Collision Theory

For a chemical reaction to occur between A +B

• Molecule must collide

• Molecule must collide with right orientation(geometry) so effective collision take place

• Molecule collide with total energy greater > activation energy

• Effective collision will lead to product formation

Collision between A + B

• No product formation • Energy collision < activation energy • Collision not in correct orientation • Collision not energetic enough to overcome activation energy

• Product formation • Energy collision > activation energy • Collision in correct orientation • Collision energetic enough to overcome activation energy

Rate of Reaction

• Concentration reactants increases ↑ • Frequency of collision increases ↑ • Frequency of effective collision increases ↑

• Temperature increases ↑ by 10C • Rate increases ↑ 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 increases ↑ by 10C • Fraction of molecules with energy > than Ea doubles • Rate increases ↑ by 100%

Ineffective collision Effective collision

Concentration Temperature

Effect of temperature on Rate of Reaction

• Area under curve proportional to number of 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 > than Ea increases

• T2 > T1 by 10C – Area under curve for T2 (DEF) = 2 X Area T1 (BEC)

• Y axis – fraction molecules having a given kinetic energy

• X axis – kinetic energy/speed for molecule

• Total area under curve for T1 and T2 are the same, BUT T2 is shifted to right

with greater proportion of molecules having higher kinetic energy

Average Kinetic Energy α Absolute Temperature

• Average kinetic energy for all molecules at T1 lower

• Fraction of molecule with energy > than Ea is less ↓

• Average kinetic energy for all molecules at T2 is higher (shifted to right)

• Fraction of molecule with energy > than Ea is higher ↑

Using Maxwell Boltzmann Distribution to explain how Temp affect rate of reaction

At T1 (lower Temp)

At T2 (Higher Temp)

http://www.chemistry.wustl.edu/~courses/genchem/Tutorials/Airbags/kinetic.htm

Relationship between Rate constant and Temperature is explained by Arrhenius Equation

Maxwell Boltzmann and Arrhenius Equation to explain relationship between rate constant and temperature

Arrhenius constant

Fraction of molecules with energy greater than Ea

Rate constant, k increases ↑ exponentially with temperature Temp increases ↑ , -ve exponent decrease ↓, k increases ↑, Rate increases↑

Rate constant

R = Gas constant, 8.314 J K-1 mol-1

Arrhenius constant • Made up of collision frequency and orientation factor

T = Absolute Temp in K

Ea = Activation energy (kJmol-1)

Maxwell Boltzmann Distribution Arrhenius Equation

Fraction of molecules with energy > than Ea Fraction of molecules = e –Ea/RT

= e -55000/8.314 x 298

= 2.28 x10 -10

Fraction of molecules with energy > than Ea Fraction of molecules = e –Ea/RT

= e -55000/8.314 x 308

= 4.60 x 10 -10

Total number of molecules with energy > than Ea = Fraction of molecules x total number of molecules in 1 mole = 2.28 x 10-10 x 6.0 x 10 23

= 1.37 x 1014

Total number of molecules with energy > than Ea = Fraction of molecules x total number of molecules in 1 mole = 4.60 x 10-10 x 6.0 x 10 23

= 2.77 x 1014

Total number molecules with energy >E a Fraction of molecules with energy > Ea = Total number of molecules in 1 mole of gas

Why rate doubles (increase by 100%) when temp rises by 10C

Total number of molecules with energy > Ea at 35C 2.77 x 10 14 2 (Double)

Total number of molecules with energy > Ea at 25C 1.37 x 10 14

Rate double = area under curve double = fraction of molecules under curve doubles

Arrhenius equation to explain why rate doubles when temperature rises from 25C to 35C

At 35C

At 25C

Fraction of molecules with energy > than Ea

How to convert fraction of molecules to total number of molecules in 1 mole of gas?

At 35C

At 25C

= =

Temperature increase from 25C to 35C (10C rise)

Rate of reaction doubles (100% increase)

Collision more energetic

Kinetic Energy α Absolute temperature • Average Kinetic Energy = 1/2mv2 α Temp • As Temp increase from 25C to 35C

1/2mv2 T2(35C) 1/2mv2 T1(25C) V2

2 308 V1

2 298 V2 √1.03 V1

V2 1.01V1

Average speed v2 = 1.01 x average speed v1

Average speed increses by only 1.6%

Rate at 35C = 1.01 x Rate 25C

Rate increase only by 1.6%

Using Arrhenius equation

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

e -23.82

k (35C) 2.2 k (25C) k (35C) = 2.2 x k (25C)

Rate at 35C = 2 x Rate at 25C

Rate increase by 100%

Collision frequency Increases

Using average kinetic equation - 1/2mv2 α Temp

=

=

=

= =

=

=

=

X

Temperature increase from 25C to 35C (10C rise)

Rate of reaction doubles (100% increase)

Collision more energetic

Kinetic Energy α Absolute temperature • Average Kinetic Energy = 1/2mv2 α Temp • As Temp increase from 25C to 35C

Collision frequency Increases

X

Conclusion • Temp increase ↑ – Rate increase ↑ is due to more energetic collision and NOT due to increase in frequency in collision

• Temp – cause more molecules having energy > Ea- leading to increase ↑ in rate

• Rate of reaction increases ↑ exponentially with increasing Temp ↑

• Rate doubles X2 for every 10C rise in temperature

• Rate increase ↑ is mainly due to more energetic collision

(collision between molecules more energetic, with energy > than Ea, leads to product formation)

Rate at 35C = 2 x Rate at 25C

Rate increase by 100% Rate at 35C = 1.01 x Rate 25C

Rate increase only by 1.6%

How Temperature and rate constant are linked together by Arrhenius Equation

Reaction between X colliding with Y X + Y → Z

Rate of rxn = (Total number of collision) x ( fraction of collision, energy >Ea) x ( [X]1[Y]1 )

Arrhenius Constant A

Fraction of molecules with energy > than Ea e –Ea/RT

Concentration [X][Y]

Rate of rxn = A x e –Ea/RT x [X]1[Y]1

Rate of rxn = k [X]1[Y]1

If conc of X and Y are constant BUT temperature changes, combine equation 1 and 2

Rate of rxn = k [X]1[Y]1 = A x e –Ea/RT x [X]1[Y]1

k = A x e –Ea/RT

Rate of rxn can be written in TWO equation forms

Rate of rxn = A x e –Ea/RT x [X]1[Y]1

Equation 1 Equation 2

Arrhenius Equation

Cancel conc on both sides

How Temperature and rate constant are linked together by Arrhenius Equation

Fraction of molecules with energy > than Ea e –Ea/RT

Concentration [X][Y]

Temperature increase

• Fraction of molecules with energy >Ea increases • Rate increases exponentially

Concentration increase

• Frequency of effective collision increase • Rate increase

Rate of rxn = A x e –Ea/RT x [X]1[Y]1

Using Arrhenius Equation to determine Ea by graphical Method

Plot of ln k vs 1/T gives a linear line • Gradient = -Ea/R ( Ea can be calculated) • ln A = intercept at y axis

Plot of log k vs 1/T gives a linear line • Gradient = -Ea/R (Ea can be calculated) • log A = intercept at y axis

ln both sides log both sides

Ea can be determine from the gradient

EX 1 : Decomposition of hydrogen iodide 2HI ↔ H2 + I2 was performed at different temp and k was measured

Plot of ln k against 1/T was performed

•Gradient = -Ea/R

• Gradient = -2.25 x 104K

-2.25 x 104K = -Ea/R

Ea = 2.25 x 104 x 8.314JK-1mol-1 = 1.87 x105Jmol-1

Step 1 Using Arrhenius Equation and ln both sides Step 2

Step 3 Find Ea from its gradient

Using Arrhenius Equation to determine Ea using ln k and 1/T by graphical Method

EX 2 : Decomposition of hydrogen iodide 2HI ↔ H2 + I2 was performed at different temp and k was measured

Plot of log k against 1/T was performed Step 1 Using Arrhenius Equation and log both sides Step 2

Step 3 Find Ea from its gradient

Gradient = -Ea/2.303R

Gradient = -9.23 x 103K

- 9.23 x 103K = -Ea/2.303R

Ea = 9.23 x 103 x 8.314 x 2.303 JK-1mol-1 = 176.6 kJ mol-1

Convert to 1/T and log k

Using Arrhenius Equation to determine Ea using log k and 1/T by graphical Method

Using Arrhenius Equation to determine Ea without graphical method

EX 1 : Reaction between 2HI ↔ H2 + I2 Rate constant at 600K is 2.7 x 10-14 Rate constant at 650K is 3.5 x 10-3

ln k1 = ln A - Ea

RT1

ln k2 = ln A - Ea

RT2

At 600K At 650K

ln k2 – lnk1 = - Ea + Ea

RT2 RT1

log k1 = log A - Ea

2.303RT1

log k2 = log A - Ea

2.303RT2

log k2 – logk1 = - Ea + Ea

2.303RT2 2.303RT1

Equation 2 – Equation 1

Equation 1 Equation 2

EX 2 : Reaction between 2HI ↔ H2 + I2 Rate constant at 600K is 2.7 x 10-14 Rate constant at 650K is 3.5 x 10-3

At 600K At 650K

Equation 1 Equation 2

Equation 2 – Equation 1

Using ln k and 1/T

Using log k and 1/T