1. Copyright Sautter 2003

2. REACTION RATES What does “rate” mean ? Can you think of an everyday measurement of rate ? How about a car speed in miles per hour! How about water flow in gallons per minute! How about an audience entering a stadium in people per hour! What do all these measures have in common?

3. REACTION RATES (cont’d) Each one contains a time unit and the word “per”.”Per” means divide! How then is the rate value set up mathematically? In each case an amount unit (miles, gallons or people) is divided by a time unit (hours, minutes or possibly seconds). Generally then rates are ratios (divisions) with an amount divided by time. Rate =  amount /  time

4. REACTION RATES (cont’d) In chemistry, the amount unit may vary but is often in moles, moles per liter (molarity), grams or even liters. Rates of chemical reactions then, are most often measured as moles per second, molarity per second. Rates of reaction can be measured in terms of reactants consumed ( negative rates) or products produced (positive rates).

5. REACTION RATES (cont’d) The progress of reactions, as reactants are converted to products are often represented by a graph. When interpreting the graph it is important to recognize whether the amount of products or reactants is being measured ! If the reactants are being measured the direction of the reaction and the sign of the rate will be different than if the products are measured. Study the following graphs and see if you can determine the direction of the reaction and the appropriate rates.

6. RXN RATE? FORWARD OR REVERSE RXN? RXN RATE? FORWARD OR REVERSE? RATE = 0 RXN RATE? FORWARD OR REVERSE? REACTION RATES TIME MOLESAMOLESA MOLESAMOLESA MOLESAMOLESA MOLESAMOLESA A  B + C FORWARD RXN RATE = CONSTANT FORWARD RXN RATE = VARIABLE REVERSE RXN RATE = VARIABLE SLOPE OF A TANGENT LINE TO AN AMOUNT VS. TIME GRAPH = RATE GRAPH 1GRAPH 2 GRAPH 3GRAPH 4

7. REACTION RATES (cont’d) In graph 1, the slope of the graph (rise divided by run) equals zero. The rate is therefore zero and no net reaction is occurring. In graph 2, as time goes on the moles of reactant A decrease, therefore reactant is being consumed and the reaction is progressing in the forward direction. The slope of the line is constant (rise divided by run is the same at all points) and negative with respect to reactant A (it is being consumed) and therefore the rate is negative (sloping downward to the right) and constant.

8. REACTION RATES (cont’d) In graph 3, as time goes on the moles of reactant A decrease, therefore reactant is being consumed and the reaction is progressing in the forward direction. The slope of the line is not constant (rise divided by run is not the same at all points) but is negative (sloping downward to the right) and therefore the rate is negative with respect to reactant A (it is being consumed) and variable as the changing slopes of the tangent lines show.

9. REACTION RATES (cont’d) In graph 4, as time goes on the moles of reactant A increase, therefore reactant is being formed and the reaction is progressing in the reverse direction. The slope of the line is not constant (rise divided by run is not the same at all points) and positive (sloping upward to the right) and therefore the rate is positive with respect to reactant A (it is being formed) and variable as the changing slopes of the tangent lines show.

10. FACTORS THAT EFFECT REACTION RATES FIVE FACTORS CAN CHANGE THE RATE OF REACTION. CAN YOU NAME THEM? THEY ARE : (1) CONCENTRATION OF REACTANTS (REMEMBER THAT CONCENTRATION IN GASES IS DIRECTLY RELATED TO PRESSURE, PV =nRT) (2) TEMPERATURE (3) SURFACE AREA OF THE REACTANTS (4) CATALYSTS (5) THE NATURE OF THE REACTANTS

11. FACTORS THAT EFFECT REACTION RATES (CONT’D) HOW DOES THE CONCENTRATION (PRESSURE) OF THE REACTANTS EFFECT REACTON RATE? AS CONCENTRATION INCREASES, REACTON RATE INCREASES. WHY? THE ANSWER LIES IN TWO FUNDAMENTAL CONCEPTS IN CHEMISTRY. THE KINETIC MOLECULAR THEORY AND THE COLLISION THEORY OF REACTION! CAN YOU EXPLAIN THESE TWO IDEAS?

12. FACTORS THAT EFFECT REACTION RATES (CONT’D) A PRIMARY CONCEPT OF THE KINETIC MOLECULAR THEORY IS THAT MOLECULES ARE IN CONSTANT RANDOM MOTION AND THAT AS TEMPERATURE INCREASES THE AVERAGE SPEED (KINETIC ENERGY) OF THOSE MOLECULES INCREASES. THE COLLISION THEORY OF REACTION TELLS US THAT MOLECULES MUST FIRST COLLIDE WITH EACHOTHER BEFORE THEY CAN REACT. IT ALSO SAYS THAT INCREASING THE NUMBER OF COLLISIONS PER SECOND (FREQUENCY) AND INCREASING THE ENERGY OF COLLISION WILL INCREASE REACTION RATE.

13. RANDOM MOTION OF GAS MOLECULES MOLECULES MOVE IN STRAIGHT LINES WITH NO PATTERN AS TEMPERATURE INCREASES THEIR MOTION BECOMES MORE ENERGETIC (RAPID)

14. HOW CONCENTRATION EFFECTS REACTION RATES AS MORE MOLECULES OCCUPY A GIVEN SPACE (CONCENTRATION INCREASES) COLLISIONS BETWEEN MOLECULES BECOME MORE FREQUENT AND REACTION RATE INCREASES. RATE OF REACTION IS DIRECTLY PROPORTIONAL TO THE CONCENTRATION OF THE REACTANTS

15. HOW CONCENTRATION EFFECTS REACTION RATES (THE RATE EQUATION) RATE ~ CONCENTRATION PROPORTIONALITIES CAN BE MADE INTO EQUALITIES BY USING A CONSTANT FOR EXAMPLE: FEET CAN ALWAYS BE CONVERTED TO INCHES BY USING THE NUMBER 12. (INCHES = FEET x 12) RATE = A CONSTANT x CONCENTRATION HOWEVER, ALL EQUATIONS ARE NOT LINEAR (FIRST POWER), SOME ARE SQUARES OR CUBES OR ETC.

16. HOW CONCENTRATION EFFECTS REACTION RATES (THE RATE EQUATON) THEREFORE OUR EQUATION MAY BE WRITTEN AS: RATE = CONSTANT x CONCENTRATIN RAISED TO SOME POWER OR RATE = k x [A] n k = A CONSTANT CALLED THE SPECIFIC RATE CONSTANT (IT IS CONSTANT FOR A SPECIFIC REACTION AT A SPECIFIC TEMPERATURE) [A] = THE CONCENTRATION OF REACTANT A IN MOLES PER LITER (BRACKETS MEAN IN MOLES PER LITER) n = THE POWER TO WHICH CONCENTRATION MUST BE RAISED (ALSO CALLED REACTION ORDER)

17. HOW CONCENTRATION EFFECTS REACTION RATES (REACTION ORDER) REACTIONS WITH RATE EQUATIONS HAVING n = 0 ARE ZERO ORDER REACTIONS. THOSE WITH n = 1 ARE FIRST ORDER AND THOSE WITH n = 2 ARE SECOND ORDER. IN ZERO ORDER REACTIONS, CHANGING THE CONCENTRATION OF THE REACTANT HAS NO EFFECT ON THE RATE. IN FIRST ORDER REACTIONS, RATE CHANGES ONE FOR ONE WITH CONCENTRATION CHANGE. FOR EXAMPLE, DOUBLING CONCENTRATION DOUBLES THE RATE.

18. HOW CONCENTRATION EFFECTS REACTION RATES (REACTON ORDER) IN SECOND ORDER REACTIONS, RATE CHANGES RELATIVE TO THE SQUARE OF THE CONCENTRATION CHANGE. FOR EXAMPLE, DOUBLING THE CONCENTRATION OF THE REACTANT RESULTS IN THE RATE INCREASING TIMES. THIS KNOWLEDGE OF HOW RATE CHANGES WITH CONCENTRATION DEPENDING ON THE ORDER LETS US FIND REACTION ORDERS BY AN EXPERIMENTAL PROCESS CALLED “METHOD OF INITIAL RATES” REACTION ORDERS MUST BE DETERMINED EXPERIMENTALLY. THEY CAN NEVER BE DETERMINED FROM THE CHEMICAL EQUATION.

19. HOW CONCENTRATION EFFECTS REACTION RATES (INITIAL RATES) USING THE METHOD OF INITIAL RATES REQUIRES THAT A REACTION BE RUN AT SERIES OF DIFFERENT STARTING CONCENTRATIONS AND THE RATE BE DETERMINED FOR EACH. GIVEN THE FOLLOWING DATA FOR THE REACTION A  B + C (TABLE 1) EXPT [A]RATE (M/SEC) 1 1 x 10 -3 4 x 10 -1 2 2 x 10 -3 8 x 10 -1 3 4 x 10 -3 16 x 10 -1 AS CONCENTRATION OF A DOUBLES, RATE DOUBLES. THE REACTION IS FIRST ORDER IN REACTANT A RATE = k[A] 1 OR RATE = k[A]

20. HOW CONCENTRATION EFFECTS REACTION RATES (INITIAL RATES) FOR THE REACTION: A + B  C + D (TABLE 2) 1 1 x 10 -3 1 x 10 –3 4 x 10 -1 2 2 x 10 -3 1 x 10 -3 8 x 10 -1 3 1 x 10 -3 2 x 10 -3 16 x 10 –1 USING EXPT 1 AND 2, [A] DOUBLES AND [B] IS CONSTANT. THE DOUBLING OF THE RATE IS THEREFORE CAUSED BY REACTANT A AND THE ORDER WITH RESPECT TO A IS FIRST. USING EXPT 1 AND 3, [A] IS CONSTANT AND [B] IS DOUBLED. THE FOUR TIMES RATE INCREASE IS THEREFORE CAUSED BY REACTANT B AND THE ORDER WITH RESPECT TO B IS SECOND. RATE = k[A] 1 [B] 2 OR RATE = k[A][B] 2

21. HOW CONCENTRATION EFFECTS REACTION RATES (RATE CONSTANTS) FROM INITIAL RATES DATA TABLE 1, RATE = k [A], k CAN BE CALCULATED BY SUBSTITUTING ANY DATA SERIES INTO THE RATE EQUATION, FOR EXAMPLE, FROM EXPT 1 ON TABLE 1 [A] = 1 x 10 –3 M, RATE = 4 x 10 –1 M/SEC 4 x 10 –1 M/SEC = k (1 x 10 –3 M ) k = 1 x 10 2 SEC -1 OR 1 x 10 2 / SEC

22. HOW CONCENTRATION EFFECTS REACTION RATES (RATE CONSTANTS) FROM INITIAL RATES DATA TABLE 2, RATE = k [A] x [B] 2, k CAN BE CALCULATED BY SUBSTITUTING ANY DATA SERIES INTO THE RATE EQUATION, FOR EXAMPLE, FROM EXPT 1 ON TABLE 2 [A] = 1 x 10 –3 M, [B] = 1 x 10 –3 M RATE = 4 x 10 –1 M/SEC 4 x 10 –1 M/SEC = k (1 x 10 –3 M ) x (1 x 10 –3 M ) 2 k = 4 x 10 5 M -1 SEC -1 OR 4 x 10 5 / M x SEC

23. CONCENTRATION OF REACTANT AND REACTION TIME RATE =  CONCENTRATION /  TIME AND RATE = k [A] n, THEREFORE CONCENTRATION, TIME, THE SPECFIC RATE CONSTANT AND THE REACTION ORDER ARE RELATED TO EACHOTHER! THE EXACT RELATIONSHIP IS DEPENDENT ON THE REACTION ORDER AND CAN BE DETERMINED USING CALCULUS (WHICH WE WILL NOT SHOW HERE).

24. CONCENTRATION OF REACTANT AND REACTION TIME ZERO ORDER REACTIONS: CONCENTRATION VS TIME RELATIONSHIPS FOR THE REACTION A  B + C [A] i = CONCENTRATION OF REACTANT AT ANY INSTANT (TIME) DURING THE REACTION [A] 0 = ORIGINAL CONCENTRATION OF REACTANT k = SPECIFIC RATE CONSTANT t = TIME [A] i = - k t + [A] 0, (TIME UNITS FOR THE RATE CONSTANT AND THE TIME VALUES MUST BE THE SAME)

25. CONCENTRATION OF REACTANT AND REACTION TIME ZERO ORDER REACTION TIME CONCACONCA [A] i = - k t + [A] 0 SLOPE = k [A] 0 = VERTICAL INTERCEPT [A] 0

26. CONCENTRATION OF REACTANT AND REACTION TIME FIRST ORDER REACTIONS: CONCENTRATION VS TIME RELATIONSHIPS FOR THE REACTION A  B + C [A] i = CONCENTRATION OF REACTANT AT ANY INSTANT (TIME) DURING THE REACTION [A] 0 = ORIGINAL CONCENTRATION OF REACTANT k = SPECIFIC RATE CONSTANT t = TIME ln [A] i = - k t + ln [A] 0, (TIME UNITS FOR THE RATE CONSTANT AND THE TIME VALUES MUST BE THE SAME)

27. ABOUT LOG AND LN VALUES A LOGIRIMITH IS A POWER. LOG REFERS TO A POWER OF 10. FOR EXAMPLE, THE LOG OF 100 IS THE LOG OF 10 2 WHICH IS 2 OR LOG (100) = 2. THE LOG OF 0.001 IS THE LOG OF 10 -3 OR –3. THE LOG OF 237 IS BETWEEN 2 (THE LOG OF 100) AND 3 (THE LOG OF 1000). HERE A CALCULATOR IS NEEDED TO GET LOG (237) = 2.37 OR 10 2.37 =237. BASE e (2.71) ARE IDENTIFIED BY LN. USING THE LN BUTTON ON THE CALCULATOR LN (237) = 5.47 OR 2.71 5.47 = 237.

28. ABOUT LOG AND LN VALUES WHEN NUMBERS HAVING THE SAME BASE ARE MULTIPLIED, THE POWERS ARE ADDED. SO TO WITH LOGIRIMTHS. WHEN NUMBER HAVING THE SAME BASE ARE DIVIDED, THEIR POWER ARE SUBTRACTED. SO TO WITH LOGARITHMS. WHEN LOGS OR LNS ARE ADDED THE RESULT IS THE LOG OF THE MULTIPIED VALUE OF THE NUMBERS. WHEN LOGS OR LNS ARE SUBTRACTED THE RESULT IS THE LOG OF THE DIVIDED VALUE OF THE NUMBERS. ln (A x B) = ln (A) + ln (B), ln (A/B) = ln (A) – ln (B)

29. CONCENTRATION OF REACTANT AND REACTION TIME FIRST ORDER REACTION TIME L C N O N C A ln [A] i = - k t + ln [A] 0 SLOPE = k [A] 0 = ln –1 ( VERTICAL INTERCEPT) Ln [A] 0

30. CONCENTRATION OF REACTANT AND REACTION TIME SECOND ORDER REACTIONS: CONCENTRATION VS TIME RELATIONSHIPS FOR THE REACTION A  B + C [A] i = CONCENTRATION OF REACTANT AT ANY INSTANT (TIME) DURING THE REACTION [A] 0 = ORIGINAL CONCENTRATION OF REACTANT k = SPECIFIC RATE CONSTANT t = TIME 1/ [A] i = + k t + 1/ [A] 0, (TIME UNITS FOR THE RATE CONSTANT AND THE TIME VALUES MUST BE THE SAME)

31. CONCENTRATION OF REACTANT AND REACTION TIME SECOND ORDER REACTION TIME 1 / [A] 1/ [A] i = + k t + 1/ [A] 0 SLOPE = k 1 / [A] 0 = ( VERTICAL INTERCEPT) 1 / [A] 0

32. CONCENTRATION OF REACTANT AND REACTION TIME (HALF LIFE) HALF LIFE IS AN IMPORTANT CONCEPT IN STUDYING REACTION RATES. IT IS THE TIME REQUIRED FOR HALF THE ORIGINAL AMOUNT OF REACTANT TO BE CONSUMED IN THE REACTION. FOR EXAMPLE, STARTING WITH 100 GRAMS OF REACTANT, IT REQUIRES ONE HALF LIFE TO BE REDUCED TO 50 GRAMS. IT REQUIRES ANOTHER EQUAL TIME PERIOD (ANOTHER HALF LIFE) TO BE REDUCED TO 25 GRAMS, AND ANOTHER TO REACH 12.5 GRAMS, ETC. THE CALCULATION OF THE HALF LIFE FOR A REACTION DEPENDS ON THE REACTION ORDER. THE RELATIONSHIP IS DIFFERENT FOR ZERO ORDER, FIRST ORDER AND SECOND ORDER.

33. CONCENTRATION OF REACTANT AND REACTION TIME (HALF LIFE) ZERO ORDER HALF LIFE EQUATION: BY SUBSTITUTING [A] o / 2 FOR [A] i INTO THE EQUATION: [A] i = - k t + [A] 0 THE EQUATION FOR THE HALF LIFE TIME VALUE t 1/2 CAN BE FOUND AS: t 1/2 = [A] o / (2 x k) FIRST ORDER HALF LIFE EQUATION: BY SUBSTITUTING [A] o / 2 FOR [A] i INTO THE EQUATION: ln [A] i = - k t + ln [A] 0 THE EQUATION FOR THE HALF LIFE TIME VALUE t 1/2 CAN BE FOUND AS: t 1/2 = ln 2 / k = 0.693 / k

34. CONCENTRATION OF REACTANT AND REACTION TIME (HALF LIFE) SECOND ORDER HALF LIFE EQUATION: BY SUBSTITUTING [A] o / 2 FOR [A] i INTO THE EQUATION: 1 / [A] i = + k t + 1 / [A] 0 THE EQUATION FOR THE HALF LIFE TIME VALUE t 1/2 CAN BE FOUND AS: t 1/2 = 1 / ([A] o x k) 100 g 50 g 25 g 12.5 g 1 HALF LIFE 1 HALF LIFE 1 HALF LIFE 3 HALF LIFES

35. SOLVING CONCENTRATION – TIME PROBLEMS PROBLEM: A first order reaction, A  B + 2C, with a rate constant of 3.2 x 10 -3 sec -1 has an initial concentration of 0.0250 M of A. What is the concentration of A after 20.0 seconds? SOLUTION: The first relationship between concentration and time is: ln [A] i = - k t + ln [A] 0, [A] o = 0.0250 M, k = 3.2 x 10 -3 sec -1, t = 20.0 seconds SOLUTION: ln [A] i = (-3.2 x10 -3 )(20) + ln (0.0250), ln [A] i = -3.75 [A] i = ln -1 (-1.66) = 0.0234 M What is the half of this reaction ? t 1/2 = ln 2 / k = 0.693 / k t 1/2 = 0.693 / ( 3.2 x 10 -3 sec -1 ) = 216 seconds

36. SOLVING CONCENTRATION – TIME PROBLEMS PROBLEM: A first order reaction, A  B + 2C, with a rate constant of 5.6 x 10 -2 sec -1. If the original concentration of A is 0.50 M, what time is required for the concentration of A to become 0.0625 M? SOLUTION: Dividing 0.50 M by 0.0625 M we get 8. This is 2 3 therefore three half lifes have elapsed. (100%  50%  25%  12.5% or 0.5 M  0.25 M  0.125 M  0.0625 M) t 1/2 = log 2 / k = 0.693 / k = 0.693 / 5.6 x 10 -2 sec -1 = 12.4 sec multiplying 3 half lifes x 12.4 sec = 37.1 seconds

37. SOLVING CONCENTRATION – TIME PROBLEMS PROBLEM: A first order reaction, A  B + 2C, with a rate constant of 5.6 x 10 -2 sec -1. What time is required for the concentration of A to be reduced to 27% of the original ? SOLUTION: [A] o = 1 x [A] o, [A] i = 0.27 [A] o ln [A] i = - k t + ln [A] 0, t = -( ln [A] I – ln[A] o ) / k or t = ln ( [A] o / 0.27 [A] o ) / k = ln (1/0.27) / k t = 1.31/ (5.6 x 10 -2 sec –1 ) = 23.4 seconds CHECK: t 1/2 = 0.693 / k = 0.693 / (5.6 x 10 -2 sec -1 ) = 12.4 sec 27% means approximately 2 half lifes have elapsed or 2 x 12.4 seconds = 24.8 seconds. Our answer is only slightly less, 23.4 seconds and is correct !

38. TEMPERATURE & REACTION RATE AT ANY TEMPERATURE THE MOLECULES IN A SYSTEM HAVE A DISTRIBUTION OF KINETIC ENERGIES (SIMILAR TO THE DISTRIBUTION OF THE SPEEDS OF CARS ON A HIGHWAY). AS THE TEMPERATURE INCREASES, THE AVERAGE KINETIC ENERGY OF THE MOLECULES INCREASE (THEY MOVE FASTER AT HIGHER TEMPERATURES) AND THEREFORE COLLIDE WITH EACHOTHER MORE FREQUENTLY AND HIT HARDER WHEN THEY DO COLLIDE. AS A RESULT, REACTION RATE INCREASES WITH TEMPERATURE.

39. KINETIC ENERGY DISTRIBUTION CURVE TEMPERATURE = T 1 KINETIC ENERGY NUMBEROFMOLESNUMBEROFMOLES LOW ENERGY MOLECULES HIGH ENERGY MOLECULES AVERGE ENERGY MOLECULES

40. KINETIC ENERGY DISTRIBUTION CURVE TEMPERATURE = T 1 TEMPERATURE = T 2 KINETIC ENERGY NUMBEROFMOLESNUMBEROFMOLES TEMP 2 > TEMP 1 AT HIGHER TEMPERATURES, MOLECULES HAVE HIGHER ENERGIES ON AVERAGE

41. TEMPERATURE & REACTION RATE IN ORDER TO REACT, MOLECULES MUST COLLIDE WITH SUFFICIENT ENERGY. THIS MININIUM ENERGY FOR REACTION IS CALLED ACTIVATION ENERGY. AS THE TEMPERATURE OF A SYSTEM IS INCREASED, THE NUMBER OF MOLECULES WITH THE NECESSARY ENERGY FOR REACTION (THE ACTIVATION ENERGY) INCREASES.

42. KINETIC ENERGY DISTRIBUTION CURVE TEMPERATURE = T 1 TEMPERATURE = T 2 KINETIC ENERGY NUMBEROFMOLESNUMBEROFMOLES ACTIVATION ENERGY MOLECULES WITH SUFFICIENT ENERGY TO REACT AT T 1 AS TEMPERATURE, REACTION RATE MOLECULES WITH SUFFICIENT ENERGY TO REACT AT T 2

43. TEMPERATURE & REACTION RATE THE ENERGY CHARACTERISTICS OF A CHEMICAL REACTION CAN BE SHOWN ON A POTENTIAL ENERGY DIAGRAM. THIS GRAPH SHOWS THE ENERGY STATE OF THE SYSTEM AS REACTANTS PROCEED THROUGH THE ACTIVATED COMPLEX TO FORM THE PRODUCTS. THE ACTIVATED COMPLEX IS THE INTERMEDIATE STATE (MOLECULAR FORM) WHICH REACTANTS GO THROUGH AS THEY CONVERT INTO THE PRODUCTS. THE ACTIVATION ENERGY IS THE ENERGY REQUIRED TO FORM THE INTERMEDIATE ACTIVATED COMPLEX MOLECULE.

44. POTENTIAL ENERGY DIAGRAM REACTION COORDINATE POTETIALENERGYPOTETIALENERGY REACTANTS PRODUCTS ACTIVATED COMPLEX A + B AB  H ACTIVATION ENERGY EXOTHERMIC  H = (-) ENERGY IS RELEASED

45. TEMPERATURE & REACTION RATE A QUANTITATIVE RELATIONSHIP BETWEEN TEMPERATURE, REACTION RATE AND ACTIVATION ENERGY WAS DISCOVERED BY A SWEDISH CHEMIST NAMED ARRENHIUS IN THE LATE 1800s. THIS EQUATION IS: k = A x e E/RT k = SPECIFIC RATE CONSTANT A = A PROPORTIONALITY CONSTANT e = BASE e (2.71) E OR E a = ACTIVATION ENERGY ( in joules/ mole) T = KELVIN TEMPERATURE (Kelvin) R = UNIVERSAL GAS CONSTANT (8.31 joules /mole K)

46. TEMPERATURE & REACTION RATE THE ARRENHIUS EQUATION IS OFTEN EXPRESSED IN TERMS OF LOGS RATHER THAN IS EXPONENTIAL FORM. k IS THE SPECIFIC RATE CONSTANT AND AS IT INCREASES, REACTION RATE INCREASES. ln k 2 – ln k 1 = E a / R [(1/T 1 ) – (1/T 2 )] OR ln (k 2 / k 1 ) = (E a / R) [(1/T 1 ) – (1/T 2 )] FREQUENTLY BASE 10 LOG ARE USED: log ( k 2 / k 1 ) = (E a / 2.303 x R) [(1/T 1 ) – (1/T 2 )]

47. TEMPERATURE & REACTION RATE SAMPLE PROBLEM: The specific rate constant for A  B + C is 3.5 x 10 -5 at 200 o C. What is the rate constant at 300 o C if the activation energy is 40.2 KJ/ mole? SOLUTION: k 1 = 3.5 x 10 -5, T 1 = (200 + 273) = 473K k 2 = ?, T 2 = (300 + 273) = 573 K E a = 40.2 KJ / mole = 40.2 x 10 3 joules / mole ln k 2 – ln k 1 = E a / R [(1/T 1 ) – (1/T 2 )] ln k 2 – ln ( 3.5 x 10 -5 ) = 40.2 x 10 3 / 8.31 [(1/473) – (1/573)] ln k 2 = 1.78 – (-1.03) = 2.81, ln –1 (2.81) = 16.6

48. TEMPERATURE & REACTION RATE SAMPLE PROBLEM: At 25 0 C the rate constant for a certain reaction is 3.0 x10 -4 sec -1. At 40 0 C the constant is 6.0 x 10 -3 s -1. What is the activation energy in KJ per mole? SOLUTION: ln (k 2 / k 1 ) = (E a / R) [(1/T 1 ) – (1/T 2 )] E a = R x ln((k 2 / k 1 ) / [(1/T 1 ) – (1/T 2 )] T 1 = (25 + 273) = 298 K, k 1 = 3.0 x10 -4 sec -1 T 2 = (40 + 273) = 313 K, k 2 = 6.0 x 10 -3 s -1 E a = 8.31 x ln(6.0 x 10 -3 /3.0 x10 -4 ) / [(1/298) –(1/313)] E a = 154,800 joules/ mole = 154.8 KJ/ mole

49. SURFACE AREA OF REACTANTS AND REACTION RATE AS THE SURFACE AREA OF REACTANTS EXPOSED TO EACHOTHER INCREASES, REACTION RATE INCREASES. REACTANT MATERIALS WHICH ARE AT THE INTERIOR OF A SUBSTANCE, ONCE PULVERIZED, NO LONGER HAVE TO WAIT TO COME INTO CONTACT WITH THE OTHER REACTING COMPONENT AND THEREFORE REACTION IS MORE IMMEDIATE. A COMMON EXAMPLE OF THIS PHENOMENIA IS A DUST EXPLOSION

50. SURFACE AREA & REACTION RATE REACTANT THAT IS INTERIOR CANNOT BE ATTACKED UNTIL THE EXTERIOR REACTANT IS CONSUMED. THE REACTION RATE IS SLOW!

51. SURFACE AREA & REACTION RATE WHEN SURFACE AREA IS INCREASED MORE REACTANTS ARE EXPOSED TO EACHOTHER SIMULTANEOUSLY AND THE REACTION IS RAPID.

52. REACTION RATES & CATALYSTS CATALYSTS ARE SUBSTANCES THAT INCREASE REACTION RATES WITHOUT BEING CONSUMED IN THE REACTION. THEY PROVIDE A LOWER ENERGY PATHWAY FOR THE REACTION TO FOLLOW. THEY DO SO BY PRODUCING LOWER ENERGY INTERMEDIATE COMPONENTS WHICH MORE READILY CONVERT OF THE PRODUCTS. FROM A POTENTIAL ENERGY STANDPOINT, THEY REDUCE THE ACTIVATION ENERGY NEEDED FOR THE REACTION BY PROVIDING AN ALTERNATE PATHWAY FOR THE REACTION.

53. POTENTIAL ENERGY DIAGRAM REACTION COORDINATE POTETIALENERGYPOTETIALENERGY REACTANTS PRODUCTS A + B AB ACTIVATION ENERGY WITHOUT CATALYST LOWER ACTIVATON ENERGY PATH WITH CATALYST

54. REACTION MECHANISMS ALL REACTIONS, NO MATTER HOW SIMPLE FOLLOW A PARTICULAR REACTION PATHWAY CALLED A MECHANISM. IT IS A SERIES OF STEPS WHICH LEAD TO THE FORMATION OF PRODUCTS. DURING THE PROCESS OFTEN MOLECULES CALLED INTERMEDIATES ARE FORMED AND SUBSEQUENTLY CONSUMED. THE SLOWEST STEP IN THE SERIES OF STEPS THAT MAKE UP THE MECHANISM IS CALLED THE RATE DETERMINING STEP. THE SPEED OF THE RATE DETERMINING STEP DEPENDS ON THE COMPLEXITY OF THE STEP (NUMBER OF MOLECULES INVOLVED CALLED THE MOLECULARITY) AND THE BOND STRENGTHS OF THE REACTING COMPONENTS

55. NO NO 2 O2O2 REACTION MECHANISM FOR; 2 NO + O 2  2 NO 2 STEP I 2 NO  N 2 O 2 (fast) STEP II N 2 O 2 + O 2  2 NO 2 (slow) N 2 O 2 INTERMEDIATE

56. REACTION MECHANISMS UNLIKE OVERALL REACTIONS, THE REACTON ORDERS FOR THE STEPS IN A MECHANISM ARE BASED ON THE NUMBER OF MOLECULES REACTING IN THAT STEP. IN THE REACTION MECHANISM FOR: 2 NO + O 2  2 NO 2 STEP I 2 NO  N 2 O 2 (fast) STEP II N 2 O 2 + O 2  2 NO 2 (slow) SINCE THERE IS ONLY ONE N 2 O 2 AND ONE O 2 IN THE RATE DETERMING STEP (THE SLOW STEP), THE RATE EQUATION FOR THE REACTION IS: RATE = k [N 2 O 2 ] x [O 2 ] THE REACTION IS FIRST ORDER IN BOTH REACTANTS