This document discusses collision theory and how temperature affects the rate of chemical reactions. It explains that for a reaction to occur, molecules must collide with sufficient energy to overcome the activation energy barrier. Raising the temperature increases the fraction of molecules with kinetic energy above this threshold, leading to more effective collisions and a higher reaction rate. This relationship between temperature and rate is quantitatively described by the Arrhenius equation, from which the activation energy can be determined graphically by plotting the natural log of rate constants versus inverse temperature.

1. Collision Theory
Chemical rxn to occur bet A +B
• Molecule collide with right orientation(geometry)
( effective collisiontake place)
• Molecule collide with total energy greater > activationenergy
• Effectivecollisionwill lead to product formation
Collision bet A + B
No product formation
• Energy collision< activationenergy
• Collisionnot in correct orientation
or energetic enough to overcome Ea
Product formation
• Energy collision> activationenergy
• Collisionenergetic enough to overcome Ea
Rate of Reaction
• Conc reactant increase ↑
• Freq of collisionincrease ↑
• Freq of effectivecollisionincrease ↑
• Temp increase ↑ by 10 C
• Rate increase ↑ by 100%
• Exponential relationship
bet Temp and Rate
Rate = k [A]1[B]1 1st order – A
1st order – B
Conc A doubles x2 – Rate x2
Conc A triples x3 – Rate x3
Collision Theory
• Temp increase ↑ by 10C
• Fractionof molecules with
energy > Ea doubles
• Rate increases ↑ by 100%
Ineffective collision Effective collision
Concentration
Temperaturedepends

2. Effect Temp on Rate of rxn
• Area under curve proportional to number molecules
• Wide range of molecules with diff kinetic energy at particular temp
• Temp increases ↑ - curve shifted to right
• Temp increases ↑ - fraction of molecules with energy > Ea increase
• T2 > T1 by 10C – Area under curve for T2 = 2 X Area T1
• Y axis – fraction molecules having a given kinetic energy
• X axis – kinetic energy/speed for molecule
• Total area under curve for T1 and T2 same, BUT T2 is shift to right
with greater proportion molecules having higher kinetic energy
Ave Kinetic Energy α Abs Temp
• Ave kinetic energyfor all moleculesat T1 lower
• Fraction of molecule with energy> Ea is less ↓
• Ave kinetic energyfor all moleculesat T2 is higher
• Fractionof moleculewith energy> Ea is higher ↑
Maxwell Boltzmann Distribution -Temp affect Rate
At T1 (Low Temp)
At T2 (HighTemp)
T1
T2
RT
Ea
Aek


Relationship k and T - Arrhenius Eqn
Rate
constant
Arrheniusconstant
• Collision freq
and orientation factor
Ea = Activation
energy
T = Abs Tempin K
R = Gas constant
8.314 J K-1 mol-1
RT
Ea
Aek


Fraction of molecules
with energy> Ea
Rate constant,k increases ↑ exponentiallywithTemp

3. Fraction molecules with energy > Ea
Fraction molecules = e –Ea/RT
= e -55000/8.314 x 298
= 2.28 x10 -10
Fraction molecules with energy > Ea
Fraction molecules = e –Ea/RT
= e -55000/8.314 x 308
= 4.60 x 10 -10
Total number molecule with energy > Ea = Fraction molecule x total number molecule in 1 mol
= 2.28 x 10-10 x 6.0 x 10 23
= 1.37 x 1014
Total number molecule with energy > Ea = Fraction molecule x total number molecule in 1 mol
= 4.60 x 10-10 x 6.0 x 10 23
= 2.77 x 1014
Total number molecule with energy >E a
Fraction molecules with energy > Ea =
Total number of molecule in 1 mole gas
Rate double (increase by 100%) when temp rise by 10C
Total number molecule with energy > Ea at 35C 2.77 x 10 14 2 (Double)
Total number molecule with energy > Ea at 25C 1.37 x 10 14
Rate double = area under curve double = fraction molecule under curve double
Arrhenius Eqn - Rate doubles when temp rises from 25C to 35C
At 35C – 308KAt 25C – 298K
Convert fractionof molecules to total number of molecules in 1 mole of gas?
At 35C
At 25C
= =
Fraction of moleculeswith energy > EaRT
Ea
eAk

 ..

4. Temp increase from 25C to 35C (10C rise)
Rate doubles (100% increase)
Collision more energetic
Kinetic Energy α Abs Temp
• Ave Kinetic Energy = 1/2mv2 α Temp
• As Temp increase from 25C to 35C
1/2mv2 T2(35C)
1/2mv2 T1(25C)
↓
V2
2
308 √1.03
V1
2
298
↓
V2 1.01V1
Ave speed v2 = 1.01 x Ave speed v1
Ave speed increase by only 1.6%
Using Arrhenius eqn
k (35C) Ae -Ea/RT
k (25C) Ae -Ea/RT
k (35C) e -58000/8.31 x 308
k (25C) e -58000/8.31 x 298
e -23.03 2.2
e -23.82
k (35C) = 2.2 x k (25C)
Rate at 35C = 2 x Rate at 25C
Rate increase by 100%
Collision frequencyIncrease
Using Ave kinetic eqn - 1/2mv2α Temp
=
=
=
=
=
= =
=
Conclusion
• Temp increase ↑ – Rate increase ↑ due to more energetic collision and NOT due to increase in freq in collision
• Temp – cause more molecule having energy > Ea- lead to increase ↑ in rate
• Rate increase ↑ exponentially with increase Temp ↑
• Rate double X2 for every 10C rise in Temp
• Rate increase ↑ mainly due to more energetic collision
(collision bet molecules more energetic, with energy > Ea

5. Temp and rate constant link by Arrhenius Eqn
X + Y → Z
Rate of rxn = (Total number collision)x ( fractioncollision, energy >Ea) x ( [X] [Y] )
Arrhenius Constant
A
Fractionmolecule energy > Ea
e –Ea/RT
Conc
[X][Y]
Rate of rxn = A e –Ea/RT [X][Y]
Rate of rxn = k [X] [Y]
If conc constant BUT Temp changes, combine eqn 1 and 2
Rate of rxn = k [X]1
[Y]1
= A e –Ea/RT [X] [Y]
k = A e –Ea/RT
Rate rxn written in TWO forms
Rate of rxn = A e –Ea/RT [X] [Y]
Eqn 1 Eqn 2
Cancel both sides
Rate of rxn = A e –Ea/RT [X][Y]
Fractionmolecule energy > Ea
e –Ea/RT
Conc
[X][Y]
Temp increase
Fraction with energy> Ea increase
Rate increaseexponentially
Conc increase
Freq collision increase
Rate increase
ArrheniusEqn - Ea by graphicalMethod
RT
Ea
eAk

 ..







TR
E
Ak a 1
lnln 






TR
E
Ak a 1
303.2
lglg
Plot ln k vs 1/T
• Gradient = -Ea/R
• ln A = intercept y axis
Plot log k vs 1/T
• Gradient = -Ea/R
• log A = intercept yaxis
ln both sides log both sides
ln k lg k
1/T1/T
-Ea/R -Ea/R

6. Decomposition of 2HI ↔ H2 + I2 at diff tempand k was measured.Find Ea
Arrhenius Eqn
Ea from its gradient
Arrhenius Eqn - Ea by graphical Method
Temp/K 1/T k/dm3 mol-1 s-1 ln k
633 1.58 x 10-3 1.78 x 10-5 -10.94
666 1.50 x 10-3 1.07 x 10-4 -9.14
697 1.43 x 10-3 5.01 x 10-4 -7.60
715 1.40 x 10-3 1.05 x 10-5 -6.86
781 1.28 x 10-3 1.51 x 10-2 -4.19
1
RT
Ea
eAk

 .. 






TR
E
Ak a 1
lnln
Plot ln k vs 1/T
2
ln both sides
-Ea/R
Gradient = -Ea/R
Gradient = -2.25 x 104
-2.25 x 104 = -Ea/R
Ea = 2.25 x 104 x 8.314
= 1.87 x105 Jmol-1
ln k
1/T
Rxn 2HI ↔ H2 + I2 Find Ea
1
1 lnln
RT
E
Ak a

2
2 lnln
RT
E
Ak a

12
4
3
121
2
1066.1
600
1
650
1
314.8107.2
105.3
ln
11
ln






























kJmolE
E
TTR
E
k
k
a
a
a












121
2 11
ln
TTR
E
k
k a
600K ---k = 2.7 x 10-14 650K --- k= 3.5 x 10-3

7. Decomposition of Mass X at diff temp and k was measured.Find Ea
Arrhenius Eqn - Ea by graphical Method
3
1
1 lnln
RT
E
Ak a

2
2 lnln
RT
E
Ak a

1
121
2
7.52
273
1
286
1
314.855.0
4.1
ln
11
ln


























kJmolE
E
TTR
E
k
k
a
a
a












121
2 11
ln
TTR
E
k
k a
Time/m Mass left/g Mass loss
0 130 0
20 120 10
60 98 32
80 86 44
273K ---k = 0.55 286K ---k = 1.4
Time/m Mass left/g Mass loss
0 130 0
10 115 15
15 106 24
30 86 44
Plot mass loss vs time
Gradient will give rate constant, k
Mass loss
273K 286K
Mass loss
time
time
Gradient
↓
rate constant, k = 0.55
Gradient
↓
rate constant, k = 1.4