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- 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 <ul><li>We have seen that, generally, as temperature increases so does the reaction rate. </li></ul><ul><li>But the powers in the rate law do NOT change </li></ul><ul><li>This means that k is temperature dependent . </li></ul>k Temperature We need a microscopic model to explain this
- 4. <ul><li>Three conditions must be met at the molecular level if a reaction is to occur: </li></ul><ul><li>The molecules must collide ; </li></ul><ul><li>They must be positioned so that the reacting groups are together in a transition state between reactants and products; </li></ul><ul><li>The collision must have enough energy to form the transition state and convert it into products. </li></ul>Collision Rate Model
- 5. <ul><li>Molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. </li></ul>Right Orientation
- 6. Activation Energy <ul><li>There is a minimum amount of energy required for reaction: the activation energy , E a . </li></ul><ul><li>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. </li></ul>
- 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
- 8. Transition State
- 9. Reaction Coordinate Diagrams <ul><li>Reaction coordinate diagrams help visualize energy changes throughout a process. </li></ul>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. <ul><li>Catalysts speed up reactions by altering the mechanism to lower the activation energy barrier, but they are not consumed. </li></ul>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. <ul><li>Imagine trying to get a bunch of cattle to where you want them to go without any help… </li></ul>“ 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. <ul><li>What if there was someone there to herd the cattle and guide them along a more efficient path? </li></ul>“ Cattle-ists” The end result is the same – but the reaction takes a more efficient path.
- 22. The metal-catalyzed hydrogenation of ethylene
- 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 <ul><li>3) Enzymes — biological catalysts </li></ul>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.
- 25. Enzyme Catalysis
- 26. Enzymes uncatalyzed enzyme catalyzed
- 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? <ul><li>Yes </li></ul><ul><li>No </li></ul>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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