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General Chemistry II CHEM 152 Week 3
Week 3 Reading Assignment Chapter 13 – Sections 13.5 (temperature), 13.7 (catalysts)
Temperature and Rate We have seen that, generally, as  temperature   increases  so does the reaction rate. But the powers in the rate law do NOT change This means that  k  is temperature dependent . k Temperature We need a microscopic model to explain this
Three conditions must be met at the molecular level if a reaction is to occur: The molecules must  collide ;   They must be  positioned  so that the    reacting groups are together in a  transition   state  between reactants and products;  The collision must have  enough  energy  to   form the transition state and convert    it into products. Collision Rate Model
Molecules must collide with the correct  orientation  and with enough  energy  to cause bond breakage and formation. Right Orientation
Activation Energy There is a minimum amount of energy required for reaction:  the  activation energy ,  E a . Just as a ball cannot get over a hill if it does not have enough energy, a reaction cannot occur unless the molecules possess  sufficient energy  to get over the  activation energy  barrier.
The higher the temperature, the more molecules have energy to overcome the activation energy barrier. Low T High T Extra molecules at high T that exceed the E a Enough Energy
Transition State
Reaction Coordinate Diagrams Reaction coordinate diagrams  help visualize energy changes throughout a process. Reaction Coordinate CH 3 NC CH 3 CN rearrangement of  methyl isonitrile.
Reaction Coordinate Diagrams E Reactants E Products ∆ E reaction E activated complex E a Notice E a  is not related to  ∆E
Find  E a  and  ∆E  in each case.  Identify endothermic and exothermic processes. Your Turn
SOLUTION A key reaction in the upper atmosphere is The  E a(fwd)  is 19 kJ, and the   H rxn  ( ∆E)  for the reaction is -392 kJ. Draw a reaction energy diagram for this reaction, postulate a transition state, and calculate  E a (rev) . Your Turn O 3 ( g ) + O( g )  2O 2 ( g ) transition state E a = 19kJ  H rxn  = -392kJ E a (rev) = (392 + 19)kJ = 411kJ
The Arrhenius Equation ln  k  = ln A - E a /RT where  k   is the rate constant at  T E a  is the activation energy R  is the energy gas constant = 8.3145 J/(mol K) T  is the Kelvin temperature A  is the collision frequency factor Temperature Effects ln k 2 k 1 = E a R - 1 T 2 1 T 1 -
ln  k  = -E a /R (1/T) + ln A Graphical Analysis . Y  =  m  X  +  b
Typical Problems Suppose a chemical reaction has an activation energy of 76 kJ/mol.  By what factor is the rate of reaction at 50  o C increased over its rate at 25  o C? k 2 =rate constant @ 50ºC k 1 =rate const @ 25ºC k 2 /k 1  = 10.7 over 10 times faster! = -76000 J 8.3145 J/mol K ln k 2 k 1 = 2.37 ln k 2 k 1 1 323.15K 1 298.15K -
The decomposition of hydrogen iodide, E a  = ___________ kJ/mol 2HI( g )     H 2 ( g ) + I 2 ( g ) has rate constants of 9.51x10 -9  L/mol*s at 500. K and 1.10x10 -5  L/mol*s at 600. K.  Find  E a .
SOLUTION The decomposition of hydrogen iodide, has rate constants of 9.51x10 -9 L/mol*s at 500. K and 1.10x10 -5  L/mol*s at 600. K.  Find  E a . E a  = 1.76 x 10 5  J/mol  = 176 kJ/mol Answer / 2HI( g )  H 2 ( g ) + I 2 ( g ) ln k 2 k 1 = E a - R 1 T 2 1 T 1 - ln 1.10x10 -5 L/mol*s 9.51x10 -9 L/mol*s 1 600K 1 500K - E a  = - (8.314J/mol*K)
Series of plots of concentra-tion vs. time Initial rates Reaction orders Rate constant ( k ) and actual rate law Integrated rate law (half-life, t 1/2 ) Rate constant and reaction order Activation energy,  E a Plots of concentration vs. time Overview Find  k   at varied T Determine slope of tangent at t 0  for each plot Compare initial rates when [A] changes and [B] is held constant and vice versa Substitute initial rates, orders, and concentrations into general rate law:  rate =  k  [A] m [B] n Use direct, ln or inverse plot to find order Rearrange to linear form and graph Find  k   at varied T
Catalysts speed up reactions by altering the mechanism to lower the activation energy barrier, but  they are not consumed. MnO 2  catalyzes decomposition of H 2 O 2 2 H 2 O 2      2 H 2 O  + O 2 Catalysis Uncatalyzed reaction Catalyzed reaction
Imagine trying to get a bunch of cattle to where you want them to go without any help… “ Cattle-ists” Will they get there very quickly on their own?  Will they take the shortest path do get there?  Is there a way to help?
What if there was someone there to herd the cattle and guide them along a more efficient path? “ Cattle-ists” The end result is the same – but the reaction takes a more efficient path.
The metal-catalyzed hydrogenation of ethylene
Catalytic Converters 13.6 CO + Unburned Hydrocarbons + O 2 CO 2  + H 2 O catalytic converter 2NO + 2NO 2 2N 2  + 3O 2 catalytic converter
Catalysis 3) Enzymes  — biological catalysts Enzymes are specialized organic substances, composed of polymers of amino acids ( proteins ), that act as catalysts to regulate the speed of the many chemical reactions involved in the metabolism of living organisms.
Enzyme Catalysis
Enzymes uncatalyzed enzyme catalyzed
Enzymes Saturation Effects Denaturization
Summary Activity: Fireflies flash at a rate that is temperature dependent.  At 29 ˚C the average firefly flashes at a rate of 3.3 flashes every 10. seconds. At 23 ˚C the average rate is 2.7 flashes every 10. seconds. Use the Arrhenius equation to determine the activation energy (kJ/mol) for the flashing process.
The activation energy of the firefly flashing process is: E a  = ___________ kJ/mol
Do you know the equation for the force of gravity? Yes  No Rotate your group spokesperson for today.  Spokesperson: go to  www.rwpoll.com  –  LOG IN  and enter the session ID. Then – Answer this question:

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D07 abbrev arrhenius and catalysts_alg

  • 1. General Chemistry II CHEM 152 Week 3
  • 2. Week 3 Reading Assignment Chapter 13 – Sections 13.5 (temperature), 13.7 (catalysts)
  • 3. Temperature and Rate We have seen that, generally, as temperature increases so does the reaction rate. But the powers in the rate law do NOT change This means that k is temperature dependent . k Temperature We need a microscopic model to explain this
  • 4. Three conditions must be met at the molecular level if a reaction is to occur: The molecules must collide ; They must be positioned so that the reacting groups are together in a transition state between reactants and products; The collision must have enough energy to form the transition state and convert it into products. Collision Rate Model
  • 5. Molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. Right Orientation
  • 6. Activation Energy There is a minimum amount of energy required for reaction: the activation energy , E a . Just as a ball cannot get over a hill if it does not have enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.
  • 7. The higher the temperature, the more molecules have energy to overcome the activation energy barrier. Low T High T Extra molecules at high T that exceed the E a Enough Energy
  • 9. Reaction Coordinate Diagrams Reaction coordinate diagrams help visualize energy changes throughout a process. Reaction Coordinate CH 3 NC CH 3 CN rearrangement of methyl isonitrile.
  • 10. Reaction Coordinate Diagrams E Reactants E Products ∆ E reaction E activated complex E a Notice E a is not related to ∆E
  • 11. Find E a and ∆E in each case. Identify endothermic and exothermic processes. Your Turn
  • 12. SOLUTION A key reaction in the upper atmosphere is The E a(fwd) is 19 kJ, and the  H rxn ( ∆E) for the reaction is -392 kJ. Draw a reaction energy diagram for this reaction, postulate a transition state, and calculate E a (rev) . Your Turn O 3 ( g ) + O( g ) 2O 2 ( g ) transition state E a = 19kJ  H rxn = -392kJ E a (rev) = (392 + 19)kJ = 411kJ
  • 13. The Arrhenius Equation ln k = ln A - E a /RT where k is the rate constant at T E a is the activation energy R is the energy gas constant = 8.3145 J/(mol K) T is the Kelvin temperature A is the collision frequency factor Temperature Effects ln k 2 k 1 = E a R - 1 T 2 1 T 1 -
  • 14. ln k = -E a /R (1/T) + ln A Graphical Analysis . Y = m X + b
  • 15. Typical Problems Suppose a chemical reaction has an activation energy of 76 kJ/mol. By what factor is the rate of reaction at 50 o C increased over its rate at 25 o C? k 2 =rate constant @ 50ºC k 1 =rate const @ 25ºC k 2 /k 1 = 10.7 over 10 times faster! = -76000 J 8.3145 J/mol K ln k 2 k 1 = 2.37 ln k 2 k 1 1 323.15K 1 298.15K -
  • 16. The decomposition of hydrogen iodide, E a = ___________ kJ/mol 2HI( g )  H 2 ( g ) + I 2 ( g ) has rate constants of 9.51x10 -9 L/mol*s at 500. K and 1.10x10 -5 L/mol*s at 600. K. Find E a .
  • 17. SOLUTION The decomposition of hydrogen iodide, has rate constants of 9.51x10 -9 L/mol*s at 500. K and 1.10x10 -5 L/mol*s at 600. K. Find E a . E a = 1.76 x 10 5 J/mol = 176 kJ/mol Answer / 2HI( g ) H 2 ( g ) + I 2 ( g ) ln k 2 k 1 = E a - R 1 T 2 1 T 1 - ln 1.10x10 -5 L/mol*s 9.51x10 -9 L/mol*s 1 600K 1 500K - E a = - (8.314J/mol*K)
  • 18. Series of plots of concentra-tion vs. time Initial rates Reaction orders Rate constant ( k ) and actual rate law Integrated rate law (half-life, t 1/2 ) Rate constant and reaction order Activation energy, E a Plots of concentration vs. time Overview Find k at varied T Determine slope of tangent at t 0 for each plot Compare initial rates when [A] changes and [B] is held constant and vice versa Substitute initial rates, orders, and concentrations into general rate law: rate = k [A] m [B] n Use direct, ln or inverse plot to find order Rearrange to linear form and graph Find k at varied T
  • 19. Catalysts speed up reactions by altering the mechanism to lower the activation energy barrier, but they are not consumed. MnO 2 catalyzes decomposition of H 2 O 2 2 H 2 O 2  2 H 2 O + O 2 Catalysis Uncatalyzed reaction Catalyzed reaction
  • 20. Imagine trying to get a bunch of cattle to where you want them to go without any help… “ Cattle-ists” Will they get there very quickly on their own? Will they take the shortest path do get there? Is there a way to help?
  • 21. What if there was someone there to herd the cattle and guide them along a more efficient path? “ Cattle-ists” The end result is the same – but the reaction takes a more efficient path.
  • 23. Catalytic Converters 13.6 CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter 2NO + 2NO 2 2N 2 + 3O 2 catalytic converter
  • 24. Catalysis 3) Enzymes — biological catalysts Enzymes are specialized organic substances, composed of polymers of amino acids ( proteins ), that act as catalysts to regulate the speed of the many chemical reactions involved in the metabolism of living organisms.
  • 27. Enzymes Saturation Effects Denaturization
  • 28. Summary Activity: Fireflies flash at a rate that is temperature dependent. At 29 ˚C the average firefly flashes at a rate of 3.3 flashes every 10. seconds. At 23 ˚C the average rate is 2.7 flashes every 10. seconds. Use the Arrhenius equation to determine the activation energy (kJ/mol) for the flashing process.
  • 29. The activation energy of the firefly flashing process is: E a = ___________ kJ/mol
  • 30. Do you know the equation for the force of gravity? Yes No Rotate your group spokesperson for today. Spokesperson: go to www.rwpoll.com – LOG IN and enter the session ID. Then – Answer this question:

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