposing the steady state approximation or assuming that there is a pre-equilibrium.

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Section (H) : Complications in first order reactions H-1.

 dx 

For an elementary reaction, net rate is  dt  = k[A]2 – k'[C] [B]2 then, select the correct statement :   2

(A) –

d[C] d[ A ]2 d [B] = = is the relation among (B) 2A dt dt dt

(C) both are correct H-2.

2B + C is the required reaction

(D) none is correct

Consider the elementary reaction sequence shown in figure. Which of the following equations are correct ?

H-3.

(A)

d[A] = k1[A] + k4[D] dt

(B)

d [ C] = k2[B]  k3[C] dt

(C)

d [D] = k4[D] + k3[D] dt

(D) Nothing can be said about order of reactions in this problem

For an elementary reaction of reversible nature, net rate is :  dx    = k [A]2 [B]1 – k [C], hence, given reaction is : 1 2  dt 

(A) 2A +

H-4.

1 B

C

(B) 2A + B

(C) 2A

C + B–1

(D) None of these

At a given temperature, k1 = k2 for the reaction C+D

A+B If

H-5.

C

 dx   dt  = k1 [A] [B] – k2[C] [D] in which set of the concentration reaction ceases?  

[A]

[B]

[C]

[D]

[A]

[B]

(A)

0.1 M

0.2 M

0.3 M

0.4 M

(B)

0.4 M

0.25 M 0.2 M

[C]

0.5 M

[D]

(C)

0.2 M

0.2 M

0.3 M

0.2 M

(D)

0.2 M

0.2 M

0.2 M

0.4 M

The substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as :

K1 = 1.26 × 10–4 sec–1 and K2 = 3.8 × 10–5 sec–1 The percentage distribution of B and C

H-6.

(A) 80% B and 20% C

(B) 76.83% B and 23.17%C

(C) 90% B and 10% C

(D) 60% B and 40% C

The rate constant for two parallel reactions were found reactions were found to be 1.0 × 10–2dm3 mol-1 s–1 and 3.0 × 10–2 dm3 mol–1 s–1. If the corresponding energies of activation of the parallel reactions are 60.0 kJ mol–1 and 70.0 kJ mol–1 respectively, what is the apparent overall energy of activation ? (A) 130.0 kJ mol–1

(B) 67.5 kJ mol–1

(C) 100.0 kJ mol–1

(D) 65.0 kJ mol–1

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CHEMICAL KINETICS # 16

PART - II : MISCELLANEOUS QUESTIONS COMPREHENSION Comprehension # 1 A(g)  2B(g) + C(g) Initially at t = 0 gas A was present along with some amount of gas (C). At t = 0 mole fraction of gas C is 1/3. After some time t = t1, total pressure is half of the final total pressure at t = tx (a very long time). Assume this decomposition is a first order, at a constant temperature. It is also given at t = tx, final total pressure is 35 bar. 1.

At t = t1 pressure of gas B is : (A) 2.5 bar

2.

(B) 1.25 bar

(D) data is insufficient

Rate constant (k) = (log 64 – log 49) s–1. Value of t1 in seconds is : (A) 2.15 s

3.

(C) 5.0 bar

(B) 1.5 s

(C) 2.3 s

(D) 1.15 s

(C) 1 : 3 : 5

(D) 1 : 3 : 5

Ratio of rate constant at t = 0 to t = t1 to t = tx is : (A) 2 : 3 : 4

(B) 1 : 1 : 1

Comprehension # 2

4.

For the (Set-1) : (A) if T1 > T2, k1 > k2 always

(B) if T1 > T2, k1 > k2 (for exothermic reaction)

(C) if T1 > T2, k1 < k2 (for endothermic reaction) (D) Ea1  Ea2 5.

For the (Set-1) : (A) Ea1 > Ea2

if T1 > T2

(C) Ea1 = Ea2 6.

(B) Ea1 < Ea2

if T1 > T2

(D) Ea1 = 0.5 Ea2

Comparing set-I and II : (A) k4 > k3 & k2 > k1 ,

if T2 > T1 (endothermic) (B) k4 < k3 & k2 > k1 ,

if T2 < T1 (endothermic)

(C) k4 > k3 & k2 > k1 ,

if T2 < T1 (exothermic) (D) k4 < k3 & k2 < k1,

if T2 > T1 (exothermic)

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Comprehension # 3 A reaction is said to be first order if it's rate is proportional to the concentration of reactant. Let us consider a reaction A(g)  B(g) + C(g) At t = 0 a 0 0 At time t a–x x x The rate of reaction is given by the expression

represented as k =

dx = k(a – x) and integrated rate equation for a given reaction is dt

1  a   where a = initial concentration and (a – x) = concentration of A after time t. ln  t ax

7.

Thermal decomposition of compound X is a first order reaction. If 75% of X is decomposed in 100 min. How long will it take for 90% of the compound to decompose? Given : log 2 = 0.30 (A) 190 min (B) 176.66 min (C) 166.66 min (D) 156.66 min

8.

Consider a reaction A(g)  3B(g) + 2C(g) with rate constant 1.386 × 10–2 min–1. Starting with 2 moles of A in 12.5 litre vessel initially, if reaction is allowed to takes place at constant pressure & at 298K then find the concentration of B after 100 min. (A) 0.04 M (B) 0.36 M (C) 0.09 M (D) None of these

Comprehension # 4 Competing First-Order Reactions Frequently a species can react in different ways to give a variety of products. For example, toluene can be nitrated at the ortho, meta, or para positions, We shall consider the simplest case, that of two competing irreversible first-order reactions : k k2 D A 1  C and A   where the stoichiometric coefficients are taken as unity for simplicity. The rate law is  d[A]    = – k1[A] – k2[A] = – (k1 + k2) [A]  dt 



[A] = [A]0 e (k1 k 2 )t .

 d [C]  For C, we have   = k1[A] = k1[A]0 e (k1 k 2 )t . Multiplication by dt and integration from time 0  dt  k [A] ( k  k ) t (where [C]0 = 0) to an arbitary time t gives [C] = 1 0 (1  e 1 2 ) k1  k 2  d [D]   gives Similarly, integration of   dt 

[D] =

k 2 [ A ]0 (1 – e (k 1  k 2 )t ) k1  k 2

The sum of the rate constants k1 + k2 appears in the exponentials for both [C] and [D]. At any time we also have

9.

A

k1 [C] = k [D] 2

starting initially with only A Which of the following is correct at time t

(A) [A]0 = [A]t +[B]t + [C]t (C) [A]0 = [A]t +

[C] t [B] t + 3 2

(B) [A]0 = [A]t + 2 [B]t + 3 [C]t (D) [A]0 =

2 [A]t +[B]t + [C]t 3

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CHEMICAL KINETICS # 18

10.

[ X] t starting with only 'X', ratio [ Y ]  [ Z] t t

X

1

11.

(A) Independent of time

(B)

(C) Depends upon initial concentration of X

(D) [A]0 (ekt –1)

(e kt  1)

At high temperature acetic acid decomposes into CO2 & CH4 and simultaneously into CH2CO (ketene) and H2O 1

k1 3s (i) CH3COOH   CH4 + CO2

1

k 2  4s (ii) CH3COOH   CH2CO + H2O

What is the fraction of acetic acid reacting as per reaction (i) ? (A)

12.

3 4

(B)

3 7

(C)

4 7

(D) none of these

starting with pure A ratio of rate of production of B to C is

For A

(A) Independent of time (C) Depends upon initial concentration of A

(B) Independent of temperature (D) Independent of mechanism of reaction

Comprehension # 5 For the given sequential reaction k

k

2 1 A  B  C the concentration of A, B & C at any time 't' is given by

[A]t =

[A]0 e  k1t

;

[B]t =

k1[ A ]0 e  k1t  e  k 2 t ( k 2  k1 )





[C]t = [A0] – ( [A]t + [B]t ) The time at which concentration of B is maximum is (A)

(B)

1 k ln 1 k 2  k1 k 2

(C)

1 k ln 1 k1  k 2 k 2

(D)

k2 k 2  k1

Select the correct option if k1 = 1000 s–1 and k2 = 20 s–1.

[B]t [A]t

(B)

[C]t

Conc.

(A)

[C]t

conc.

[B] t

time

[A]t

time

(C)

[C]t

(D) [B]t

Conc.

14.

k1 k 2  k1

Conc.

13.

[C]t

[B]t [A] t

[A]t

time

time ETOOSINDIA.COM

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MATCH THE COLUMN 15.

16.

17.

The polarimeter readings in an experiment to measure the rate of inversion of cane suger (1st order reaction) were as follows time (min) : 0 30  angle (degree) : 30 20 – 15 Then match the following. (Use : log 35 = 1.54, log 3 = 0.48, log 2 = 0.3) Column - I

Column - II

(A)

The half life of the reaction

(p)

120 min.

(B)

The solution is optically inactive at

(q)

7.5°

(C)

The equimolar mixture of the products

(r)

75.2 min.

(D)

The angle at half time

(s)

laevorotatory

Match the following : Column - I

Column - II

(A)

A + B  C + D r = k1 [A] [B]

(p)

Unit of rate constant possess concentration unit

(B)

A + B  C + D r = k2 [A] [B]º

(q)

Rate constant for the reaction of both the reactants are equal

(C)

A + B  C + D r = k3 [A]º [B]º

(r)

Rate of consumption of at least one of the reactants is equal to rate of production of at least one of the products

(D)

2A + B  2C + 3D r = k3 [A]º [B]º

(s)

If both reactants are taken in stoichiometric ratio, half life for both reactants are equal

For the reaction of type A(g)  2B(g) Column-I contains four entries and column-II contains four entries. Entry of column-I are to be matched with ONLY ONE ENTRY of column-II Column - I

Column - II

(A)

d[ B]  d[A ] vs for first order dt dt

(p)

(B)

[A] vs t for first order

(q)

(C)

[B] vs t for first order

(r)

(D)

[A] vs t for zero order

(s)

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18.

Match the following : Column-I

Column-II

(A)

If the activation energy is 65 kJ then how much time faster a reaction proceed at 25°C than at 0°C

(p)

2

(B)

Rate constant of a first - order reaction is 0.0693 min–1. If we start with 20 mol L–1, it is reduced to 2.5 mol L–1 in how many minutes

(q)

Zero

(C)

Half - lives of first - order and zero order reactions (r) are same. Ratio of rates at the start of reaction is how many times of 0.693 Assume initial concentration to be same for the both.

11

(D)

The half-life periods are given , [A]0 (M) 0.0677 t1/2 (sec) 240 order of the reaction is

30

(s) 0.136 480

0.272 960

ASSERTION / REASON Directions : Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. (E) Statement-1 and Statement-2 both are False. 19.

Statement-1 Statement-2

: A fractional order reaction must be a complex reaction. : Fractional order of RDS equals to overall order of a complex reaction.

20.

Statement-1

: The time of completion of reactions of type A  product (order