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IB Chemistry on Collision Theory, Maxwell Boltzmann Distribution Curve and Arrhenius Equation
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IB Chemistry on Collision Theory, Maxwell Boltzmann Distribution Curve and Arrhenius Equation

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IB Chemistry on Collision Theory, Maxwell Boltzmann Distribution curve and Arrhenius Equation

IB Chemistry on Collision Theory, Maxwell Boltzmann Distribution curve and Arrhenius Equation

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  • 1. Tutorial on Collision Theory, ArrheniusEquation and Maxwell BoltzmannDistribution Curve. Prepared by Lawrence Kok http://lawrencekok.blogspot.com
  • 2. Collision TheoryFor 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 Ineffective collision Effective collision • No product formation • Product formation • Energy collision < activation energy • Energy collision > activation energy • Collision not in correct orientation • Collision in correct orientation • Collision not energetic enough to • Collision energetic enough to overcome activation energy overcome activation energy
  • 3. 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 Ineffective collision Effective collision • No product formation • Product formation • Energy collision < activation energy • Energy collision > activation energy • Collision not in correct orientation • Collision in correct orientation • Collision not energetic enough to • Collision energetic enough to overcome activation energy overcome activation energy Rate of Reaction Concentration Temperature• Concentration reactants increases ↑ • Temperature increases ↑ by 10C• Frequency of collision increases ↑ • Rate increases ↑ by 100%• Frequency of effective collision increases ↑ • Exponential relationship bet Temp and Rate Rate = k [A]1[B]1 1st order – A 1st order – B • Temp increases ↑ by 10C Conc A doubles x2 – Rate x2 • Fraction of molecules with Conc A triples x3 – Rate x3 energy > than Ea doubles • Rate increases ↑ by 100% Collision Theory
  • 4. Effect of temperature on Rate of Reaction Using Maxwell Boltzmann Distribution to explain how Temp affect 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 At T1 (lower Temp) • Average kinetic energy for all molecules at T1 lower • Fraction of molecule with energy > than Ea is less ↓ At T2 (Higher Temp) • Average kinetic energy for all molecules at T2 is higher (shifted to right) • Fraction of molecule with energy > than Ea is higher ↑http://www.chemistry.wustl.edu/~courses/genchem/Tutorials/Airbags/kinetic.htm
  • 5. Relationship between Rate constant and Temperature is explained by Arrhenius Equation Rate constant Ea = Activation energy (kJmol-1) T = Absolute Temp in KArrhenius constant• Made up of collision frequency R = Gas constant, 8.314 J K-1 mol-1 and orientation factor Maxwell Boltzmann and Arrhenius Equation to explain relationship between rate constant and temperature Arrhenius Equation Maxwell Boltzmann Distribution
  • 6. Relationship between Rate constant and Temperature is explained by Arrhenius Equation Rate constant Ea = Activation energy (kJmol-1) T = Absolute Temp in KArrhenius constant• Made up of collision frequency R = Gas constant, 8.314 J K-1 mol-1 and orientation factor Maxwell Boltzmann and Arrhenius Equation to explain relationship between rate constant and temperature Arrhenius Equation Maxwell Boltzmann Distribution 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↑
  • 7. Arrhenius equation to explain why rate doubles when temperature rises from 25C to 35C Fraction of molecules with energy > than Ea
  • 8. Arrhenius equation to explain why rate doubles when temperature rises from 25C to 35C At 35C Fraction of molecules with energy > than Ea Fraction of molecules = e –Ea/RT = e -55000/8.314 x 308 = 4.60 x 10 -10 At 25C 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
  • 9. Arrhenius equation to explain why rate doubles when temperature rises from 25C to 35C At 35C Fraction of molecules with energy > than Ea Fraction of molecules = e –Ea/RT = e -55000/8.314 x 308 = 4.60 x 10 -10 At 25C 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 EaHow to convert fraction of molecules to total number of molecules in 1 mole of gas? Total number molecules with energy >E a Fraction of molecules with energy > Ea = Total number of molecules in 1 mole of gas
  • 10. Arrhenius equation to explain why rate doubles when temperature rises from 25C to 35C At 35C Fraction of molecules with energy > than Ea Fraction of molecules = e –Ea/RT = e -55000/8.314 x 308 = 4.60 x 10 -10 At 25C 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 How to convert fraction of molecules to total number of molecules in 1 mole of gas? Total number molecules with energy >E a Fraction of molecules with energy > Ea = Total number of molecules in 1 mole of gasAt 25C 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 1014At 35C 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 Why rate doubles (increase by 100%) when temp rises by 10C
  • 11. Arrhenius equation to explain why rate doubles when temperature rises from 25C to 35C At 35C Fraction of molecules with energy > than Ea Fraction of molecules = e –Ea/RT = e -55000/8.314 x 308 = 4.60 x 10 -10 At 25C 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 How to convert fraction of molecules to total number of molecules in 1 mole of gas? Total number molecules with energy >E a Fraction of molecules with energy > Ea = Total number of molecules in 1 mole of gasAt 25C 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 1014At 35C 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 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
  • 12. Temperature increase from 25C to 35C (10C rise) Rate of reaction doubles (100% increase) XCollision more energetic Collision frequency Increases Kinetic Energy α Absolute temperature • Average Kinetic Energy = 1/2mv2 α Temp • As Temp increase from 25C to 35C
  • 13. Temperature increase from 25C to 35C (10C rise) Rate of reaction doubles (100% increase) XCollision more energetic Collision frequency Increases Kinetic Energy α Absolute temperature • Average Kinetic Energy = 1/2mv2 α Temp • As Temp increase from 25C to 35C Using average kinetic equation - 1/2mv2 α TempUsing Arrhenius equation 1/2mv2 T2(35C)k (35C) Ae -Ea/RT = = 1/2mv2 T1(25C)k (25C) Ae -Ea/RT V 22 308k (35C) e -58000/8.31 x 308 = = V12 298k (25C) e -58000/8.31 x 298 e -23.03 V2 √1.03 = = e -23.82 V1k (35C) = 2.2k (25C) V2 = 1.01V1k (35C) = 2.2 x k (25C)
  • 14. Temperature increase from 25C to 35C (10C rise) Rate of reaction doubles (100% increase) XCollision more energetic Collision frequency Increases Kinetic Energy α Absolute temperature • Average Kinetic Energy = 1/2mv2 α Temp • As Temp increase from 25C to 35C Using average kinetic equation - 1/2mv2 α TempUsing Arrhenius equation 1/2mv2 T2(35C)k (35C) Ae -Ea/RT = = 1/2mv2 T1(25C)k (25C) Ae -Ea/RT V 22 308k (35C) e -58000/8.31 x 308 = = V12 298k (25C) e -58000/8.31 x 298 e -23.03 V2 √1.03 = = e -23.82 V1k (35C) = 2.2k (25C) V2 = 1.01V1k (35C) = 2.2 x k (25C) Average speed v2 = 1.01 x average speed v1 Average speed increses by only 1.6% Rate at 35C = 2 x Rate at 25C Rate at 35C = 1.01 x Rate 25C Rate increase by 100% Rate increase only by 1.6%
  • 15. Temperature increase from 25C to 35C (10C rise) Rate of reaction doubles (100% increase) X Collision more energetic Collision frequency Increases Kinetic Energy α Absolute temperature • Average Kinetic Energy = 1/2mv2 α Temp • As Temp increase from 25C to 35C Rate at 35C = 2 x Rate at 25C Rate at 35C = 1.01 x Rate 25C Rate increase by 100% Rate increase only by 1.6%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)
  • 16. How Temperature and rate constant are linked together by Arrhenius EquationReaction between X colliding with Y X+Y→ZRate of rxn = (Total number of collision) x ( fraction of collision, energy >Ea) x ( [X]1[Y]1 ) Arrhenius Constant Fraction of molecules with energy > than Ea Concentration A e –Ea/RT [X][Y]
  • 17. How Temperature and rate constant are linked together by Arrhenius EquationReaction between X colliding with Y X+Y→ZRate of rxn = (Total number of collision) x ( fraction of collision, energy >Ea) x ( [X]1[Y]1 ) Arrhenius Constant Fraction of molecules with energy > than Ea Concentration A e –Ea/RT [X][Y] Rate of rxn = A x e –Ea/RT x [X]1[Y]1
  • 18. How Temperature and rate constant are linked together by Arrhenius EquationReaction between X colliding with Y X+Y→ZRate of rxn = (Total number of collision) x ( fraction of collision, energy >Ea) x ( [X]1[Y]1 ) Arrhenius Constant Fraction of molecules with energy > than Ea Concentration A e –Ea/RT [X][Y] Rate of rxn = A x e –Ea/RT x [X]1[Y]1 Rate of rxn can be written in TWO equation forms Equation 1 Equation 2 Rate of rxn = k [X]1[Y]1 Rate of rxn = A x e –Ea/RT x [X]1[Y]1 If conc of X and Y are constant BUT temperature changes, combine equation 1 and 2
  • 19. How Temperature and rate constant are linked together by Arrhenius EquationReaction between X colliding with Y X+Y→ZRate of rxn = (Total number of collision) x ( fraction of collision, energy >Ea) x ( [X]1[Y]1 ) Arrhenius Constant Fraction of molecules with energy > than Ea Concentration A e –Ea/RT [X][Y] Rate of rxn = A x e –Ea/RT x [X]1[Y]1 Rate of rxn can be written in TWO equation forms Equation 1 Equation 2 Rate of rxn = k [X]1[Y]1 Rate of rxn = A x e –Ea/RT x [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 Cancel conc on both sides k =Axe –Ea/RT
  • 20. How Temperature and rate constant are linked together by Arrhenius EquationReaction between X colliding with Y X+Y→ZRate of rxn = (Total number of collision) x ( fraction of collision, energy >Ea) x ( [X]1[Y]1 ) Arrhenius Constant Fraction of molecules with energy > than Ea Concentration A e –Ea/RT [X][Y] Rate of rxn = A x e –Ea/RT x [X]1[Y]1 Rate of rxn can be written in TWO equation forms Equation 1 Equation 2 Rate of rxn = k [X]1[Y]1 Rate of rxn = A x e –Ea/RT x [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 Cancel conc on both sides k =Axe –Ea/RT Arrhenius Equation
  • 21. How Temperature and rate constant are linked together by Arrhenius Equation Rate of rxn = A x e –Ea/RT x [X]1[Y]1 Fraction of molecules with energy > than Ea Concentration e –Ea/RT [X][Y] Temperature increase Concentration increase • Fraction of molecules with • Frequency of effective energy >Ea increases collision increase • Rate increases exponentially • Rate increase
  • 22. How Temperature and rate constant are linked together by Arrhenius Equation Rate of rxn = A x e –Ea/RT x [X]1[Y]1 Fraction of molecules with energy > than Ea Concentration e –Ea/RT [X][Y] Temperature increase Concentration increase • Fraction of molecules with • Frequency of effective energy >Ea increases collision increase • Rate increases exponentially • Rate increase Using Arrhenius Equation to determine Ea by graphical Method ln both sides log both sides
  • 23. How Temperature and rate constant are linked together by Arrhenius Equation Rate of rxn = A x e –Ea/RT x [X]1[Y]1 Fraction of molecules with energy > than Ea Concentration e –Ea/RT [X][Y] Temperature increase Concentration increase • Fraction of molecules with • Frequency of effective energy >Ea increases collision increase • Rate increases exponentially • Rate increase Using Arrhenius Equation to determine Ea by graphical Method ln both sides log both sides Plot of ln k vs 1/T gives a linear line Plot of log k vs 1/T gives a linear line• Gradient = -Ea/R ( Ea can be calculated) • Gradient = -Ea/R (Ea can be calculated)• ln A = intercept at y axis • log A = intercept at y axis Ea can be determine from the gradient
  • 24. Using Arrhenius Equation to determine Ea using ln k and 1/T by graphical Method EX 1 : Decomposition of hydrogen iodide 2HI ↔ H2 + I2 was performed at different temp and k was measuredStep 1 Using Arrhenius Equation and ln both sides Step 2 Plot of ln k against 1/T was performedStep 3 Find Ea from its gradient •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
  • 25. Using Arrhenius Equation to determine Ea using log 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 Convert to 1/T and log kStep 1 Using Arrhenius Equation and log both sides Step 2 Plot of log k against 1/T was performedStep 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
  • 26. 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 At 600K At 650K ln k1 = ln A - Ea Using ln k and 1/T ln k2 = ln A - Ea Equation 2 Equation 1 RT1 RT2 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 Using log k and 1/Tlog k1 = log A - Ea Equation 1 log k2 = log A - Ea Equation 2 2.303RT1 2.303RT2
  • 27. 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 At 600K At 650K ln k1 = ln A - Ea Using ln k and 1/T ln k2 = ln A - Ea Equation 2 Equation 1 RT1 RT2 Equation 2 – Equation 1 ln k2 – lnk1 = - Ea + Ea RT2 RT1 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 Using log k and 1/Tlog k1 = log A - Ea Equation 1 log k2 = log A - Ea Equation 2 2.303RT1 2.303RT2 Equation 2 – Equation 1 log k2 – logk1 = - Ea + Ea 2.303RT2 2.303RT1
  • 28. 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 At 600K At 650K ln k1 = ln A - Ea Using ln k and 1/T ln k2 = ln A - Ea Equation 2 Equation 1 RT1 RT2 Equation 2 – Equation 1 ln k2 – lnk1 = - Ea + Ea RT2 RT1 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 Using log k and 1/Tlog k1 = log A - Ea Equation 1 log k2 = log A - Ea Equation 2 2.303RT1 2.303RT2 Equation 2 – Equation 1 log k2 – logk1 = - Ea + Ea 2.303RT2 2.303RT1
  • 29. AcknowledgementsThanks to source of pictures and video used in this presentationThanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/Prepared by Lawrence KokCheck out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com

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