File No.24/11/18/12/2014
IX CLASS - IIT/N.T.S.E FOUNDATION - OLYMPAID
ANDHRA PRADESH - TELANGANA
2014-2015 PROGRAM M E
CHEMISTRY_ 3RD CHAPTER - SOLUTION
CHEMICAL KINETICS
1.
Given reaction 4A  B  2C  2D
 from this equation, rate of dissapearence of B is 1/4th of rate of dissapearance of A;
 The rate of formation of D is half the rate of consumption of A and
 rate of formation of C and D are equal.
2.
for a reaction 2A+B  C+D
d{A}
 K{A 2 }{B}
dt
d{B} 1
 K{A 2 }{B}
dt
2
from this
3.
The rate of chemical reaction is directly proportional to active masses of reactants.
4.
Precipitation of Agcl by mixing AgNo3 and Nacl is fastest reaction
5.
Given N2+3H2  2NH3
rate of reaction w.r.t Hydrogen - d
- 2015
{H2 }2014
1
 K{N2 }{H2 }3
dt
3
Rate of reaction w.r.t. hydrogen = decrease in concentration of H2 in unit time.
6.
Given reaction is 2A+B  C+D
rate = K{A2} {B}
given concentration of reactants are increased by 3 times then
K1 = rate = K {3A}2 {3B}
= 27K1 {A2}{B}
K1 = 27K
1
O2
2
d{N2O5 }
rate w.r.t reactant, is
 K1(N2O5 )
dt
d{NO2 }
1
rate w.r.t.
NO2 is
 K 2 {N2O5 }
dt
2
1 d{O2 }
O2
 2K 3 {N2O5 }
2 dt
K
K1  2  2K 3
2
 2K 1  K 2  4K 3
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7. Given N2 O5 2NO2+
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8.
IX CLASS - IIT/N.T.S.E FOUNDATION - OLYMPAID
If doubling the concentration of reactant A increases the rate 4 times and tripling the concentrartion of
A increases the rate 9 times then
from given information rate A 2
9.
Given 2a+b  A2B
According to rate law (or) rate equation
rate of reaction w.r.t reactants
10.
Given A+B  products
rateofreaction
11.
d{A}
 K{A}2 {B}
dt
d{A}
 K{A}{B}
dt
Given H2 +I2  2HI
d{H 2}
 K1{H2 }{I2 } (i)
dt
1 d{HI}
for 2HI H2  I2 , Rate 
 K2 {HI}2  (ii)
2 dt
given K1 49
Rate of reaction
Given equation is reversible then
from(1) &(2)
K2 
1 1

k1 49
12.
Activation energy of any reaction depends on nature of reactants.
13.
A catalyst has no effect on the state of equilibrium
14.
Given 2A+B  product
2014 - 2015
unit of K are L/mole/Sec.
15.
 dc 
The term    in a rate equation refers to the decrease in concentration of the reactant with time.
 dt 
16.
Given A+B  products
given conditions, {A} = double then K = doubled
and {A} = double, {B} = doubled then K = 8 times.
then rate = K{A} {B2}
17.
 H = 189.0KJ


2 x O2 +O2 
 2SO3,
this reaction yields at high pressure and low temperature.
The state of equilibrium is dynamic in nature.
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18.
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19.
IX CLASS - IIT/N.T.S.E FOUNDATION - OLYMPAID

 2AB rate constant =K1
A 2 +B2 
1
1


AB 
 A 2  B2 rate cos t  ?
2
2


Rate of (i)st equation, A 2  B 2 
 2AB
{dAB}
 K 1{AB}2
dt
1
1 d{A}2
d{AB} 1
Rate equationof (ii)nd {
 K 2 {A 2 }1/ 2 {B2 } 2 

2 dt
dt
k
20.
For the equilibrium PCl5  PCl3+Cl2 the increase in the increase in volume favours dissociation of PCL5
1.
2A+B  products.
According to def Rate of reaction (or) rate law = Y = K{A2} {B}
2.


2NO+O2 
 2NO2
Volume of reaction vessel is reduced to 1/3
rate of reaction 
3.
1
so, rate of reaction = 3 times.
v
1 d ( A) 1 d ( B )
d ( B ) 2 d ( A)
=
=.

3 dt
2 dt
dt
3 dt
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4.
When volume is reduced to
1
, concentrations become four times and reaction rate become 16 times
4
because it is 2 order reaction.
6.
Given Kp = 1.44  10–5
W.K.T. R = 0.0821
Kp = Kc K c (RT) n

 2N3 n = 2 – 4 = –2
For reaction N2 + 3H2 
T = 5000C = 273 + 500 = 773K
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Kc 
7.
Kp
(RT)n

1.44 10 5
10.082  773)2

 P, K = 0.01, the concentration of product is higher than the con centration of
For the reaction R 
reactant at equilibrium.
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8.

 2NH3
N2 + 3H2 
2
9.
IX CLASS - IIT/N.T.S.E FOUNDATION - OLYMPAID
d[NH3 ]
d[H2 ]
3
dt
dt
Given equation 2H2O2  2H2O + O2
Order of the reaction = sum of the powers of concentration terms of reactants in the rate equation.
concentration terms of reactants [H2O2] = 1
order of given reaction = 1
10.
2NO2 + F2  2NO2F
order of reaction = [NO2] [F2] = 2
11.

 2NO, equilibrium means that, concentration of all substances is
In the given reaction N2 + O2 
constant.
12.
For every 100 reaction in temperature the rate of reaction is doubled. But given temperature is increased
from 100C to 1000C, the rate of rection will be come –29 = 512 times.
13.
A + 2B  C+D
dA
 5 10 4 mol / lit / sec.
dt
For the reaction,
dA 1 dB

dt 2 dt
dB
dA
2
 2(5  104 ) = 1  10–3
dt
dt
14.
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H2 + 2ICl  2HCl + I2
Mechanism A: H2 + 2ICl  HCl + I2
Mechanism B: H2 + ICl  HCl + HI Slow HI + ICl  HCl + I2
Mechanism B consists with the given information.
15.
Active mass of 64 gm of HI
Givent wt. of HI = 64 gm
V = 2 lit.
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M.W.T of HI = 128
active mass =
16.
64 1
  0.25
128 2
Given aX + bY = bZ + aw is a homogeneous reaction.
for homogenous reaction Kp = Kc
17.

 C+D
In the reaction A + B 
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at equilibrium
IX CLASS - IIT/N.T.S.E FOUNDATION - OLYMPAID
[A] = 0.5, [B] = 0.8
[C] = 0.4, [D] = 1.0
[C][D]
1 0.4
Kc = [A][B]  0.5  0.8 1.0
18.
Given 2A + 3B  A2B3
Rate = K[A2] [B]2
[A]= 16X; [B] Y
rate = K
given concentration of a doubles and that of B triples then.
Rate = K [A2]2 [B3]3
= K1 [A4] [B]6
R1 = rate = (2X)2
(3Y)3
R1 = (108)R
19.
For endothermic reaction A  B has activation energy 15 KCal / mole.
and energy reaction is 5 KCal / mol.
activation energy for B  a = 15 – 5 = 10 KCal / mol.
20.
A + B  C+D H = –20KJ/mol.
activation energy for the for ward reaction is 85 KJ
activation energy for backward reaction
C+D  A+B
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H = (Activation energy of forward reaction - activation energy of back ward reaction)
Given –20 = 85–x  x = 105 KJ/mol.
BRAIN TWISTERS
1.
The reaction, CH 3 COOC 2 H 5  NaOH  CH 3 COONa  C 2 H 5 OH is bimole cular reaction, second
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order reaction
2.
The trimolecular reactions are 2SO2  O2  2SO3 , 2CO  O2  2CO2 & 2FeCl3  SnCl2  SnCl4  2FeCl2
3.
Order of reaction may have zero values, fractional values, positive values & negative values.
4.
For a reversible vexy it can’t be influenced by a catalyst, it can never proceed to completion and it can be
attained in open vessel
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5.
1)
IX CLASS - IIT/N.T.S.E FOUNDATION - OLYMPAID
2)
3)
6.
7.
a) Increase of temperature
Favours endothermic reaction
b) Decrease of pressure
Favours the reaction in the direction of increase in reactants
volume
c) Increase of concentration of
Favours the reaction in the direction of increase in reactants
volume
d) Addition of products to the system
Favours the backward reaction
a) Forward reaction
Reaction proceeds from left to right
b) Backward reaction
Reaction proceeds from right to left
c) Reversible reaction
Formation of HI from hydrogen and iodine
Formation of NO from nitrogen and oxygen
d) Irreversible reaction
8.
9.
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10.
a) 2Mg  O2  2MgO
Precipitation reaction
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Irreversible reaction
b) N2O4  NO2
Homogeneous reversible reaction
c) rf  rb
Equilibrium state


 CaO s   CO2g
d) CaCO3 s  
Hetergeneous reversible reaction
a) Concentration equilibrium constant
Kc
b) Rate constant for forward reaction
Kf
c) Rate constant for backward reaction
Kb
d) Partial pressure equilibrium contant
Kp
a) Effect of change in concentration
Increases
b) Effect of change of pressure
Decreases
c) Effect of change in temperature
Favours endothermic reaction
d) Effect of role of catalyst
Doesn’t effect
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11.
IX CLASS - IIT/N.T.S.E FOUNDATION - OLYMPAID
Kc = [A] [B2] on doubling concentration of [A], [B2] should be to maintain constancy of Kc.

 A(g)  B 2(g) the concentration of B is
On doubling the concentration of A in the equilibrum, AB2(g) 
2
haved.
Both Assertion and Reason are correct, and Reason explanation Assertion
12.
For equilibrium reaction, rate of forward reaction = rate of back ward reaction.
Both Assertion and Reason are correct, and Reason explanation Assertion
13.
A catalyst influences the rates of both forward and backward reactions to the same extent, So, catalyst has
no effect on the state of equilibrium
Both Assertion and Reason are correct, and Reason explanation Assertion
14.
For the reaction A  B , rate of forward reaction =10
K c = 100, rate of back ward reaction = 0.1
Ka
So, K c  K but given reason is correction.
b
15.
K p  K c  RT 
Δn
Kp , Kc depends on the change in the number of moles of gaseous reactants and products.


 Kc
Kp 

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16.
At equilibrium state concentrations of reactants and products remain unchanged
17.
At equilibrium rate of forward reaction is equal to rate of backward reaction.
18.
The state of equilibrium is dynamic in nature.
19.
Ka = 0.25 m/l/s
Kb = ?
Kc 
Ka
0.25
 Kb 
5
Kb
0.05
2O3  3O2 n  3 – 2 =1.
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20.
Kc = 0.05
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1.
IX CLASS - IIT/N.T.S.E FOUNDATION - OLYMPAID
For the reaction, RCl  H2O  ROH  HCl
It is a pseudo forst order, reaction.
2.
Hydrolysis of ester is presence of acid medium is appeared as 2nd order but it is of first order reaction.
3.
Rate K = [A] [B] [C] order reaction =     
given 2H2 O2  2H2 O  O2
Rate = K [H2O2 ]2 [H2O] 2 [O 2 ]1
Order of reaction = 1
4.
Order of 2NO2  F2  2NO2F is 2
5.
The order of CH3CHO  CH4  CO is 1.5
6.
1) Pb NO3 2  aq  2 NaI  aq   PbI2  s   2 NaNO3  aq – irreversible reaction
 2 NaOH (aq) + H2 (g) – irreversible reaction
2) 2 Na (s) + 2 H2O (l) 
 AgCl (s) + HNO3(aq) – irreversible reaction
3) AgNO3 (aq) + HCl (aq) 
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 KCl (aq) + NaNO3 (aq) – reversible reaction
4) KNO3 (aq) + NaCl (aq) 
 KCl (aq) + NaNO3 (aq)
An example of a reversible reaction is KNO3 (aq) + NaCl (aq) 
7.
i) CDiamond s   CGraphite  s 
– Homeogeneous reaction.
ii) H 2 O  s   H 2 O l 
– Leterogeneous reaction
iii) N2g +3H2g  2NH3 g
– Homogeneous reaction
iv) MgCO3 s   MgO s + CO2g – Heterogeneous reaction
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v) P C l 3  g  + C l 2  g   P C l 5  g  – Homogeneous reaction
CDiamond s   CGraphite  s  , N2g +3H2g  2NH3 g , P C l 3  g  + C l 2  g   P C l 5  g 
are homo gen eou s,
H 2 O  s   H 2 O l  , MgCO3 s   MgO s + CO2g are heterogeneous systems.
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8.
IX CLASS - IIT/N.T.S.E FOUNDATION - OLYMPAID

 2HI
Given 10H2 + 5I2 
A+ equilibrium, 10% of H2 dissociated then
9 moles of H2, 4 moles of I2, 2 moles of HI.
9.

i) CH3COOH + C2H5OH 
 CH3COOC2H5 + H2O Rate R  K CH3COOHC2H5OH

ii) N2O4 
 2NO2 Rate R  K N2O4 

iii) mX + nY 
 Xm Yn Rate R  K  X 
m
10.
n
Y
2Cl2O  2Cl2 + O2 Rate R = K[Cl2O]2.
i) Given [Cl2O] is reduced to 1/3 of its original value they
R
R1 = K[1/3 Cl2O]2  9
ii) Given [Cl2O] Rate is doubled.
then [Cl2O] = ?
R = K[Cl2O]2
 given K [Cl2O]2 = 2R
then [Cl2O] =
2R
iii) Given [Cl2O] 3 times [Cl2O]
then R1 = K [3Cl2O]2
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R1 = 9R
11.
Given reaction 2A + 2B  A2B3,
12.

PCls 
 PCl3 + Cl2
2  60 2  40 2  40
100 100 100
Volume of container = 2 litre.
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2  40 2  40

100
 2 100  2  0.266
Kc 
2  60
100  2
2
13.
 10 3 
2 

2 
[NO2 ]2 
106
K


 105
[N2O4 ]
101
 2
 2
 
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14.
IX CLASS - IIT/N.T.S.E FOUNDATION - OLYMPAID
N2O4  2NO2,
total mole at equilibrium = (1–  )+2  = 1 + 
15.
K P  1.110 2 ; K b  1.5  10 3
Kc 
16.
Kf
1.110 2

 7.33
K b 1.5  103
H2 + I2  2HI
x
x
–
0
x - 6.34 x-4.28 - 42.85
KC 
17.
[HI]2
 45.20
[H2 ][I2 ]
Given H2O2(aq) + 2KI (aq) + H2SO4(aq)  2H2O (l) + I2(aq) + K2SO4(aq)
Rate w.r.t KI is
1 dKI
2 dt
initial concentrat = 0.2 m/lit
t = comin = 10  60 sec.
finial concentration = 0.05 m/lit
Rate of reaction =
1 initial concentration  final concentration

2
time
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
1 0.2  0.05

2
60  10
 1.25  104 moles / lit/ sec.
18.
Pt (NH3)2 Cl2(s) + water  [Pt (NH3)2Cl]+(aq) + Cl–(aq)
Given initial conc. = 0.0100 m/ lit
t = 20 min final conc. = 0.00970 m/lit.
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Rate of reaction 
initial conc.  final conc.
time

0.0100  0.00970
20

0.0003
 1.5105 m / lit / min
20
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19.
IX CLASS - IIT/N.T.S.E FOUNDATION - OLYMPAID
Zymase
Given i) C6H12O6 
 2C2H5OH+2CO2
t1
ii) NaCl  AgNO3  AgCl  NaNO3
t2
heat
ii) NH4NO2 (aq) 
 N2(g)  2H2O( )
t3
i
ii) Reaction is fastreaction so, +2 is very low.
iii) NH4NO2 on heating so, it will take time and (i) decomposition of gluose takes more time.
i.e., t1 > t3 > t2
20.
2HI  H2  I2 is Bimolecular reaction
2CO  O2  2CO2 it is trimolecular reaction
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2014 - 2015
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