Name:- SIDDHESH MAHENDRA KAREKAR
Roll No. :- 02
M.Sc.-1
It is explained in two ways:-
Thermal H2-Br2 reaction
Photochemical H2-Br2 reaction
Reaction
𝐻2 + 𝐵𝑟2 → 2HBr
It is studied by Bodenstein & Lind
Over the temperature range 205°- 302° C
Result fitted with the following expression for
the rate of consumption of H2 & Br2
𝑑 [𝐻𝐵𝑟]
𝑑𝑡
=
𝑘 𝐻2 [𝐵𝑟2 ]1/2
1+𝑘´
[𝐻𝐵𝑟]
Christiansen, Herzfeld & Polanyi suggested the
following mechanism:-
𝐵𝑟2 →
𝑘1
2Br ……..[chain initiation]
Br + 𝐻2 →
𝑘2
HBr + H ……[chain propagation]
H + 𝐵𝑟2 →
𝑘3
HBr + Br ……[chain propagation]
H + HBr →
𝑘4
𝐻2 + Br ……[chain inhibition]
Br + Br →
𝑘5
𝐵𝑟2 ……[chain termination]
The rate of formation of HBr is given in steps (2)
& (3) is
𝑑[𝐻𝐵𝑟]
𝑑𝑡
= 𝑘2[Br] [𝐻2] + 𝑘3[H] [𝐵𝑟2]
The rate of consumption of HBr is given in steps
(4)
-
𝑑[𝐻𝐵𝑟]
𝑑𝑡
= 𝑘4[H] [HBr]
The net rate is given by
𝑑[𝐻𝐵𝑟]
𝑑𝑡
= 𝑘2[Br] [𝐻2] + 𝑘3[H] [𝐵𝑟2] - 𝑘4[H] [HBr]
………..(2)
The steady state equation for H is
𝑑[𝐻]
𝑑𝑡
= 𝑘2[Br] [𝐻2] - 𝑘3[H] [𝐵𝑟2] - 𝑘4[H] [HBr] = 0
………(3)
The steady state equation for Br is
𝑑[𝐵𝑟]
𝑑𝑡
= 𝑘1[𝐵𝑟2] - 𝑘2[Br] [𝐻2] + 𝑘3[H] [𝐵𝑟2] +
𝑘4[H] [HBr] - 𝑘5[𝐵𝑟]2
= 0 ……(4)
from equations (3) & (4)
[𝐵𝑟]2
=
𝑘1
𝑘5
[𝐵𝑟2]
[Br] =[
𝑘1
𝑘5
]1/2
[𝐵𝑟2]1/2
by substituting the value of Br in equation (3)
get the value of H
[H] =
𝑘
2 (
𝑘1
𝑘5
)
1/2
[𝐵𝑟2]
1/2
𝑘3 𝐵𝑟2 + 𝑘4 [𝐻𝐵𝑟]
Substiuting the equations of H & Br in equation
(2) we get
𝑑[𝐻𝐵𝑟]
𝑑𝑡
=
2 𝑘2 (
𝑘1
𝑘5
)1/2 [𝐵𝑟2]
1
2 [𝐻2]
1+
𝑘4 [𝐻𝐵𝑟]
𝑘3 [𝐵𝑟2]
………(5)
Where k = 2 𝑘2 (
𝑘1
𝑘5
)1/2
& k‘ =
𝒌 𝟒
𝒌 𝟑
Therefore equation (5) can be written as follow
𝑑[𝐻𝐵𝑟}
𝑑𝑡
=
𝒌 [𝑯 𝟐] [𝑩𝒓 𝟐] 𝟏/𝟐
𝟏+ 𝒌′
[𝑯𝑩𝒓]
[𝑩𝒓 𝟐]
The photochemical reaction is
𝐻2 + 𝐵𝑟2
ℎν
2HBr
The suggested mechanism:-
𝐵𝑟2 + ℎν →
𝑘1
2Br ……… [chain initiation]
Br + 𝐻2 →
𝑘2
HBr + H ……[chain propagation]
H + 𝐵𝑟2 →
𝑘3
HBr + Br ……[chain propagation]
H + HBr →
𝑘4
𝐻2 + Br ……[chain inhibition]
Br + Br →
𝑘5
𝐵𝑟2 ……[chain termination]
Steady state equation for Br is
𝑑[𝐵𝑟]
𝑑𝑡
= 𝑘1 𝐼 𝑎𝑏𝑠 - 𝑘2[Br] [𝐻2] + 𝑘3[H] [𝐵𝑟2] +
𝑘4[H] [HBr] - 𝑘5[𝐵𝑟]2
= 0 …….(6)
Steady state equation for H is
𝑑[𝐻]
𝑑𝑡
= 𝑘2[Br] [𝐻2] - 𝑘3[H] [𝐵𝑟2] - 𝑘4[H] [HBr] = 0
…….. (7)
from above equations
𝑘5 [𝐵𝑟]2
= 𝑘1 𝐼 𝑎𝑏𝑠
Br = (
𝑘1
𝑘5
)1/2
(𝐼 𝑎𝑏𝑠)1/2
The value of H
H =
𝑘2 (
𝑘1
𝑘5
)1/2 (𝐼 𝑎𝑏𝑠)1/2 [𝐻2]
𝑘3 𝐵𝑟2 + 𝐾4 [𝐻𝐵𝑟]
Substituting the equations of Br & H in equation
(2) we get
𝑑[𝐻𝐵𝑟]
𝑑𝑡
=
2 𝑘2 (
𝑘1
𝑘5
)1/2 (𝐼 𝑎𝑏𝑠)
1
2 [𝐻2]
1+
𝑘4[𝐻𝐵𝑟]
𝑘3[𝐵𝑟2]
…….. (8)
Equation (8) can be written as
𝑑[𝐻𝐵𝑟]
𝑑𝑡
=
2 𝑘2 (
𝑘1
𝑘5
)1/2 (𝐼 𝑎𝑏𝑠)
1
2 𝐻2 [𝐵𝑟2]
[𝐵𝑟]2+
𝑘4[𝐻𝐵𝑟]
𝑘3
……(9)
Let, k = 2𝑘2(
𝑘1
𝑘5
)1/2
& k´ =
𝒌 𝟒
𝒌 𝟑
Hence equation (9) can be written as
𝑑[𝐻𝐵𝑟]
𝑑𝑡
=
𝒌 (𝐼 𝑎𝑏𝑠)
1
2 𝐻2 [𝐵𝑟2]
𝑩𝒓 𝟐 +k´ [𝑯𝑩𝒓]
…..(10)
The overall reaction is
CO + 𝐶𝑙2 →
←
𝐶𝑂𝐶𝑙2
The rate of formation of phosgene
𝑑[𝐶𝑂𝐶𝑙2]
𝑑𝑡
= k [𝐶𝑙2]3/2
[CO] ……(1)
The rate of decomposition of phosgene
-
𝑑[𝐶𝑂𝐶𝑙2]
𝑑𝑡
= k‘ [𝐶𝑙2]1/2
[𝐶𝑂𝐶𝑙2] …….(2)
Equating these rates give the expression for
equilibrium constant
k=
𝑘
𝑘´
=
[𝐶𝑂𝐶𝑙2]
𝐶𝑂 [𝐶𝑙2]
…….(3)
Bodenstein & Plautgivestwoalternativemechanisms
𝟏 𝒔𝒕 Mechanism :-
𝐶𝑙2 →
←
2Cl
Cl + CO →
←
COCl
COCl + 𝐶𝑙2 →
←
𝐶𝑂𝐶𝑙2 + Cl
𝟐 𝒏𝒅 Mechanism
𝐶𝑙2 →
←
2Cl
Cl + 𝐶𝑙2 →
←
𝐶𝑙3
𝐶𝑙3 + CO →
←
𝐶𝑂𝐶𝑙2 + Cl
Equation for the reaction (1) can be written as
𝑘1=
[𝐶𝑙]2
[𝐶𝑙2]
…….(1)
Equation for the reaction (2) can be written as
𝑘2 =
[ 𝐶𝑂𝐶𝑙]
𝐶𝑙 [𝐶𝑂]
……(2)
Solving for [COCl] & [Cl] gives
[Cl] = 𝑘1
1/2
[𝐶𝑙2]1/2
…….(3)
[COCl] = 𝑘2 𝑘1
1/2
[𝐶𝑙2]1/2
[CO] ……..(4)
Rate constant for the reaction 3 may be written
as
The overall rate is given by
𝑑[𝐶𝑂𝐶𝑙2]
𝑑𝑡
= 𝑘3 [COCl] [𝐶𝑙2]
= 𝑘3 𝑘2 𝑘1
1/2
[CO] [𝐶𝑙2]3/2
= 𝑘∗ [CO] [𝐶𝑙2]3/2
……(5)
Equation 5 is same form of equation 1
Rate of decomposition of phosgene
-
𝑑[𝐶𝑂𝐶𝑙2]
𝑑𝑡
= 𝑘−3[𝐶𝑂𝐶𝑙2] [Cl]
= 𝑘−3 𝑘1
1/2
[𝐶𝑂𝐶𝑙2] [𝐶𝑙2]1/2
= 𝑘∗
[𝐶𝑂𝐶𝑙2] [𝐶𝑙2]1/2
……..(6)
Equation for the reaction 2 is
𝑘2 =
[𝐶𝑙3]
𝐶𝑙 [𝐶𝑙2]
[𝐶𝑙3] = 𝑘2 [Cl] [𝐶𝑙2] = 𝑘2 𝑘1
1/2
[𝐶𝑙2]3/2
Rate of formation of [𝐶𝑂𝐶𝑙2]
𝑑[𝐶𝑂𝐶𝑙2]
𝑑𝑡
= [𝐶𝑙3] [CO]
=𝑘2 𝑘1
1/2
[𝐶𝑙2]3/2
[CO]
= 𝑘∗ [CO] [𝐶𝑙2]3/2
Rate of decomposition of [𝐶𝑂𝐶𝑙2]
-
𝑑[𝐶𝑂𝐶𝑙2]
𝑑𝑡
= 𝑘−3[𝐶𝑂𝐶𝑙2] [Cl]
= 𝑘−3 𝑘1
1/2
[𝐶𝑂𝐶𝑙2] [𝐶𝑙2]1/2
= 𝑘∗
[𝐶𝑂𝐶𝑙2] [𝐶𝑙2]1/2
Jahn proposed the mechanism
𝑂3 →
←
𝑂2 + O
O + 𝑂3 → 2𝑂2
Benson & Axworthy modified these mechanism
as follows:-
𝑂3 + M →
←
𝑂2 + O+ M
O + 𝑂3 → 2𝑂2
The rate of decomposition of ozone is
The rate of decomposition of ozone is
-
𝑑[𝑂3]
𝑑𝑡
= 𝑘1[𝑂3][M] - 𝑘−1[𝑂2][O][M]+ 𝑘2[O][𝑂3]
…….(1)
Applying the steady state approximation to the
concentration of oxygen atom gives
𝑑[𝑂]
𝑑𝑡
= 𝑘1[𝑂3][M] - 𝑘−1[𝑂2][O][M] - 𝑘2 O [𝑂3]= 0
[o] =
𝑘1[ 𝑂3][M]
𝑘−1[ 𝑂2][M]+ 𝑘2[ 𝑂3]
Substitution the value of [O] in equation (1)
-
𝑑[𝑂3]
𝑑𝑡
=
2𝑘1 𝑘2 [𝑂3]2 [𝑀]
𝑘−1[ 𝑂2][M]+ 𝑘2[ 𝑂3]
……..[2]
Basics Chemical Kinetics, G. L. Agarwal
Chemical Kinetics, 3 𝑟𝑑
ed., K. J. Laidler

Siddhesh karekar roll no. 02

  • 1.
    Name:- SIDDHESH MAHENDRAKAREKAR Roll No. :- 02 M.Sc.-1
  • 2.
    It is explainedin two ways:- Thermal H2-Br2 reaction Photochemical H2-Br2 reaction Reaction 𝐻2 + 𝐵𝑟2 → 2HBr
  • 3.
    It is studiedby Bodenstein & Lind Over the temperature range 205°- 302° C Result fitted with the following expression for the rate of consumption of H2 & Br2 𝑑 [𝐻𝐵𝑟] 𝑑𝑡 = 𝑘 𝐻2 [𝐵𝑟2 ]1/2 1+𝑘´ [𝐻𝐵𝑟]
  • 4.
    Christiansen, Herzfeld &Polanyi suggested the following mechanism:- 𝐵𝑟2 → 𝑘1 2Br ……..[chain initiation] Br + 𝐻2 → 𝑘2 HBr + H ……[chain propagation] H + 𝐵𝑟2 → 𝑘3 HBr + Br ……[chain propagation] H + HBr → 𝑘4 𝐻2 + Br ……[chain inhibition] Br + Br → 𝑘5 𝐵𝑟2 ……[chain termination]
  • 5.
    The rate offormation of HBr is given in steps (2) & (3) is 𝑑[𝐻𝐵𝑟] 𝑑𝑡 = 𝑘2[Br] [𝐻2] + 𝑘3[H] [𝐵𝑟2] The rate of consumption of HBr is given in steps (4) - 𝑑[𝐻𝐵𝑟] 𝑑𝑡 = 𝑘4[H] [HBr] The net rate is given by 𝑑[𝐻𝐵𝑟] 𝑑𝑡 = 𝑘2[Br] [𝐻2] + 𝑘3[H] [𝐵𝑟2] - 𝑘4[H] [HBr] ………..(2)
  • 6.
    The steady stateequation for H is 𝑑[𝐻] 𝑑𝑡 = 𝑘2[Br] [𝐻2] - 𝑘3[H] [𝐵𝑟2] - 𝑘4[H] [HBr] = 0 ………(3) The steady state equation for Br is 𝑑[𝐵𝑟] 𝑑𝑡 = 𝑘1[𝐵𝑟2] - 𝑘2[Br] [𝐻2] + 𝑘3[H] [𝐵𝑟2] + 𝑘4[H] [HBr] - 𝑘5[𝐵𝑟]2 = 0 ……(4) from equations (3) & (4) [𝐵𝑟]2 = 𝑘1 𝑘5 [𝐵𝑟2] [Br] =[ 𝑘1 𝑘5 ]1/2 [𝐵𝑟2]1/2
  • 7.
    by substituting thevalue of Br in equation (3) get the value of H [H] = 𝑘 2 ( 𝑘1 𝑘5 ) 1/2 [𝐵𝑟2] 1/2 𝑘3 𝐵𝑟2 + 𝑘4 [𝐻𝐵𝑟]
  • 8.
    Substiuting the equationsof H & Br in equation (2) we get 𝑑[𝐻𝐵𝑟] 𝑑𝑡 = 2 𝑘2 ( 𝑘1 𝑘5 )1/2 [𝐵𝑟2] 1 2 [𝐻2] 1+ 𝑘4 [𝐻𝐵𝑟] 𝑘3 [𝐵𝑟2] ………(5) Where k = 2 𝑘2 ( 𝑘1 𝑘5 )1/2 & k‘ = 𝒌 𝟒 𝒌 𝟑 Therefore equation (5) can be written as follow 𝑑[𝐻𝐵𝑟} 𝑑𝑡 = 𝒌 [𝑯 𝟐] [𝑩𝒓 𝟐] 𝟏/𝟐 𝟏+ 𝒌′ [𝑯𝑩𝒓] [𝑩𝒓 𝟐]
  • 9.
    The photochemical reactionis 𝐻2 + 𝐵𝑟2 ℎν 2HBr The suggested mechanism:- 𝐵𝑟2 + ℎν → 𝑘1 2Br ……… [chain initiation] Br + 𝐻2 → 𝑘2 HBr + H ……[chain propagation] H + 𝐵𝑟2 → 𝑘3 HBr + Br ……[chain propagation] H + HBr → 𝑘4 𝐻2 + Br ……[chain inhibition] Br + Br → 𝑘5 𝐵𝑟2 ……[chain termination]
  • 10.
    Steady state equationfor Br is 𝑑[𝐵𝑟] 𝑑𝑡 = 𝑘1 𝐼 𝑎𝑏𝑠 - 𝑘2[Br] [𝐻2] + 𝑘3[H] [𝐵𝑟2] + 𝑘4[H] [HBr] - 𝑘5[𝐵𝑟]2 = 0 …….(6) Steady state equation for H is 𝑑[𝐻] 𝑑𝑡 = 𝑘2[Br] [𝐻2] - 𝑘3[H] [𝐵𝑟2] - 𝑘4[H] [HBr] = 0 …….. (7) from above equations 𝑘5 [𝐵𝑟]2 = 𝑘1 𝐼 𝑎𝑏𝑠 Br = ( 𝑘1 𝑘5 )1/2 (𝐼 𝑎𝑏𝑠)1/2
  • 11.
    The value ofH H = 𝑘2 ( 𝑘1 𝑘5 )1/2 (𝐼 𝑎𝑏𝑠)1/2 [𝐻2] 𝑘3 𝐵𝑟2 + 𝐾4 [𝐻𝐵𝑟] Substituting the equations of Br & H in equation (2) we get 𝑑[𝐻𝐵𝑟] 𝑑𝑡 = 2 𝑘2 ( 𝑘1 𝑘5 )1/2 (𝐼 𝑎𝑏𝑠) 1 2 [𝐻2] 1+ 𝑘4[𝐻𝐵𝑟] 𝑘3[𝐵𝑟2] …….. (8) Equation (8) can be written as 𝑑[𝐻𝐵𝑟] 𝑑𝑡 = 2 𝑘2 ( 𝑘1 𝑘5 )1/2 (𝐼 𝑎𝑏𝑠) 1 2 𝐻2 [𝐵𝑟2] [𝐵𝑟]2+ 𝑘4[𝐻𝐵𝑟] 𝑘3 ……(9)
  • 12.
    Let, k =2𝑘2( 𝑘1 𝑘5 )1/2 & k´ = 𝒌 𝟒 𝒌 𝟑 Hence equation (9) can be written as 𝑑[𝐻𝐵𝑟] 𝑑𝑡 = 𝒌 (𝐼 𝑎𝑏𝑠) 1 2 𝐻2 [𝐵𝑟2] 𝑩𝒓 𝟐 +k´ [𝑯𝑩𝒓] …..(10)
  • 13.
    The overall reactionis CO + 𝐶𝑙2 → ← 𝐶𝑂𝐶𝑙2 The rate of formation of phosgene 𝑑[𝐶𝑂𝐶𝑙2] 𝑑𝑡 = k [𝐶𝑙2]3/2 [CO] ……(1) The rate of decomposition of phosgene - 𝑑[𝐶𝑂𝐶𝑙2] 𝑑𝑡 = k‘ [𝐶𝑙2]1/2 [𝐶𝑂𝐶𝑙2] …….(2) Equating these rates give the expression for equilibrium constant k= 𝑘 𝑘´ = [𝐶𝑂𝐶𝑙2] 𝐶𝑂 [𝐶𝑙2] …….(3)
  • 14.
    Bodenstein & Plautgivestwoalternativemechanisms 𝟏𝒔𝒕 Mechanism :- 𝐶𝑙2 → ← 2Cl Cl + CO → ← COCl COCl + 𝐶𝑙2 → ← 𝐶𝑂𝐶𝑙2 + Cl 𝟐 𝒏𝒅 Mechanism 𝐶𝑙2 → ← 2Cl Cl + 𝐶𝑙2 → ← 𝐶𝑙3 𝐶𝑙3 + CO → ← 𝐶𝑂𝐶𝑙2 + Cl
  • 15.
    Equation for thereaction (1) can be written as 𝑘1= [𝐶𝑙]2 [𝐶𝑙2] …….(1) Equation for the reaction (2) can be written as 𝑘2 = [ 𝐶𝑂𝐶𝑙] 𝐶𝑙 [𝐶𝑂] ……(2) Solving for [COCl] & [Cl] gives [Cl] = 𝑘1 1/2 [𝐶𝑙2]1/2 …….(3) [COCl] = 𝑘2 𝑘1 1/2 [𝐶𝑙2]1/2 [CO] ……..(4)
  • 16.
    Rate constant forthe reaction 3 may be written as The overall rate is given by 𝑑[𝐶𝑂𝐶𝑙2] 𝑑𝑡 = 𝑘3 [COCl] [𝐶𝑙2] = 𝑘3 𝑘2 𝑘1 1/2 [CO] [𝐶𝑙2]3/2 = 𝑘∗ [CO] [𝐶𝑙2]3/2 ……(5) Equation 5 is same form of equation 1 Rate of decomposition of phosgene - 𝑑[𝐶𝑂𝐶𝑙2] 𝑑𝑡 = 𝑘−3[𝐶𝑂𝐶𝑙2] [Cl] = 𝑘−3 𝑘1 1/2 [𝐶𝑂𝐶𝑙2] [𝐶𝑙2]1/2 = 𝑘∗ [𝐶𝑂𝐶𝑙2] [𝐶𝑙2]1/2 ……..(6)
  • 17.
    Equation for thereaction 2 is 𝑘2 = [𝐶𝑙3] 𝐶𝑙 [𝐶𝑙2] [𝐶𝑙3] = 𝑘2 [Cl] [𝐶𝑙2] = 𝑘2 𝑘1 1/2 [𝐶𝑙2]3/2 Rate of formation of [𝐶𝑂𝐶𝑙2] 𝑑[𝐶𝑂𝐶𝑙2] 𝑑𝑡 = [𝐶𝑙3] [CO] =𝑘2 𝑘1 1/2 [𝐶𝑙2]3/2 [CO] = 𝑘∗ [CO] [𝐶𝑙2]3/2
  • 18.
    Rate of decompositionof [𝐶𝑂𝐶𝑙2] - 𝑑[𝐶𝑂𝐶𝑙2] 𝑑𝑡 = 𝑘−3[𝐶𝑂𝐶𝑙2] [Cl] = 𝑘−3 𝑘1 1/2 [𝐶𝑂𝐶𝑙2] [𝐶𝑙2]1/2 = 𝑘∗ [𝐶𝑂𝐶𝑙2] [𝐶𝑙2]1/2
  • 19.
    Jahn proposed themechanism 𝑂3 → ← 𝑂2 + O O + 𝑂3 → 2𝑂2 Benson & Axworthy modified these mechanism as follows:- 𝑂3 + M → ← 𝑂2 + O+ M O + 𝑂3 → 2𝑂2 The rate of decomposition of ozone is
  • 20.
    The rate ofdecomposition of ozone is - 𝑑[𝑂3] 𝑑𝑡 = 𝑘1[𝑂3][M] - 𝑘−1[𝑂2][O][M]+ 𝑘2[O][𝑂3] …….(1) Applying the steady state approximation to the concentration of oxygen atom gives 𝑑[𝑂] 𝑑𝑡 = 𝑘1[𝑂3][M] - 𝑘−1[𝑂2][O][M] - 𝑘2 O [𝑂3]= 0 [o] = 𝑘1[ 𝑂3][M] 𝑘−1[ 𝑂2][M]+ 𝑘2[ 𝑂3]
  • 21.
    Substitution the valueof [O] in equation (1) - 𝑑[𝑂3] 𝑑𝑡 = 2𝑘1 𝑘2 [𝑂3]2 [𝑀] 𝑘−1[ 𝑂2][M]+ 𝑘2[ 𝑂3] ……..[2]
  • 22.
    Basics Chemical Kinetics,G. L. Agarwal Chemical Kinetics, 3 𝑟𝑑 ed., K. J. Laidler