Chemical Kinetics

Reconstruct Your Chemistry With Prince Sir

Chemical Kinetics is the branch of chemistry which deals with the rate of change of conecentration from intial state to final state under non-equilibrium conditions and the factors affecting the rate of reaction. RATE OF REACTION

The rate of reaction is defined as the change in conecentration of reactants or products in unit time. Rate of reaction = Change in concentrat ion of reactants/ Products Time taken Rate can be defined in two ways : (i) Average Rate:- The rate of reaction measured over a time interval is defined as the average rate of reaction. C2  C1

r i S

C

Rav. = t  t  t 2 1 (ii) Instanteneous Rate :- The rate of reaction at any given instant is defined as the instanteneous rate .

e c n

“It is simply the slope of concentration time graph at that particular instant”

C1

rav=

C2

t1 t2

P y

–[R] –[C2– C1] = [t2– t1 ] t

B

ir

concentration

concentration

[R]0

[P]

slope =rinst=

time

d[P] dt

time

(b)

(a)

Dependence of rate on Stoichiometry

For any Reaction

Rate of Reaction  

  

n1A +n2 B  n3C + n4D

1 d [A ]

n1 dt



1 d [B ]

n2 dt



1 d [C ]

n3 dt



1 d [D ]

n4

dt

For reactants negative sign is used because the concentration of reactants decreases with time. For products positive sign is used because the concentration of products increases with time. Specific care should be taken while dealing with this expression here ,

Reconstruct Your Chemistry With Prince Sir



1 d [A] n1 dt

= rate of reaction where as 



1 d [C ] n1 dt

= rate of reaction where as 

d [A] = rate of disappearence of A dt

d [C ] = rate of appearence of C dt

Example 1. For the reaction: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) The concentration of NO changes from 1.8M to 5.4 M in 100 seconds than calculate (i) Rate of disappearence of Ammonia (ii) Rate of reaction Solution - Rate of appearence of NO = 

Now from the rate expression Rate of Reaction  

Rate of disappearence of NH3 =  (ii) Rate of reaction = 

ri

d[NO] (5.4  1.8)   3.6  102 molL-1 sec1 dt 100

S e

1 d [NH3 ] 1 d [O2 ] 1 d [NO ] 1 d [H 2O ]    4 dt 5 dt 4 dt 6 dt

c in

d [NH3 ] 4 d [NO ] d [NO ]    3.6  102 mol L-1 sec1 dt 4 dt dt

1 d [NO ] 1   3.6  10  2  9  10  3 mol L- 1 sec 1 4 dt 4

r P

LAW OF MASS ACTION According to this law the rate of a chemical reaction is directly proportional to the product of active masses of reactants raised to the power their stochiometric coefficients. For any reaction n1A + n2B  n3C + n4D

y B

Rate  [ A ] n [ B ] n 1

2

Rate  K [ A ] n [ B ] n 1

2

This expression is called rate expression or rate law. Active Mass - For aqeous phase active mass can be replaced by concentration (Molarity). For gaseous reactants active mass can be replaced by either concentration or partial pressure.

Here ‘K’ is called the rate constant or specific rate of the reaction .Greater is the value of rate constant more will be the rate of reaction. Note :-1. The reagent which is present in excess is not included in the rate law, since the change in its concentration is insignificant. 2. Intermediates are not involved in the rate law they has to be eleminated if are present in the rate law.

Reconstruct Your Chemistry With Prince Sir ORDER AND MOLECULARITY MOLECULARITY :- The number of reactant molecules which collide with each other in a chemical reaction to form products is called the molecularity of the reaction.On the basis of Molecularity reactions can be classified as Br2  2Br - Unimolecular H2 + I2  2HI - Bimolecular 2NO + O2  2NO2- trimolecular Note :- 1. Molecularity can niether be zero , negative , fractional and infinite. 2. Moleuclarity is a theoretical quantity. 3. Molecularity of a reaction cannot be greater than 3 because more than 3 molecules can’t collide effectively to form products. 4. Molecularity is independent of pressure and temperature. 5. Molecularity of a single step reaction i.e. simple reaction is the number of molecules participating in the reaction. 6.But for a complex multistep reaction the molecularity of each step is different and there is no meaning of overall molecularity of the reaction ORDER

ir

e c n

r i S

It is defined as “The sum of the powers of all the concentration terms included in the rate law” n1A + n2 B  products

P y

rate  k [ A ] n [ B ] n 1

2

Order of reaction = (n1+n2) Note:-

1. Order of a reaction is an experimentally determined quantity. 2. Order can be zero, negative and fractional but cannot be infinity. 3. Order depends on pressure and temperature. 4. For a multistep complex reaction the overall order of the reaction is determined by RDS(Rate Determinind Step).

B

Example 2 : For the reaction 2H2 + 2NO  N2 + 2H2O, the following mechanism has been suggested: (i) 2NO  N2O2 equilibrium constant K1 (fast) (ii)

k2 N2O2 + H2   N2O+ H2O (slow)

k3 (iii) N2O + H2   N2 + H2O (fast) Determine the order of the reaction

Sol:- The slowest step is the rate RDS(Rate Determining Step) k2 N2O2 + H2   N2O+ H2O (slow) the rate law for RDS rate = k[N2O2] [H2]

............(1)

Reconstruct Your Chemistry With Prince Sir But since N2O2 is an inetermediate it cannot be included in rate law so it has to be elemintaed So for the equilibrium reaction 2NO  N2O2 equilibrium constant K1 K1 

[N 2O 2 ] [NO ]

2

 [N 2O 2 ]  K1 [NO ] 2

replacing [N2O2] in equation ..(1) Rate = KK1[NO]2[H2] = K’ [NO]2[H2] So the order of the reaction = 2+1 = 3 The molecularity of each step is as follows: (i) 2 (ii) 2 (iii) 2

ri

Significance of the Order Of Reaction The order of a reaction is different with repect to each reactant. The order with respect to a particular reactant tells that how the rate of reaction will change on changing the concentration of that reactant. For example consider the rate law rate = k [A]2[B]3 So if concentration of A is doubled than rate of reaction will become four times as the order with respect to A is 2. if concentration of B is doubled than rate of reaction will become 8 times as the order with respect to B is 3.

r P

c in

S e

Types of reactions on the basis of order

ZERO ORDER REACTIONS Consider a general reaction A  Products at t=0 [A0] at t=t [At] Since the order iszero so the rate of reaction can be written as

y B

Rate = 

d[At ] dt

 d[At ] =kdt

=k[At]0

On integrating

[At] = -kt + c

At t=0, [At] = [A0]

, So c = [A0]

So for zero order reaction

[A]t = [A]0 – kt

Half Life “the time taken for the concentration of reactants to decrease to half of its initial value” [ A ]0

t1/2 = 2 k

Reconstruct Your Chemistry With Prince Sir Rate

conc.

[A] time

time

Examples : (i) Photochemical Reactions (ii)

h H2(g) + Cl2 (g)   2HCl (g) Decomposition of substances over metal surface 1 hot Pt.  N2 (g) + N2O(g)  O (g) Surface 2 2 Mo or W

2NH3 (g)  N2 + 3H2 surface

(iii)

Enzyme catalysed reaction

e c n

(E ) Substrate(S) Enzyme    product(P). FIRST ORDER REACTION Consider a general reaction A  Products at t=0 [A0] = a

at t=t

ir

r i S

[At]= (a-x) Since the reaction is of first order the rate of reaction can be

written as

P y Rate = 

B

d[At ] dt

=k[At]1

[ A0 ] 1 On integrating we get , k  t ln [ A ] t

k 

2.303

t

log10

[ A0 ] [ At ]

or k 

2.303

t

log10

a (a  x )

The Exponential form of 1st order equation is [A]t = [A]0e kt Half -life : t1/2 =

0 .693

k

For a first order reaction half life is independent of initial concentration of reactant.

Reconstruct Your Chemistry With Prince Sir t1/2

a

The equation k



1

t

ln

[ A0 ]

[ A t ] , can be written as

ln[Ao] - ln[At] = kt or ln[At] = -kt + ln[Ao] The graph between ln[At] and t will be a straight line with negative slope, and slope = rate constant (k)

In [A]t

log10 [A]t

t

r P

c in

S e

ri

t

If the graph is plotted between ln[At] and t the slope will be equal to k and if the graph is plotted between log10[At] and ‘t’ the slope will be Examples: (i) Radioactive disintegration is a first order reaction. (ii) (iii) (iv)

y B

k . 2.303

H  catalysed hydrolysis

 C6H12O6 + C6H12O6. C12H22O11 + H2O  Inversion

(glucose) Mineral acid catalyzed hydrolysis of esters. Decomposition of H2O2 in aqueous solution.

(fructose)

Methods To Calculate the rate constant of First Order Reaction 1. By Titration:- In this method an unknown solution (say ‘A’) is titrated against a standard solution of known concentration(Say ‘B’). The volume of standard solution required at different time intervals is measured from which the concentration of unknown solution can be calculate . Vo = Volume of standard solution used at time t = 0 Vt = Volume of standard solution used at time t

V = Volume of standard solution used at completion of titration i.e. at t = 

So,

Reconstruct Your Chemistry With Prince Sir  Initial amount of A present

V - Vo

V - Vt  1

k 

t

k=

ln

Amount of A present at any time t

[ A0 ] [ At ]

2.303

t

V  V o   log  V  V  t   

2. By Measuring angle of rotation:-

r i S

In this method the total angle of rotation of the sample is measured at different time intervals from which we can calculate the amount present at any given time. Consider a reaction: A  B If r0 = initial angle of rotation of the sample (when only A is present) rt = angle of rotation after time t (when both A and B are present) r = angle of rotation after completion of reaction (when only B is present) r0- r  Initial amount of A present

r  - rt 

2.303  r0  r  k = t log r  r    t

P y

3. By Measuring Pressure

B

Total pressure at time t

k=

ir

A(g)  B (g) + C(g) Po Po- x x x

at t=0 at t=t

k=

e c n

Amount of A present at any time t

2.303

t

2.303

t

log

Pt = Po- x + x + x = Po+ x  x  Pt  Po

P0 P0  x

log

P0

2P0  PT

where P0 = Initial partial pressure of A PT = Total pressure of gaseous system at time 't'

(i)

Reconstruct Your Chemistry With Prince Sir Kinetics Of Some Miscelleneous Reactions Parallel Reactions :- here both the side reactions are of first order. B X

A at any time t , ( A - x )

C X

Let us assume the initial concentration of A is ‘a’ and after time t ‘x’ of it decomposes to form B and C.So we can write Rate of diappearence of A = Rate of formation of B + rate of formation of C

dx  k1 (a  x )  k 2 (a  x ) =(k1+k2) (a-x) dt 

dx

(a  x )

 (k 1  k 2 ) t

S e

a 1 a  ln  (k1  k2 )t  (k1  k2 )  ln (a  x ) t (a  x ) Here ,

(K1 + K2) = Kav.

r P

K1 % Yield of B = K  K  100 2 1

So,

y B

[A] = [ Ao ]e

( K1 + k 2 )t

;

c in

[B ] k1  [C ] k2

and

ri

K2 % Yield of C = K  K  100 2 1

k1[A0 ]

[B] = k + k (1 - e 2 1

-(k1 +k2 )t

);

k2[A0 ]

[C] = k + k (1 - e 2 1

- (k1 +k2 )t

)

Reversible reaction Consider the reaction where both the forward and backward reactions are of first order. (ii)

A

kf

kb

B

At time t = 0 a b At time t = t a–x x At time t = teq a – xeq xeq Rate of the forward reaction = Kf(a-x) Rate of the backward reaction = Kbx at equilibrium , Rate of forward reaction = Rate of backward reaction

xeq

kf

Kf(a-x) = Kbx  a  x = k eq b

Reconstruct Your Chemistry With Prince Sir Now net rate of formation of B

dx = Kf(a-x)- Kbx dt

Substituting the value of Kb and on integrating we get

 x eq    ln (kf + kb) = t  x eq  x    1

(iii)

Consecutive Reaction

ir

k1 k2 A  B  C Here we can write that , Rate of formation of B = Rate of disappearence of A - rate of formation of C

[A]t = [A]0e k1t

;

[C]t = [A0] – ( [A] + [B] ) tmax

conc.•

 k1  1 = (k  k ) ln k  2 1  2

[C]

[B] [A]

y B time

S e 

c ni

;

r P

k1[A]0 [B]t = (k  k ) ek1t  ek2t 2 1

[B]max

k2



 k  k 2  k1 = [A0].  1   k2 

In this reaction the concentration of B will first reach to a mximum and than decreases as initially the rate of formation of B is greater than rate if its decomposition. (iv) Pseudo First Order Reactions: These are the reactions which doesn’t appear to be first order but are actually of first order. in these types of reaction one of the reactants is present in excess, so it is not included in rate law and the rate of reaction does not depend on the concentration of that reactant.Consider the hydrolysis of ester CH3COOC2H5 + H2O

H CH3COOH + C2H5OH  Here water is taken in excess so its concentration doesn’t changes much and therefore it is not included in the rate law . Rate = k [CH3COOC2H5] Intially the reaction appears to be of the second order but the reaction actually is of first order and therefore is called a pseudo first order reaction. For a pseudo first order reaction we can apply the kinetics of the first order.

Reconstruct Your Chemistry With Prince Sir METHODS OF DETERMINATION OF ORDER OF REACTIONS : A few methods commonly used are given below : 1. Hit & Trial Method : It is method of using integrated rate equations, where the experimentalvalues of a, x & t are put into these equations. One which gives a constant value of k for different sets of a, x & t correspond to the order of the reaction. 2. Graphical Method : In this method graphs are plotted between the concentration of reactants and time (i) A plot of [At] Vs ‘t’ gives a straight line for a zero order reaction. conc.

[A] time

S e

ri

(i) A plot of log [At] versus 't' gives a straight line for the First order reaction.

In [A]t

r P t

y B

c in

(ii) A plot of (a – x)– (n–1) versus 't' gives a straight line any reaction of the order n (except n = 1). 3

.

Half Life Method : The half life of nth order reaction is

t 1 (a )1n , Where n is the order of the reaction

So,

2

(t 1 / 2 )1

(t 1 / 2 )2

a   1 a   2

1 n

By experimental observation of the dependence of half life on initial concentration we can determine n, the order of reaction. n = 1+

log(t1 / 2 )2  log(t1 / 2 )1 log(a0 )1  log(a0 )2

.

4.

Reconstruct Your Chemistry With Prince Sir By the unit of rate constant :

 mol  The unit of rate constant for an ‘nth’ order reaction is    Litre 

(1  n )

sec 1

For zero order reaction the unit will be mol L -1sec-1. For first order reaction the unit of rate constant is sec -1. For second order reaction the unit of rate constant is Lmol -1sec-1.

DEPENDENCE OF RATE CONSTANT ON TEMPERATURE(Arrehnius Equation) According to Arrhenius hypothesis : (i) Among all the reactant molecules only a certain number of molevules participate in a chemical reaction which are known as active molecules and the rest are known as passive molecules. (ii) there exist an equillibrium between active and passive molecules and when temperature is increased the active molecules are converted into passive molecules by absorbing energy. Active molecules Passive molecules

ir

e c n

r i S

The temperature dependence of rate constant can be explained on the basis of Arrhenius equation: E  a RT

k = A. e Where A is Arrhenius constant or frequency factor or pre-exponential factor and is characteristic of a reaction. R  Universal gas constant Ea Activation energy (in J/Mol-1)

B

P y

Straight from NCERT

Ea

* In the Arrhenius equation e RT is the fraction of molecules that have kinetic energy greater than activation energy . 

Ea

Using k = A. e RT E ln k   a  ln A RT 

Thus the plot of lnk Vs 1/T will be a straight line

Reconstruct Your Chemistry With Prince Sir

let at temperature T1 rate constant be K1 and temperature T2 rate constant be K2, than

k1  Ae

E a RT 1

E a Ea Ea RT  ln k1    ln A & k 2  Ae  ln k 2    ln A RT 2 RT 1 1

on solving we get k2 Ea log k  2.303R 1

1 1    T1 T2 

S e

Temperature coefficient()

ri

The temperature coefficient of a chemical reaction is defined as the ratio of the reat of reaction two temperatures differing by 10°C. kt  10 Temperature coefficient() = k . t Generally for every 100C rise in temperature the rate of reaction becomes by twice or thrice , therefore the value of temperature coefficient lies between 2 & 3. This change in reaction rate can be explained on the basis of distribution curve proposed by Boltzman and Maxwell. They plotted the kinetic energy with fraction of molecules

y B

r P

c in

NE NT ( Where NE  number of molecules with energy E and

NT  total number of molecules).

Reconstruct Your Chemistry With Prince Sir

Let the temperature coefficient of a reaction be ‘ ’ when temperature is increased from T1 to T2 than the ratio of rtae constants can be calculated as kT

KT

2

 ( )

(T 2 T 1 ) 10

2

COLLISION THEORY According to this theory , the reactant molecules are assumed to be hard spheres and a reaction will occurr only if the molecules collide with each other.  Not all collisions result in product formation, those collisions which result in product formation are known as Effective Collisions.For a collision to be effective following two barrier has to be crossed (i) Energy Barrier (ii) Orientation Barrier

r i S

Fraction of molecules

(i) Energy Barrier :- For an effective collision to occurr the colliding reactant molecules must have kinetic energy greater than the activation energy.

ir

e c n

Fraction of molecules having energy greater than activation energy and can prdouce effective collisions

P y

Energy

B

A collision between high energy molecules overcomes the force of repulsion and forms an unstable activated complex which has maximum energy and is highly unstable.This activated than breaks into products or reactants. Thershold Energy(ET) :- The minimum energy which the reactant molecules must have in order to form products in a chemical reaction is called Threshold Energy . Activation Energy (Ea) :- The extra energy which the reactant molecules require to collide effectively and form products in a chemical reaction is called activation energy . Here, Eaf  Activation energy of forward reaction Eab  Activation energy of backward reaction

Reconstruct Your Chemistry With Prince Sir

Eb

Energy

Ea

ET R eactants

ER

Pro duct

Ep

R eactio n C o-ordinates

Overall activation energy of reaction = Eaf-Eab = H

ri

(ii) Orientation Barrier :- For a collision to be effective the reactant molecules must collide with proper orientation, such that they cover the maximum surface area of each other. ADD DIAGRAM FROM NCERT Thus in collision theory Activation energy and proper orientation together determine the criteria for an effective collision and the rate of chemical reaction.

r P

c in

S e

 There are two factors on which the rate of reaction depends: (i) Collision Frequency (ii) Activation energy Collision Frequency(Z) :- The number of collisions per second per unit volume of the reaction mixture is known as collision frequency. For a reaction A + B  Products

y B

The rate of reaction can be written as rate = Z AB Where, ZAB is the collision frequency and

e

 Ea RT

e

 Ea RT

is the fraction of molecules

having energy equal to or greater than Ea.

REACTION MECHANISM

The mechanism of a reaction is the path through which the conversion of reactants take place into products. The reaction mechanism could consist of several steps . these steps can be fast and slow reactions. We are concerned with the slowest reaction which we call RDS (Rate Determining Step) , because we can observe the change in concentration of reactants and products only for slow reactions.  To determine the order of the reaction we will first write the rate law for the RDS and if the RDS consist of any intermediate specie we will eleminate it .

Reconstruct Your Chemistry With Prince Sir  following two methods can be used to determine the rate law for a multistep reaction (i) By equilibrium Consideration consider the following mechanism for the reaction 2O3(g)  3O2(g) Mechanism : Step-1 step-2

O3(g)

kf

kb

O2(g) + O(g)

fast

O(g) + O3(g) k2 2O2(g)

slow

r i S

Since step-2 is the RDS so the rate law will be : Rate = k [O] [O3] but here [O] is the intermediate and it can not be included in the rate law so we have to eleminate it therefore for the step -1

kf

equilibrium constant (Kc) = k = b So the rate law will be

[O 2 ][O ]

e c n [O 3 ]

 [O ] 

[O3 ]2 k . k  k '[O3 ]2[O2 ]1 c rate = [O2 ]

ir

Hence the order of the reaction is = 2+(-1) = 1

K c [O 3 ] [O 2 ]

(ii) By Steady State Approximation Here we assume that the concentration of the intermediate is constant under steady state condition so its rate of formation/ disappearence is zero. Consider the reaction NO2(g) + CO (g) CO2(g) + NO(g) The reaction mechanism can be given as

B

step 1 :

P y

NO2 + NO2 k1  NO + NO3 (fast )

k 2 CO2 + NO2 ( slow) Since , NO3 is the intermediate , from steady state approximation

step 2 :

NO3 + CO

we can say that  But 

d [NO3 ] =o dt

d [NO3 ] = k1[NO2]2 -k2[NO3][CO] = 0 dt

Reconstruct Your Chemistry With Prince Sir

So,

k1 [NO 2 ] 2 [NO3] = k [CO ] 2

rate of reaction = rate of disappearence of NO 2 rate = 

d [NO 2 ] = k1[NO2]2 + k2[NO3][CO] dt

k [NO 2 ] 2 [CO] [CO ]

= k1[NO2]2 + k2  1 k2

rate = 2k1[NO2]2 So, the order of the reaction is 2.

ri

Types of reactions on the basis of influence of temperature Effect of presence of a catalyst

c in

S e E p

Rate constant in presence of catalyst k p  Ae RT E a

Rate constant in absence of catalyst k  Ae RT a So,

kp ka

e

r P

(E a  E p )

RT

Ea  Activation energy in absence of catalyst Ep  Activation energy in presence of catalyst

y B

Reconstruct Your Chemistry With Prince Sir BOOST YOUR BASICS Q.1. For the equation 2NO + 2H2  N2 + 2H2O, write the most probable rate law from the follow ing data. What is the order of reaction? Also calculate rate constant. Experiment [NO] [H2] Rate (mol L-1s-1) I 0.2 0.2 3 × 10-3 II 0.4 0.2 1.2 × 10-2 III 0.2 0.4 6 × 10-3 Q.2. For the reaction OCl- + I- + OH-  OI- + Cl- + H2O kinetic data are given below. Calculate rate law and value of rate constant.

[OCl–]

[I–]

[OH–]

(mol dm-3) 0.0017 0.0034 0.0017 0.0017

(mol dm-3) 0.0017 0.0017 0.0034 0.0017

(mol dm-3) 1.0 1.0 1.0 0.5

Exp. No.

1. 2. 3. 4.

d[IO  ] dt (mol dm-3) (s-1) 1.75×10-4 3.5×10-4 3.5×10-4 3.5×10-4

e c n

r i S

Q.3.

According to the reaction Cr2O72- + 5H+ + 3HNO2  2Cr3+ + 3NO3- + 4H2O The rate of disappearance of Cr2O72- is found to be 2.4 × 10-4 mole L-1s-1. Find the rate of appearance of Cr3+ during given time interval.

Q.4.

Fill in the blanks in the following table which treats reaction of a compound A with a compound B, that is of the first order with respect to A and zero order with respect to B. Expt. No. 1. 2. 3. 4.

P y

[A] (mol L-1) 0.1 ---0. 4 ----

B

[B] (mol L-1) 0.1 0.2 0. 4 0.2

ir

Initial rate (mol L-1 s-1) 20 × 10-2 4.0 × 10-2 ----2.0 × 10-2

Q.5. The reaction 2NO + O2  2NO2, follows the rate law = k[NO]2[O2]. What is the order of the reaction?If k = 2.0×10–6 mol–2 L2 s–1, what is the rate of the reaction when [NO] = 0.04 mol L–1 and [O2] = 0.2 mol L–1 ? Q.6. From the following data of initial concentrations and rates, calculate the order of reaction aA  Products; and its rate constant. [A] mol L–1 0.1 0.2 0.4 –1 –1 –5 –5 Rate mol L s 9 × 10 36×10 144 × 10–5 Q.7. For a reaction aA + bB  mM + nN, the rate of reaction is given as k[A]x[B]y Calculate the order of the reaction and the rate constant from the given initial concentrations and the corresponding rates. [A] mol L–1 0.1 0.1 0.2 –1 [B] mol L 0.2 0.4 0.2 Rate mol L–1 s–1 4 × 104 16 × 104 8 × 104

Q.8.

Reconstruct Your Chemistry With Prince Sir The data given in the following table pertain to the reaction, 2A + B  C Determine the form of the rate equation and the value of the rate constant Expt. No. 1. 2. 3. 4.

Q.9.

Initial concentration (mol L-1) [A] [B] -3 1 × 10 1 × 10-3 -3 1 × 10 2 × 10-3 -3 1 × 10 3 × 10-3 -3 2 × 10 3 × 10-3

Initial rate (mol L-1 s-1) 7 × 10-6 14 × 10-6 21 × 10-6 84 × 10-6

The following initial rate data were obtained for the reaction : 2NO(g) + Br2(g)  2NOBr(g) Run

[NO]/M

[Br2]/M

Rate/M min-1

1. 2. 3.

1.0 2.0 4.0

1.0 1.0 2.0

1.30 × 10-3 5.20 × 10-3 4.16 × 10-2

c in

S e

Determine the reaction rate law and the value of the rate constant.

ri

Q.10.

For the reaction AB, it was found that the concentration of B increased by 0.3 mol L–1 in 2 hours. What is the average rate of reaction ?

Q.11.

In the reaction X Y, the initial concentration of X is 2.5 mol L–1 and its concentration after 3 hours is 0.7 mol L–1. What is the average rate of the reaction ?

Q.12.

When ammonia is treated with O2 at elevated temperaturs, the rate of disappearance of ammonia is found to be 3.5×10–2 mol dm–3s–1 during a measured time interval. Calculate the rate of appearance of nitric oxide and water.

Q.13.

In a reaction

y B

r P

N2O5  2NO2 +

1 O , the rate of disappearance of 2 2

N2O2 is 6.5 × 10–3 mol L–1 s–1. Compute the rates of formation of NO2 and O2. Q.14.

The following reaction was carried out at 44°C : N2O2  2NO2 + 1/2O2 The concentration of NO2 is 6.0 × 10–3 M after 10 minutes of the start of the reaction. Calculate the rate of production of NO2 over the first ten minutes of the reaction.

Q.15.

The rate of a particular reaction doubles when temperature changes from 27° C to 37°C. Calculate the energy of activation for such reaction. (R = 8.314 J K–1 mol–1)

Q.16.

The rate constant of a reaction is 2 × 10–2 s–1 at 300 K and 8 × 102s–1 at 340 K. Calculate the energy of activation of the reaction.

Q.17.

For a chemical reaction the energy of activation is 85 kJ mol–1. If the frequency factor is 4.0 × 109 L mol–1 s–1, what is the rate constant at 400 K ?

Q.18.

The energy of activation of a reaction is 140 kJ mol–1. If its rate constant at 400 K is 2.0 × 10–6 s–1, what is the value at 500 K ?

Q.19.

Reconstruct Your Chemistry With Prince Sir Calculate the ratio of the catalysed and uncatalysed rate constant at 20° C if the energy of activation of a catalysed reaction is 20 kJ mol–1 and for the uncatalysed reaction is 75 kJ mol–1.

Q.20.

The decomposition of methyl iodide, 2CH3I(g)  C2H6(g) + I2(g) at 273° C has a rate constant of 2.418 × 10–5 s–1. If activation energy for the reaction is +179.9 mol–1, what is the value of collision factor 'A' at 273°C ?

Q.21.

The reaction, 2N2O5  2N2O4 + O2 occurs in carbon tetrachloride. The rate constant is 2.35×10–4 sec–1 at 30°C. Calculate the activation energy of the reaction.

Q.22.

For the inversion of cane sugar, C12H22O11 + H2O  C5H12O6 + C6H12O6 the rate constant is 2.12 × 10–4 L mol–1sec–1 at 27°C. The activation energy of the reaction is 1.07 × 105 J mol–1. What is the rate constant of the reaction at 37° C ?

e c n –1

r i S

Q.23.

For a first order reaction, the rate constant is 0.1 s . How much time will it take to reduce the concentration from initial value of 0.6 mol L–1 to 0.06 mol L–1 ?

Q.24.

A substance decompose following first order reaction. If the half life period of the reaction is 35 minutes, what is the rate constant of this reaction ?

Q25.

For a certain first order reaction, it take 5 minutes for the initial concentration of 0.6 mol L–1 to become 0.2 mol L–1. What is the rate constant for this reaction ? [log 3 = 0.4771]

Q.26.

Find the two -thirds life (t2/3) of a first order reaction in which k = 5.48 × 10–1 sec–1 .

Q.27.

A first order reaction has a specific rate of 10–3sec–1. How much time will it take from 10 g of the reactant to reduce to 7.5 g (log 2 = 0.3010; log 4 = 0.6020 and log 6 = 0.7782) ?

Q.28.

In a reaction 5 g ethyl acetate is hydrolysed per litre in presence of dilute HCl in 300 min. If the reaction is of first order and initial concentration is 22 g per litre, calculate the rate constant for the reaction.

Q.29.

Calculate the half life of the reaction A  B, when the initial concentration of A is 0.01 mol L–1 and initial rate is 0.00352 mol L–1 min–1. The reaction is of the first order.

Q.30.

In reaction A  B + C, the following data were obtained: t in seconds 0 900 1800 Concentration of A 50.8 19.7 7.62 Prove that it is a first order reaction.

Q.31.

The first order reaction has k = 1.5 × 10–6 per second at 200°. If the reaction is allowed to run for 10 hours at the same temperature, what percentage of the initial concentration would have changed into the product ? What is the half life period of this reaction ?

Q.32.

A first order reaction is 20% completed in 10 minutes. Calculate (i) the specific rate con stant of the reaction and (ii) the time taken for the reaction to go to 75% completion.

Q.33.

The rate constant of a reaction with respect to reactant A is min–1. If we start with [A] = 0.8 mol L–1, when would [A] reach the value of 0.08 mol L–1 ?

B

P y

ir

Reconstruct Your Chemistry With Prince Sir Q.34.

What will be the initial rate of a reaction if the constant is 10–3 min–1 and the concentration is 0.2 mol dm–3?How much of the reactant will be converted into the product in 200 minutes ?

Q.35..

A substance A decomposes by the first order reaction. Starting initially with [A] = 2.00 M, after 200 minutes [A] = 0.25M. For this reaction what is t1/2 and k ?

Q.36.

Q.37.

Q.38.

Q.39.

Q.40.

In this case we have 2A 3B + C Time Total pressure of A + B+C Find k. A 2B + 3C Time Total pressure of ( B+C) Find k.

t P2

 P3

 P3

t P2

S e

ri

A 2B + 3C Time t  Volume of reagent V2 V3 Reagent reacts with all A, B and C. Find k. [Assuming n-factor of A, B & C are same]

r P

c in

[C] after one hour from the start of [A] the reaction. Assuming only A was present in the beginning. k1 = x hr–1; k1 : k2 = 1 : 10. Calculate

y B

How much time would be required for the B to reach maximum concentration for the reaction . k1 k2 A  B  C. Given k1 =

ln 2 ln 2 , k2 = . 4 2

Reconstruct Your Chemistry With Prince Sir Insight to the concepts Q.1.

The reaction cisCr(en)2(OH)+2 transCr(en)2(OH)+2 is first order in both directions. At 25°C the equilibrium constant is 0.16 and the rate constant k1 is 3.3 × 10 4s 1. In an experiment starting with the pure cis form, how long would it take for half the equilibrium amount of the trans isomer to be formed ?

Q.2.

Two reations (i) A  products (ii) B  products, follow first order kinetics. The rate of the reaction (i) is doubled when the temperature is raised from 300 K to 310K. The half life for this reaction at 310K is 30 minutes. At the same temperature B decomposes twice as fast as A. If the energy of activation for the reaction (ii) is half that of reaction (i), calculate the rate constant of the reaction (ii) at 300K.

Q.3.

The half life for a reaction between fixed concentration of reactants veries with temperature as follows : t°C 520 533 555 574 t1/2 sec 1288 813 562 477 Calculate the activation energy of this reaction.

e c n

r i S

Q.4. What percentage of reactant molecules will crossover the energy barrier at 325 K ? Heat of reaction is 0.12 kcal and activation energy of backward reaction is 0.02 kcal.

ir

Q.5. A decomposition reaction has following mechaism 2N2O5  4NO2 + O2 (overall) (i) N2O5 NO2 + NO3 (Fast decomposition) (ii) NO2 + NO3  NO + NO2 + O2 (slow) (iii) NO + NO3 2NO2 (fast) Determine rate law and the order of reaction.

P y

B(g), H for the forward reaction is –33 kJ/mol (Note : H = E in this [B] case). Show that equilibrium constant K = = 5.572 × 105 at 300 K. If the activation energies [A] Ef & Eb are in the ratio 20 : 31, calculate Ef and Eb at this temperature. Assume that the preexponential factor is the same for the forward and backward reactions.

B

Q.6. For the system A(g)

1 O is 2.4 hrs. a 30oC. What time would be 2 2 required to reduce 5 × 1010 molecules of N2O5 to 108 molecules ?

Q.7. The half life period for the reaction N2O5  2NO2 + .

Q.8. 10 gram atoms of an  -active radio isotope are disintegating in a sealed container. In one hour the helium gas collected at STP is 11.2 cm3. Calculate the half-life of the radio-isotope. Q.9. In a reaction, aA  products, the rate is doubled when the concentration of A is increased 4 times. If 50 % of the reaction occurs in 1414s, how many seconds would it take for the completion of 75% reaction?

Reconstruct Your Chemistry With Prince Sir Q.10.Mechanism for the reaction 2A  C is given as follows. k1 (i) A + A  (Activation)  A* + A k2 (ii) A* + A  (Deactivation)  A+A k3 (iii) A*  (Reaction)  C Deduce general rate law for the reaction, using steady state approximation.

Q.11 (a) The decomposition of HI to I2 at 508oC has a half-life of 135 minutes when PHI = 0.1 atm (initial) which comes down to one-tenth of that value when the initial pressure is 1 atm. Calculate the rate constant. Q.12 Above 500oC, the reaction NO2 + CO  CO2 + NO obeys the rate law rate = k[NO3][CO]. Below 500oC the rate law for this reaction is rate = k[NO2]2. Suggest mechanism for each of the cases

ri

Q.13. Decomposition of non-volatile solute 'A' into another non-volatile solute B and C, when dissolved in water follows first order kinetics as: A  2B + C When one mole of A is dissolved in 180 gm of water and left for decomposition, the vapour pressure of solution was found to be 20 mm Hg after 12 hrs. Determine the vapour pressure of the solution after 24 hrs. Assume constant temperature of 25°C, throughout. The vapour pressure of pure water at 25°C is 24 mm Hg. [Give your answer by adding all the digits ] Q.14. Consider a reversible reaction :

A

k1

B

k2

y B

r P

c in

Which is a first order in both the directions (k1 =

S e

1.38 × 10–2 min–1). The variation in concentration 3

is plotted with time as shown below. 0.3

conc. (M)

0.2

[A]

0.1

[B]

0

time 

Calculate the time (in minute) at which 25 % of A would be exhausted. [n 2 = 0.69 ]

Reconstruct Your Chemistry With Prince Sir OBJECTIVES Q.1 A zero order reaction is one(A) In which reactants do not react (B) In which one of the reactants is in large excess (C) Whose rate does not change with time (D) Whose rate increase with time Q.2 If concentration are measured in mole/li9tre and time in minutes, the unit for the rate constant of a 3rd order reaction are (A) mol lit-1min-1 (B) lit2mol-2min-1 (C) lit mol-1min-1 (D) min-1 Q.3 Following reaction was carried out at 300 K. 2SO2(g) + O2(g)  2SO3(g) How is the rate of formation of SO3 related to the rate of disappearance of O2 ? (A) –

[O 2 ] 1 [SO3 ] =+ t 2 t

(B) –

[O 2 ] [SO3 ] = t t

(C) –

[O 2 ] 1 [SO3 ] =– t 2 t

(D) None of these

r i S

Q.4 According collision theory of reaction(A) Every collision between reactiont molecules leads to a chemical reaction. (B) Rate of reaction is proportional to the velocity of the molecules (C) Rate of reaction is proportional to the average energy of the molecules (D) Rate of reaction is proportional to the number of collision per second.

ir

e c n

Q.5 Which of the following rate laws has an overall order of 0.5 for the reaction, A + B + C  product(A) R = k[A].[B].[C] (B) R = k[A]0.5[B]0.5[C]0.5 (C) R = k[A]1.5[B]-1[C]0 (D) R = k[A] [B]0[C]0.5

P y

Q.6 For the system A2(g) + B2(g)  2AB(g), H =–160 kJ If the activation energy for the forword step is 100 kcal/mol. What is the ratio of temperature at which the forword and backward reaction shows the same % change of rate constant per degree rise of temperature ? (1 cal = 4.2 J) (A)

0.72

(B)

B

0.84

(C)

0.42

(D)

1

Q.7 The t1/2 of first order reaction is 10 minutes. Starting with 100 grams/lit of the reaction, the amount remaining after one hour is (A) 25.0 g (B) 3.130 g (C) 12.50 g (D) 1.563 g 1 d[NO] represents 2 dt (B) The average rate of the reaction (D) All the above

Q.8 For the reaction, 2NO  N2 + O2, the experssion (A) The rate of formation of NO (C) The instantaneous rate reaction

Q.9 For a reaction of the type 2A + B   2C, the rate of the reaction is given by k[A]2[B]. When the volume of the reaction vessel is reduced to 1/4 th of the original volume, the ratev of reaction by a factor of (A) 0.25 (B) 16 (C) 64 (D) 4

Reconstruct Your Chemistry With Prince Sir Q.10 What is order withrespect to A, B, C respectively [A] [B] [C] rate(M/sec.) 0.2 0.1 0.02 8.08×10-3 0.1 0.2 0.02 2.01×10-3 0.1 1.8 0.18 6.03×10-3 0.2 0.1 0.08 6.464×10-2 (A) –1,1, 3/2 (B) –1,1, 1/2 (C) 1, 3/2, –1 (D) 1, –1, 3/2 Q.11 For reaction, rate = k[A] [B]2/3 the order of reaction is(A) 1 (B) 2 (C) 5/3 (D) Zero Q.12 The elementary reaction A + B  products, has k = 2 × 10-5 M-1 s-1 at a temperature of 27oC. Severl experimentary runs carried out using stoichiometric proportion. The reaction has a temperature corfficient value of 2.0. At what temperature should the reaction be carried out if inspite of halving the concentrations, the rate of reaction is desired to be 50% higher than a previous run. (Given n6 = 2.585) n2 (A) 47oC

(B) 53oC

(C) 57oC

S e (D) 37oC

ri

Q.13 According to collision theory(A) All collisions are sufficiently violent (B) All collisions are responsible formation. (C) All colisions are effective. (D) Only a fraction of collisions are effective which have enough energy to form products.

r P

c in

Q.14 When the temperature of a reaction increases from 270C to 370C, the rate increases by 2.5 times, the activation energy in the temperature range is (A) 53.6 kJ (B) 12.61 kJ (C) 7.08 kJ (D) 70.8 kJ Q.15 For the reaction R – X + OH–  ROH + X–The Rate is given as Rate = 5.0 × 10-5 [R – X][OH–] + 0.20 × 10-5[R–X] what percentage of R – X reacted by SN2 mechanism when [OH -] = 1.0 × 102 M (A) 96.1% (B) 3.9% (C) 80% (D) 20%

y B

Q.16 Which of the following curves represents a Ist order reaction-

(A)

(B)

(C)

(D)

Q.17 The rate of a reaction increases 4-fold when conecentration of reaction is increased 16 times. If the rate of reaction is 4 × 10-6 mole L-1 when concentration of the reaction is 4 × 10-4 mole L-1, the rate constant of the reaction will be (A) 2 × 10-4mole1/2L-1/2sec-1 (B) 1 × 10-2sec-1 (C) 2 × 10-4mole-1/2,L1/2sec-1 (D) 25 mole-1Lmin-1 Q.18 If 'a' is the intial concentration of a substance which reacts according to zero order kinetic and k rate conatant, the time for the reaction to go to completion is(A) a/k (B) 2/ka (C) k/a (D) Infinite

Reconstruct Your Chemistry With Prince Sir Q.19 The following data are obtained from the decomposition of a gaseous compound Initial pressure, (atm) Time for 50% reaction,(min)

The order of the reaction is (A) 1.0 (B) 1.5

1.6 80

0.8 113

(C) 2.0

0.4 160

(D) 0.5

Q.20 For a given reaction the concentration of the reactant plotted against time gave a straight line negative slope. The order of the reaction is(A) 3 (B) 2 (C) 1 (D) 0 Q.21 The rate constant of a first order reaction is generally determined from a plto of (A) Concentration of reactant vs time t (B) log (concentration of reaction) vs time t (C)

r i S

1 vs time t (D) Concentration of reactant vs log time t concentration of reaction

Q.22 Which of the following correctly represents the variation of the rate of the reaction with temperature ?

(A)

(B)

(C)

P y

B

Q.23 The plot of log K versus (A)

ir

(D)

Ea R

(B)

e c n

1 is linear with a slope of T

Ea R

(C)

Ea 2.303R

(D)

Ea 2.303R

Q.24 According to the collision theory, the rate of reaction increases with temperature due to(A) Greater number of collision (B) Higher velocity of reacting molecules (C) Greater number of molecules having the activation energy (D) Decrease in the activation energy Q.25 If a reaction A + B  C is exothermic to the extent of 30 kJ/mol and the forward reaction has an activation energy 70 kJ/mol, the activation energy for the reverse reaction is (A) 30 kJ/mol (B) 40 kJ/mol (C) 70 kJ/mol (D) 100 kJ/mol Q.26 A first order reaction takes 40 min for 30% decomposition. Calculate t1/2. (A) 77.7 min. (B) 27.2 min. (C) 55.3 min. (D) 67.3 min.

Reconstruct Your Chemistry With Prince Sir Q.27 For a certain reaction involving a single reaction, it is found that C0 T is constant where C0 is the initial concentration of the reaction and T is the half-life. What is the order the reaction ? (A) 1 (B) Zero (C) 2 (D) 3 Q.28 The high temperature (  1200K) decomposition of CH3COOH(g) occurs as follows as per simultaneous 1st order reactions. k1 CH3COOH   CH4 + CO2 k2 CH3COOH   CH2CO + H2O What would be the % of CH4 by mole in the product mixture (excluding CH3COOH) ?

50k1 (A)  k  k  1 2

100k1 (B)  k  k  1 2

200k1 (C)  k  k  1 2

(C) 3

c in

N N = 0.125 from = 1 of a N0 N0

S e

Q.29 The number of half-lives (t1/2) required to bring the ratio to radioactive element is (A) 2 (B) 4

ri

(D) it depends on time

(D) 1

Q.30 The reaction, A(g) + 2B(g)  C(g) + D(g) is an elementary process. In an experiment, the initial partial pressure of A and B are PA = 0.60 and PB = 0.80 atm. When PC = 0.2 atm the rate of reaction relative to the initial rate is (A) 1/48 (B) 1/24 (C) 9/16 (D) 1/6

r P

Q.31 Half life period for a first order reaction is 20 minutes. How much time is required to change the concentration of the reactants from 0.08M to 0.01 M (A) 20 minutes (B) 60 minutes (C) 40 minutes (D) 50 minutes Q.32 The kinetic datas for the reaction: 2A + B2  2AB are as given below: [A] [B] Rate -1 -1 mol L mol L mol L-1min-1 0.5 1.0 2.5 × 10-3 1.0 1.0 5.0 × 10-3 0.5 2.0 1 × 10-2 The order of reaction w.r.t. A and B2 are, respectively, (A) 1 and 2 (B) 2 and 1 (C) 1 and 1 (D) 2 and 2

y B

Q.33 If for any reaction rate constant is equal to the rate of the reaction at all conaentrations, the order is(A) 0 (B) 2 (C) 1 (D) 3 k Q.34 For the irreversible unimolecular type reaction A   products, in a batch reactor, 80% reactant A(CA0 = 1 mole/lit.) is converted in a 480 second run and conversion is 90% after 18 mintue. The order of this reaction is(A) 1 (B) 2 (C) 1/2 (D) 3/2

Q.35 The rate constant, the activation energy and the Arrhenius parmeter of a chemical reaction at 25oC a r e 3.0 × 10-4s-1, 104.4 kJ mol-1 and 6.0 × 1014s-1 respectively.The value of the rate constant as T    is:

(A) 2.0 × 1018s-1 (B) 6.0 × 1014s-1

Reconstruct Your Chemistry With Prince Sir (C)  (D) 3.6 × 1030s-1

Q.36 For the first order reaction t99% = x × t90% the value of 'x' will be(A) 10 (B) 6 (C) 3 (D) 2 Q.37 The gas phase decomposition 2N2O5  4NO2 + O2 follows the first order rate law with rate cons t a n t K = 7.5 × 10-3 sec-1. The initial pressure of N2O5 is 0.1atm. The time of decomposition of N2O5 so that total pressure becomes 0.15 atm will be(A) 54 sec (B) 5.4 sec (C) 3.45 sec (D) 34.55 sec Q.38 If in the fermentation of sugar in an enzymatic solution that is 0.12 M, the concentration of the sugar is reduced to 0.06 M in 10h and to 0.03 M in 20h, what is the order of the reaction(A) 1 (B) 2 (C) 3 (D) 0

r i S

Q.39 In a Ist order reaction, A  products, the concentration of the reactant decreases to 6.25% of its initial value in 80 minutes. What is the rate constant reaction 100 minutes after the starts, if the initial concentration is 0.2 mole/litre ? (A) 2.17 × 10-2min-1 (B) 3.465 × 10-2min-1 -3 -1 (C) 3.465 × 10 min (D) 2.166 × 10-3min-1

e c n

Q.40 For the first order reaction A(g)  2B(g) + C(g), the initial pressure is PA = 90 mm Hg, the pressure after 10 minutes is found to be 180 mm Hg. The rate constant of the reaction is (A) 1.15 × 10-3sec-1 (B) 2.3 × 10-3sec-1 -3 -1 (C) 3.45 × 10 sec (D) 6 × 10-3sec-1

ir

Q.41 For a certain reaction the variation of the rate constant with temperature is given by the equation  1n3  In kt = 1n k0 +   t (t  0oC) The value of the temperature coefficient of the reaction rate is 10 

P y

therefore(A) 4 (B) 3 (C) 2 (D) 10 Q.42 A catalyst lowers the activation energy of a reaction from 20 kJ mole-1 to 10 kJ mole-1. The temperature at which the uncatalysed reaction will have the same rate as that of the catalysed at 27oC is (A) –123oC (B) 327oC (C) 1200oC (D) +23oC

B

Q.43 The rate of reaction triples when temperature changes from 20oC to 50oC. Calculate energy of activation for the reaction. (A) 28.81 kJ mol-1 (B) 38.51 kJ mol-1 (C) 18.81 kJ mol-1 (D) 8.31 kJ mol-1 Q.44 The half life time for the decomposition of a substance dissolved in CC14 is 2.5 hour at 30oC. How much of the substance will be left after 10 hours, if the initial weight of the substance is 160 gm ? (A) 20 gm (B) 30 gm (C) 40 gm (D) 10 gm 2+ 3+ k1 k2 Q.45 For the reaction [Cr(H2O)4C12]+(aq.)   [Cr(H2O)5C1] (aq.)   [Cr(H2O)6C1] (aq.) K1 = 1.78 × 10-3s-1 and k2 = 5.8 × 10-5s-1 If initial concentration of [Cr(H2O)4C12]+ is 0.0174 M at 273 K. Calculate time at which concentration of [Cr(H2O)5C1]2+ is maximum. (A) 1990 seconds (B) 1090 seconds (C) 2100 seconds (D) 2110 seconds.

Reconstruct Your Chemistry With Prince Sir Q.46 What is the activation energy for the decomposition of N2O5 as N2O5  2NO2 + (1/2) O2 If the value of the rate constant are 3.45 × 10-5 and 6.9 × 10-3 at 27oC and 67oC respectively. (A) 102 × 102 kJ (B) 488.5 kJ (C) 92 kJ (D) 112.3 kJ Q.47 If for a first order reaction, rate constant varies with temperature according to the graph given below. At 27oC, 1.5 × 10-4 percent of the reaction molecules are able to cross-over the potential barrier. At 52oC, the slope of this graph is equal to 0.2 K-1 sec-1, calculate the value of rate constant at 52oC, assuming that activation energy does not change in this temperature range. (A) 3.14×10-2min-1 (B) 1.35 × 10-2min-1 -2 -1 (C) 0.75 × 10 min (D) 8.75 × 10-2min-1

ri

Q.48 the rate constant for a first order reaction is 60s-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value(A) 0.046 s (B) 0.46 s (C) 0.124 s (D) 2.123 s

S e

Q.49 1 mole of gas changes linearly from initial state (2 atm, 10 lt) to final state (8 atm, 4 lt). Find the value of rate constant, at the maximum temperature, that the gas can attain. Maximum rate constant is equal to 20 sec-1 and value of activation energy is 40 kJ mole-1, assuming that activation energy does not change in this temperature range. (A) 0.56 × 10-3 sec-1 (B) 3.16 × 10-3 sec-1 (C) 1.56 × 10-3 sec-1 (D) 5.12 × 10-3 sec-1

r P

c in

Q.50 A first order reaction was started with a decimolar solution of the reactant. After 8 minutes and 20 seconds, its concentration was found to be M/100. Determine the rate constant of the reaction. (A) 4.6 × 10-3 sec-1 (B) 16.6 × 10-3 sec-1 (C) 24.6 × 10-3 sec-1 (D) 40.6 × 10-3 sec-1 Q.51 87.5% of a radioactive substance disintegrates in 40 minutes. What is the half life of the substance ? (A) 13.58 min (B) 135.8 min (C) 1358 min (D) None of these

y B

Q.52 A gaseous substance AB3 decomposes irreversibly according to the overall equation AB3 

1 A 2 2

3 B . Starting with pure AB3, the partial pressure of the reaction varies time for which the data 2 2 are given below. Time in hours 0 5.0 15.0 35.0

+

PAB3 mm Hg

660 330 165 What is the order of the reaction ? (A) 2 (B) 0.5 (C) 1

82.5 (D) 1.5

Q.53 The inversion of cane sugar proceeds with half-life of 500 minute at pH 5 for any concentration of sugar. However if pH = 6, the half-life changes to 50 minute. The rate law ecpression for the sugar inversion can be written as (A) r = K[sugar]2[H]6 (B) r = K[sugar]1[H]0 0 + 6 (C) r = K[sugar] [H ] (D) r = K[sugar]0ss[H+]1 Q.54 The decomposition of A into product has value of k as 4.5 × 103s-1 at 10oC and energy of activation

Reconstruct Your Chemistry With Prince Sir 60 kJ mol-1. At what temperature would k be 1.5 × 104s-1 ? (A) 12oC (B) 24oC (C) 48oC (D) 36oC Q.55 Decomposition on NH3 on heated tungsten yields the following data : Initial pressure (mm) 65 105 y 185 Half-life 290 x 670 820 What are the values of x and y in that order ? (A) 420 s, 110 mm (B) 500 s, 160 mm (C) 520 s, 170 mm (D) 460 s, 150 mm Q.56 The half life period of gaseous substance undergoing thermal decomposition was measured for various initial pressure 'P' with the following result. P(mm) 250 300 400 450 t1/2 (min) 136 112.5 85 75.5 Calculate the order of reaction. (A) 2 (B) 4 (C) 6 (D) 10

r i S

Q.57 For the reaction 2N2O5(g)  4NO2(g) + O2(g), the concentration of NO2 increases by 2.4 × 10-2 Mol lit-1 in 6 second. What will be the rate of appearance of NO2 and the rate of disappearance of N2O5. (A) 2 × 10-3 mol L-1sec-1, 4 × 10-3 mol L-1 sec-1 (B) 2 × 10-3 mol L-1sec-1, 1 × 10-3 mol L-1 sec-1 (C) 2 × 10-3 mol lit-1sec-1, 2 × 10-3 mol lit.-1 sec-1 (D) 4 × 10-3 mol lit.-1sec-1, 2 × 10-3 mol lit.-1 sec-1

ir

e c n

Q.58 The rate constant of a particular reaction has the dimensions of a frequency. What is the order of the reaction ? (A) Zero (B) First (C) Second (D) Fractional Q.59 The reaction of iodomethane with sodium ethoxide proceeds as : EtO  + MeI  EtOMe + 1  .

P y

  MeI  A plot of log  on the Y-axis against 's' on the X-axis gives a straight line with a positive     EtO     slope. What is the order of the reaction ? (A) Second (B) First (C) Third

B

(D) Fractional

Q.60 For the reaction,C2H5I + OH-  C2H5OH + I- the rate constant was found to have a value of 5.03×10-2 mol-1dm3s-1 at 289 K and 6.71 mol-1dm3s-1 at 333 K. What is the rate constant at 305 K ? (A) 1.35 mol-1dm3s-1 (B) 0.35 mol-1dm3s-1 -1 3 -1 (C) 3.15 mol dm s (D) 7.14 mol-1dm3s-1 Q.61 A plot of ln rate Vs ln C for the nth order reaction gives(A) a straight line with slope n and intercept ln kn. (B) a straight line with slope (n - 1) (C) a straight line with slope ln kn and intercept 'n' (D) a straight line with slope -n and intercept kn. Q.62 The first order rate constant for a certain reaction increases from 1.667×10-6 s-1 at 727oC to 1.667×104 -1 s at 1571oC. The rate constant at 1150oC, assuming constancy of activation energy over to given temperature range is (A) 3.911 × 10-5sec-1 (B) 1.139 × 10-5sec-1 -5 -1 (C) 3.318 × 10 sec (D) 1.193 × 10-5sec-1

Reconstruct Your Chemistry With Prince Sir Q.63 The solvolysis of 2-chloro-2-methyl propane is aqueous acetone : H2O + (CH3)3 C - Cl  HO C(CH3)3 + H+Cl- has a rate equation. Rate = K[(CH3)3 C-Cl]. From this it may be inferred that the energy profile of the reaction leading from reactants to products is

(A)

(B)

(C)

(D)

S e

ri

Q.64 Thermal decomposition of a compound is of first order. If 50 % of a sample of the compound is decompound in 120 minutes, show how long will it take for 90 % of the compound to decompose ? (A) 399 min (B) 410 min (C) 250 min (D) 120 min Q.65 The rate constant of a certain first order reaction increases by 11.11% per degree rise of temperature at 27oC. By what % will it increases at 127oC, assuming constancy of activation energy over the given temperature range ? (A) 5.26% (B) 5.62% (C) 6.25% (D) 7.33%

r P

c in

Q.66 The half-life for radioactive decay of 14C is 5730 y. An archaeogical artefact contained wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. [Radioactive decays follow the first order kinetic] (A) 1657.3 y (B) 1845.4 y (C) 1512.4 y (D) 1413.1 y

y B

Q.67 Two substance A and B are present such that [A0] = 4[B0] and half-life of A.is 5 minute and that of B is 15 minute. If they start decaying at the same time following first order kinetics how much time later will the concentration of both of them would be same. (A) 15 minute (B) 10 minute (C) 5 minute (D) 12 minute Q.68 The following data were obtained during the first order thermal decomposition of SO 2Cl2 at a constant volume. SO2Cl2(g)  SO2(g) + Cl2(g) Experiment Time/s Total pressure/atm 1 0 0.5 2 100 0.6 Calculate the rate of the reaction when total pressure is 0.65 atm. (A) 7.8 × 10-4s-1 atm. (B) 0.8 × 10-4s-1 atm. -2 -1 (C) 2.4 × 10 s atm. (D) 6.1 × 10-8s-1 atm.

Reconstruct Your Chemistry With Prince Sir

SUBJECT APTITUDE

One or more than one choice may be correct : Q.1 Bicyclo hexane was found to undergo two paeallel first order rearrangements, as given belowBicyclo hexane

Choose the correct options(A) % of cyclohexane = 77 (C) % of methylcyclopentance = 77

(B) % of methylcyclopentance = 23 (D) % of cyclohexane = 23

Q.2 Consider the following case of competing 1st order reaction.

e c n

r i S

After the start of the reaction at t = 0 with only A, the [C] is equal to the [D] at all times. The time in which all three concentrations will be equal is given by (A) t =

1 n3 2k1

(B) t =

1 1 1 n3 n3 (C) t = n3 (D) t = 3k 2 2k 2 3k1

ir

Q.3 At 25oC, the second order rate constant for the reaction I- + ClO-  IO3-+Cl- is 0.0606 M-1sec-1. If a solution is initially 3.5 × 10-3 M with respect to each reactants. Choose the correct options – (A) concentration of A = 3.29 × 10-3 M after 300 sec. (B) concentration of B = 3.29 × 10-3 M after 300 sec. (C) concentration of A = 0.19 × 10-3 M after 300 sec. (D) None of these

P y

Q.4 Which is correct graph :

(A)

B

 dc  (C) n    dt

 dc  (B) n     dt 

(D)

t 0.75 t 0.5

Q.5 In a pseudo first order hydrolysis of ester in water, the following results were obtained : t/s 0 30 60 90 [Ester]/M 0.55 0.31 0.17 0.085

Reconstruct Your Chemistry With Prince Sir (A) Average rate of reaction between the time interval 30 to 60 seconds is 4.7 × 10-3 mol L-1s-1 (B) The pseudo first order rate constant for the hydrolysis of ester is 1.98 × 10-2s-1 (C) The pseudo first order rate constant for the hydrolysis of ester is 4.7 × 10-3s-1 (D) Average rate of reaction between the time interval 30 to 60 seconds is 1.98 × 10-2 mol L-1s-1 Q.6 The zone layer in the earth's upper atmosphere is important in shielding the earth from very harmful ultravilet radiation. The ozone, O3, decomposes according to the equation 2O3(g)  3O2(g). The mechanism of the reaction is through to proceed through an initial fast, reversible step.   Step 1 : Fast, reversible O (g)  O (g) + O(g)  3

2

Step 2 : Slow O3(g) + O(g) 2O2(g) The which of the following is correct ? (A) Step 2 is rate determining step

(B) The rate expression for step 2 is-

(C) For Step 1, molecularity is 1

(D) For Step 2, molecularity is 2

ri

d [O ] = k[O3][O] dt 3

S e

Q.7 The substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as:

c in

K1 = 1.26 × 10-4sec-1 and K2 = 3.8 × 10-5sec-1 The percentage distribution of B and C (A) 76.83% B (B) 23.17% C (C) 90% B

r P

(D) 40% C

 products, the t1/2 as a function of [A]0 is given as below : Q.8 For a certain reaction A  [A]0 (M) : 0.1 0.025 t1/2 (min) : 100 50 Which of the following is true :

y B

(A) The order is 1/2 (C) The order is 1

(B) t1/2 would be 100 10 min for [A]0 = 1 M (D) t1/2 would be 100 min for [A]0 = 1 M

Q.9 For the reaction 2A + B   C with the rate law

d[C] = k[A]1[B]-1 and started with A and B in dt

stoichimetric proportion which is/are teur ? (A) unit of k is Mole litre-1s-1 (B) [A], [B] and [C] all will be linear functions of time (C) [C] = 2kt (D) [C] = kt Q.10 The half-period T for the decomposition of ammonia on tungsten wire, was measured for different initial pressure P of ammonia at 25oC. Then P (mm Hg) 11 21 48 73 120 T (sec) 48 92 210 320 525 (A) zero order reaction (B) First order reaction (C) rate constant for reaction is 0.114 seconds. (D) rate constant for reaction is 1.14 seconds.

Reconstruct Your Chemistry With Prince Sir Q.11 The reaction given below, involving the gases is boserved to be first order with rate constant 7.48 × 10-3 sec-1. The time required for the total pressure in a system cosntaining A at an initial pressure of 0.1 atm to rise to 0.145 atm is t0 and the total pressure after 100 sec is x atm. 2A(g)  2B(g) + C(g) (A) t0 = 47.7 sec. (B) x = 0.180 atm (C) x = 0.03 atm (D) t0 = 22.7 sec. Q.12 In a reaction bteween A and B, the initial rate of reaction was measured for different initial concentrations of A and B as given below : A/M 0.20 0.20 0.40 B/M 0.30 0.10 0.05 R0/M s-1 5.07 × 10-5 5.07 × 10-5 7.6 × 10-5 (A) The order of reaction with respect to A is 0.5 (B) The order of reaction with respect to B is 0.5 (C) The order of reaction with respect to B is 0 (D) The order of reaction with respect to A is 1.5

e c n

r i S

Q.13 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 Identify the true statement (s) log 2 = 0.3, log3 = 0.48, log7 = 0.84, loge 10 = 2.3 (A) The half life of the reaction is 82.7 min (B) The solution is optically inactive at 131.13 min. (C) The equimolar mixture of the products is dextrorotatory (D) The angle would be 7.5o at half time

P y

ir

Q.14 A certain reaction obeys the rate equation (in integeated form) [C(1 - n) - C0 (1 - n)] = (n - 1) kt where C0 is the initial concentration and C is the concentration aftert time t, then (A) the unti of k for n = 1 is sec-1 (B) the unti of k for n = 2 is litre mol-1sec-1 (C) the unti of k for n = 3 is mol litre-1 sec-1 (D) the unti of k for n = 3 is litre mol-2 sec-1

B

ASSERTION AND REASON QUESTIONS : Note : Each question contains STATMENT-1 (Assertion) and STATMENT-2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. (A) Statment-1 is Ture, Statment-2 is Ture; Statment-2 is a correct explanation for Statment-1. (B) Statment-1 is Ture, Statment-2 is Ture; Statment-2 is NOT a correct explanation for Statment-1. (C) Statment-1 is Ture, Statment-2 is False. (D) Statment-1 is Ture, Statment-2 is True. Q.15 Statment-1 : A lump of coal burns at a moderate rate in air while coal dust burns explosively. Statment-2 : Coal dust contains very fine particles of carbon. Q.16 Statment-1 : Liquid bromine reacts slowly as compared to bromine vapour. Statment-2 : In liquid bromine, the bromine molecules are held together by a force which is much weaker than the force existing between the two molecules of bromine in the vapour phase.

Reconstruct Your Chemistry With Prince Sir Q.17 Statment-1 : The reaction, N2(10atm) + 3H2(10atm)  2NH3(g) is faster.. Statment-2 : Catalyst lowers the activation energy of the reaction. Q.18 Statment-1 : Molecularity of a reaction cannot be determined experimentally. Statment-2 : Molecularity is assigned to the reactions on the basis of mechanism. Q.19 Statment-1 : For a first order reaction, the degree of dissocation is equal to (1 - e-kt) Statment-2 : For a first order reaction, the pre-exponential factor in the Arrhenius equation has the dimension of time-1. Q.20 Statment-1 : Molecularity does not influence the rate of reaction. Statment-2 : The overal kinetics of a reaction is governed by the slowest step in the reaction scheme.

ri

MATCH THE COLUMN TYPE QUESTIONS : Each question constant statments given in two columns which have to be matched. Statements (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in column II. Q.21 Match the following : Coiumn-I (A) Slope of graph c vs t (abscissa) for zero order (B) Slope of graph log c vs (abscissa) for first order dc (C) Slope of    vs c for zero order  dt 

r P

 dc  (D) Slope of    vs In c first order  dt 

y B

Q.22 Match the following : Coiumn-I (A) Half life of first order reaction (B) Arrhenius equation (C) Molar concentration

(D) half life period of zero order reaction Q.23 Match the following : Coiumn-I

c in

S e

Coiumn-II (P) unity (Q) zero (R) -k

(S) -

k 2.303

Coiumn-II (P) Active mass (Q) k = Ae  Ea / RT (R) t1/2 =

0.693 k

(S) a/2k

Coiumn-II

(A) Zero

(P) k =

x at  a  x 

(B) One

(Q) k =

x t

(C) Two

(R) k =

x  2a  x 

t.2a 2  a  x 

2

Reconstruct Your Chemistry With Prince Sir 2.303 a (S) k = log t ax

(D) Three

Q.24 Match the following : Coiumn-I Coiumn-II (A) If the activation energy is 65 kJ then how much time (P) 5 o o faster a reaction proceed at 25 C than 0 C (B) Rate constant of a first-order reaction is 0.0693 min-1. (Q) 11 If we start with 20 mol L-1, it is reduced to 2.5 mol L-1 in how many minutes (C) Half-lives of first - order and zeroth order reaction are (R) 30 same. Ratio of rates at the start of reaction is how many times of 0.693 (D) the half-life periods are given, (S) 1/4 [A]0 (M) 0.0677 t1/2 (sec) 240 order of the reaction is

0.136 480

0.272 960

(T) 2 (U) 0

e c n

PASSAGE BASED QUESTIONS : .

r i S

PASSAGE -1 Consider the inter conversion of nitrosotriacetoamine into nitrogen phorone and water. O C H3C

P y

CH3(aq)

H3C N

CH3

NO

ir

O C

(aq)-20kJ/mol

C–CH3 N2(g) + H2O( ) + H3C–C CH3 CH3

The reaction is 1st order in each dieection, with an equilibrium constant of 104, the activation energy for the forward reaction is 57.45 kJ/ mol. Assuming Arrhenius pre-exponential factor of 1012s-1. .

B

Q.25 What is the expected forward constant at 300K, if we initiate this reaction starting with only reactant (A) 102 (B) 106 (C) 108 (D) 104 .

Q.26 If the change in entropy of the reaction is 0.07 kJ K-1mol-1 at 1 atm pressure. Calculate up to which temperature the reaction would not be spontaneous. (For forward reaction) (A) T < 285.7 K

(B) T > 250 K

(C) T < 340.2 K

(D) T > 200 K

(C) 24.6 × 104 atm

(D) 2.82 × 102 atm-1

.

Q.27 Calculate Kp of the reaction at 300 K (A) 2.4 × 104 atm-1 . .

(B) 104 atm

Reconstruct Your Chemistry With Prince Sir PASSAGE-2 The reaction between A and B is of first order with respect toA and of zero order with respect to B. Fill in the blanks in the following table : Experiment [A]/M [B]/M Initial Rate/M min-1 I 0.1 0.1 2.0 × 10-2 II × 0.2 4.0 × 10-2 III 0.4 0.4 y IV z 0.2 2.0 × 10-2 .

Q.28 The value of x is(A) 0.2

(B) 0.8

(C) 0.1

(D) 0.5

(B) 0.08

(C) 0.1

(D) 0.5

(B) 0.8

(C) 0.1

.

Q.29 The value of y is(A) 0.2 .

Q.30 The value of z is(A) 0..2

PASSAGE-3 For the reaction, A + B  product Following initial rate data is given : Exp. No. [A]0 (m/1) 1. 1.5 2. 3 3. 1.5 .

r P

Q.31 Order of reaction with respect to A and B is: (A) 1, 1 (B) 1, 2 .

Q.32 Rate constant of reaction is: (A) 2 × 10-3sec-1 (B) 10-3sec-1 .

y B

c in

[B]0 (m/1) 4.5 4.5 9

S e (D) 0.5

..

ri

Initial rate (m/1/s) 3 × 10-3 6 × 10-3 3 × 10-3

(C) 2, 1

(D) 1, 0

(C) (1/2) × 10-3sec-1

(D) 2.25 × 10-3sec-1

Q.33 Time required to decrease concentration of A form 16 M to 1M is: (A) 346.5 sec (B) 639 sec (C) 1386 sec

(D) 6930 sec

Reconstruct Your Chemistry With Prince Sir

Q.1

Q.2 (i) (ii)

FEEL THE HEAT (Previous Year’s JEE Questions) The degree of dissociation is 0.4 at 400K & 1.0 atm for the gasoeus reaction PCl5  PCl3 + Cl2(g). Assuming ideal behaviour of all gases. Calculate the density of equilibrium mixture at 400K & 1.0 atm pressure. [JEE 1999] When 3.06g of solid NH4HS is introduced into a two litre evacuated flask at 27°C, 30% of the solid decomposes into gaseous ammonia and hydrogen sulphide. Calculate KC & KP for the reaction at 27°C. What would happen to the equilibrium when more solid NH4HS is introduced into the flask? [JEE 2000]

r i S

Q.3

When two reactants A and B are mixed to give products C and D, the reaction quotient Q, at the initial stages of the reaction : [JEE 2000] (A) is zero (B) decrease with time (C) independent of time (D) increases with time

Q.4

For the reversible reaction : [JEE 2000] N2(g) + 3H2(g)  2NH3(g) at 500°C. The value of Kp is 1.44 × 10–5, when partial pressure is measured in atmospheres. The corresponding value of Kc with concentration in mol L–1 is : (A) 1.44 × 10–5 /(0.082 × 500)2 (B) 1.44 × 10–5 /(8.314 × 773)2 (C) 1.44 × 10–5 /(0.082 × 500)2 (D) 1.44 × 10–5 /(0.082 × 773)2

Q.5

At constant temperature, the equilibrium constant (KP) for the decomposition reaction. N2O4  2NO2 is expressed by KP = 4x2P/(1 – x2) where P is pressure, x is extent of decomposition. Which of the following statement is true ? [JEE 2001] (A) KP increases with increase of P (B) KP increases with increase of x (C) KP increases with decrease of x (D) KP remains constant with change in P or x

Q.6

When 1-pentyne (A) is treated with 4N alcoholic KOH at 175°C, it is converted slowly into an equilibrium mixture of 1.3% 1-pentyne (A), 95.2% 2–pentyne (B) & 3.5% of 1, 2,–pentadiene (C). The equilibrium was maintained at 1750C. Calculate G° for the following equilibria. B = A G10 = ? B= C G20 = ? From the calculated value of G10 & G20 indicate the order of stability of A, B & C. Write a reasonable reaction mechanism sharing all intermediate leading to A, B & C. [JEE 2001]

Q.7

Consider the following equilibrium in a closed container : N2O4(g)  2NO2(g). At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilibrium constant (K P) and degree of dissociation () : [JEE 2002] (A) Neither KP nor  changes

B

P y

ir

e c n

Reconstruct Your Chemistry With Prince Sir (B) Both KP and  change (C) KP changes, but  does not change (D) KP does not change, but  changes Q.8

(i) (ii) Q.9

Q.10

Q.11

N2O4(g)  2NO2(g) This reaction is carried out at 298 K and 20 bar. 5 mol each of N2O4 and NO2 are taken initially. Given: G oN 2O 4 = 100 kJ mol–1; G oNO2 = 50 kJ mol–1 Find G for reaction at 298 K under given condition. Find the direction in which the reaction proceeds to achieve equilibrium. [JEE 2004] If Ag+ + NH3  [Ag(NH3)]+ ; [Ag(NH3)]+ + NH3  [Ag(NH3)2]+ ; The formation constant of [Ag(NH3)2]+ is : (A) 6.08 × 10–6 (B) 6.8 × 10–6

K1 = 1.6 × 103 and K2 = 6.8 × 103 . (C) 1.6 × 103

ri

[JEE 2006]

(D) 1.088 × 107

S e

N2 + 3H2 l 2NH3 Which is correct statement if N2 is added at equilibrium condition? (A) The equilibrium will shift to forward direction because according to II law of thermodynamics the entropy must increases in the direction of spontaneous reaction. (B) The condition for equilibrium is G N 2  3G H 2  2G NH 3 where G is Gibbs free energy per mole of the gaseous species measured at that partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward reactions to the same extent. (C) The catalyst will increase the rate of forward reaction by  and that of backward reaction by . (D) Catalyst will not alter the rate of either of the reaction. [JEE 2006] STATEMENT-1 : reaction is zero

r P

c in

For every chemical reaction at equilibrium, standard Gibbs energy of

y B

and STATEMENT-2 : At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy. (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 [JEE 2008] Q.12. For a first order reaction A  P, the temperature (T) dependent rate constant (k) was found to follow the equation log k = -(2000)

1

T

+ 6.0

The pre-exponential factor A and the activation energy Ea, respectively, are (A) 1.0 × 106 s-1 and 9.2 kJmol-1 (B) 6.0 s-1 and 16.6 kJmol-1

Reconstruct Your Chemistry With Prince Sir (C) 1.0 × 106 s-1 and 16.6 kJmol-1 (D) 1.0 × 106 s-1 and 38.3 kJmol-1 [JEE 2009] Q.13. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is [JEE 2010]

K

K

(A*)

(B)

T

T

K

e c n

K

(C)

(D)

T

ir

T

r i S

Q.14. The concentration of R in the reaction R  P was measured as a function of time and the following data is obtained: [R] (molar) 1.0 0.75 0.40 0.10 t (min.) 0.0 0.05 0.12 0.18 The order of the reaction is? Sol. 0 Q.15. For the first order reaction 2N2O5(g)  4NO2 g 2O2(g) (A) the concentration of the reactant decreases exponentially with time (B) the half-life of the reaction decreases with increasing temperature (C) the half-life of the reaction depends on the initial concentration of the reactant (D) the reaction proceeds to 99.6% completion in eight half-life duration

B

P y

Q.16. An organic compound undrgoes first order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are t1/8 and t1/10 respectively. What is [t1/ 8 ] the value of [t ] ×100 ? 1 / 10

[JEE-2012]

Reconstruct Your Chemistry With Prince Sir BOOST YOUR BASICS

Answer Key 1. Order =3, k = 0.375 mol-2L2sec-1 4. 0.2, 8×10-2, 0.1 7. order =3

2. Rate = [OCl-]1[I-]1[OH-]-1

5. rate = 6.4×10-10 molL-1 sec-1

8. Rate = k[A]2[B]

11. 6.0 ×10–1 mol L–1 hr–1

12.

3. rate = 4.8×10-4

6. k = 9×10-3mol-1Lsec-1

9. Rate = k[NO]2[Br2] 10.

0.15 mol L–1 hr–1

[NO] [H 2 O]  3.5  10 2 mol dm–3 s–1,  5.3  10 2 mol dm–3 s–1 dt dt

13.

d[NO 2 ] d[O2 ]  1.25  10 2 mol L–1 s–1,  3.125  10 3 mol L–1 s–1 14. 6.0 × 10–4 molL–1 min–1 dt dt

15.

53.59 kJ mol–1 16.29.39 kJ mol–1

17. k = 3.19 × 10–2 s–1

S e

19. 6.4 × 109

20. 3.9 × 1012s–1

23. 23.0 second

24.1.98 × 10–2 min–1 25. 0.2197 min-1

27. 287.7 s

21.

28. k =8.59 × 10–4 min–1

32. k = 2.232 × 10–2 min–1; t = 62.12 min

r P

100.343 kJ mol–1

c in

40. t = 4 min

33. 0.3838 min

34. Initial rate = 2 ×10–4

Insight To The concepts

1. 4.83 min 2. k = 0.0327 min-1 3. 86.47 kJmol-1 5. k = 1.9  10-2 min-1 6. Ef = 6  104,Eb = 9.3  104 J 9. 2412 seconds

k1k3 [ A]2 10. k3 [ A]  k3

12. (i) 2NO2  NO + O3 (fast) 13. t1/2 = 1.386 time 14. 3.4 min 1. C 13.D 25.D 37.A 49.C 61.A

2.B 14.D 26.A 38.A 50.A 62.C

3.A 15.D 27.D 39.B 51.A 63.B

26.2.005 s

31.t1/2 =128.3 hours;5.27%

4V3 l 38. k = t ln 5(V  V ) 3 2

y B

22. 8.46 Msec–1

29. 1.969 min

mol L–1 s–1; 0.036 mol 35. t1/2 = 66.66 min, k = 0.01039 min–1 4 P3 1 37. k  t ln 5( P  P ) 3 2

ri

18. 9.06 × 10–3 s–1

39.

P3 1 36. k  t ln 2( P  P ) 3 2

[C] 10 1x – 1) = (e11x [A] 11

4. 4.8  10-3 7. 21.5 hrs

8. 1.58 yrs

11. k2=7  10-3 (ii) NO3 + CO  CO2 +NO2

Objective exercise 4.D 5.C 6.B 7.D 16.B 17.A 18.A 19.B 28.A 29.C 30.D 31.B 40.A 41.B 42.B 43.A 52.A 53.B 54.B 55.D 64.A 65.C 66.B 67.A

8.C 20.D 32.A 44.D 56.A 68.A

9.C 21.B 33.A 45.A 57.D

10.D 22.B 34.B 46.D 58.B

11.C 23.D 35.B 47.D 59.A

12.B 24.C 36.D 48.A 60.B

Reconstruct Your Chemistry With Prince Sir Subject Aptitude 1. a,b 2. a,b 3. a,b 4.a,b,c 5. a,b 6. a,b,c,d 7.a,b

8. a,b 9. a,b,c 10. a,c 11.a,b 12. a,c

13. a,b,d

14. a,b,d

19. b

Cq,D p

22. Ar, Bq Cp Ds 23.Aq, Bs Cp Dr 24. Aq, Br Ct

Du 25. b

26. a

Q.1

4.54 g

dm–3

Q.2

(i) Kc= 8.1 × 10–5 mol2 L2 ; Kp = 4.91 × 10–2 atm2 (ii) Noeffect;

Q.3

D

Q.5

D

Q.6

15991 J mol–1 , 12304 J mol–1 ; B > C > A Q.7

D

Q.8

(i) 5.705 × 103 J mol–1 (ii) Since initial Gibbs free energy change of the reaction is positive, so the reverse reaction will take place

Q.9 D Q.14. 0

27. c

15. a

28.a

16. c

29. b

17. b

30. c

31. d 32. a

20. b

21. A r, B s,

33. c

Feel The Heat

Q.4

D

Q.10 B Q.15. (A, B, D) Q.16. 9

B

18. a

P y

ir

e c n

Q.11

D

Q.12. D

r i S Q.13. A