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Entry Task: Dec 4 th Tuesday

Entry Task: Dec 4 th Tuesday. Question: In the reaction A + 2 B ==> 3 C + 4 D, the initial concentration of B was 0.0416 M after 20.2 min the concentration of B was 0.0283 M. What is the average rate of appearance (M/s) of reactant C? You have 5 minutes!. Agenda:.

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Entry Task: Dec 4 th Tuesday

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  1. Entry Task: Dec 4th Tuesday Question: In the reaction A + 2 B ==> 3 C + 4 D, the initial concentration of B was 0.0416 M after 20.2 min the concentration of B was 0.0283 M. What is the average rate of appearance (M/s) of reactant C? You have 5 minutes!

  2. Agenda: • Quickly discuss backside of Rate Rxn 2 • Sign off and discuss Ch. 14 section 2 • Embedded notes and practice Rate Law • HW: Rate Law ws 1

  3. I can… • Explain the meaning of a rate constant and be able to calculate it. • Determine the rate law by comparing the changes in concentration of reactants to the reaction rate. • Use the rate law of a particular reaction to predict the rate of reaction with various concentrations of reactants.

  4. WARNING!!! Black text slides are embedded notes!!

  5. Chapter 14Chemical Kineticssections 2

  6. Two Mathematical Expressions to Describe Reaction Rate A B • Rate = -A/ t = +B/ t • Determined from stoichiometry • Uses both reactants & products • Rate Law; rate =k[A]x • Determined by experimental data • Stoichiometry of equation is irrelevant • Only reactants in rate law

  7. 14.2 The Dependence of Rate on ConcentrationHow does the concentration of the reactant affect the rate of reactions? Increase concentration of reactant the rate of reaction increases vice versa

  8. N2(g) + 2 H2O(l) NH4+(aq) + NO2−(aq) In Table 14.3, notice that if you change the concentration of either reactant you change the rate of reaction.Compare what happened in experiment 1 verse 2- what’s going on with the rate of reaction? Comparing Experiments 1 and 2, when [NH4+] doubles, the initial rate doubles.

  9. N2(g) + 2 H2O(l) NH4+(aq) + NO2−(aq) Compare what happened in experiment 5 verse 6- what’s going on with the rate of reaction? Likewise, comparing Experiments 5 and 6, when [NO2−] doubles, the initial rate doubles.

  10. How is this relationship expressed- mathematically? • This means Rate  [NH4+] Rate  [NO2−] Rate  [NH+] [NO2−] or Rate = k [NH4+] [NO2−] • This equation is called the rate law, and k is the rate constant.

  11. What is a rate law? Explain what k is. Rate = k [NH4+] [NO2−] • The rate of reaction depends on the concentrations of the reactant. This equation is called the rate law, and k is the rate constant.

  12. Look at it this way The number of students is proportional to the number of girls #of students  # of girls Change to be an equality By adding a constant (k) we can!! #of students = k # of girls What is k ? 2

  13. Look at it this way Now that we have the “key” to the relationship we can change ANY of the variables. This relationship is called a rate law #of students = 2 # of girls If there were 345 girls, how many students are there? #of students = 2 (345) 690 students If there were 4250 students, how many girls are there? 2125 girls 4250 = 2 (X)

  14. What information do we need in order to calculate the rate constant? • Rate law for a reaction • Rate for a set of reactant concentrations

  15. Looking at Chemistry-Using Table 14.3, provide the rate constant for experiment 4- SHOW YOUR WORK The rate of reaction is 32.3 x10-7 M/s The concentration of NH4+ is 0.0600M The concentration of NO2- is 0.200M

  16. Looking at Chemistry-Using Table 14.3, provide the rate constant for experiment 4- SHOW YOUR WORK Rate law is 32.3 x10-7 M/s = k (0.0600)(0.200) 32.3 x10-7 M/s = k (0.0600 M)(0.200M) 2.7 x 10-4 M-1s-1

  17. Using the rate constant to find out concentrations or rate of reactions The rate of reaction is 25.3 x10-7 M/s. The concentration of NO2- has not changed. What is the concentration of NH4+? 25.3 x10-7 M/s = 2.7 x 10-4M-1s-1 (X M) (0.200 M) 25.3 x10-7 M/s = x (5.4 x 10-5) 0.0468 M Does it make sense with graph?

  18. Looking at Chemistry-Using Table 14.3, provide the rate constant for experiment 4- SHOW YOUR WORK 25.3 x10-7 M/s = 2.7 x 10-4M-1s-1 (X M) (0.200 M) 25.3 x10-7 M/s = x (5.4 x 10-5) 0.0468 M

  19. Using the rate constant to find out concentrations or rate of reactions What if we doubled the concentration of NH4+ to 0.120 M and NO2- to 0.400M? What is the new rate of reaction? X = 2.7 x 10-4M-1s-1 (0.120 M) (0.400 M) 1.3 x 10-5 M/s= 2.7 x 10-4M-1s-1 (0.048 M)

  20. aA + bB cC + dD The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. Rate = k [A]x[B]y k is the Rate Law Constant x and y are reaction orders determined experimentally, and do not depend on stoichiometric coefficients from balanced equation 13.2

  21. aA + bB cC + dD The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. Rate = k [A]x[B]y Reaction Order reaction is xth order in A reaction is yth order in B reaction is (x +y)th order overall • Reaction order tells how quickly rate • will increase when concentration increases 13.2

  22. Rate of Reaction Orders • Reactions are classified as either zero-order, first-order, second-order, or mixed order (higher order) reactions. • The rate of chemical reaction and the size of the rate constant (k) is dependent on the “order” of the reaction.

  23. Reaction Order Tells How Changing Reactant Concentration Affects Rate 0Order; rate =k[A]0 ; double [A], rate same 1stOrder; rate = k[A]1; double [A], rate doubles 2ndOrder; rate =k[A]2 ; double [A]; rate quadruple 3rdOrder; rate =k[A]3 ; double [A]; rate 8X

  24. Reaction OrderProvide the general form of the rate law expression Rate = k [reactant 1]m [reactant 2]n

  25. What do the exponents mean? Rate = k [reactant 1]m [reactant 2]n The m and n are the reaction orders, and their sum is the overall reaction order.

  26. What order is this reaction? Look at and compare experiment 1 and 2. In respect to NH4+: When this reactant double the rate doubled so this is 1st order. So we write it as: Rate = k[NH4+]1

  27. What order is this reaction? Look at and compare experiment 5 and 6. In respect to NO2-: When this reactant double the rate doubled so this is 1st order. So we write it as: Rate = k[NO2-]1.

  28. What does it mean by first power? Explain. • A rate law shows the relationship between the reaction rate and the concentrations of reactants. • The exponents tell the order of the reaction with respect to each reactant. • This reaction is First-order in [NH4+] First-order in [NO2−]

  29. What order is this reaction? What about the reaction (Overall order)? So we write it as: Rate = k [NH4+]1 [NO2-]1 overall is 2nd order

  30. Units of Rate Constants The units of the rate constant depend on the overall reaction order of the rate law. For example, in a reaction that is second order overall, the units of the rate constant must satisfy

  31. Units of Rate Constants Unit of rates = (unit of rate constant)(units of concentration)2 Units of rate is M/s Units of concentration = M2 Units of rate constant = M-1s-1 This is so you – won’t get confused- when you see the label.

  32. Using Initial Rates to Determine Rate Laws (IMPORTANT)What is a zero order reaction? If a reaction is zero order in a particular reactant, then changing the concentration will not change the rate.

  33. What is a first order reaction? If a reaction is first order in a reactant, changes in concentration of that substance will produce proportional changes in the rate. Thus doubling the concentration, doubles the rate.

  34. What is a second order reaction? If a reaction is second order in a reactant, doubling the concentration will increase the rate by a factor of 22 = 4, tripling the concentration will increase the rate by a factor of 32 = 9.

  35. STRONGLY RECOMMEND that you go through the Sample Exercise 14.4, then belowSHOWhow to do the practice exercise: The initial rate of a reaction A + B C was measured for several different starting concentrations of A and B, and the results are as follows: Using these data, determine (a)the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [A] = 0.050 M and [B] = 0.100 M.

  36. STRONGLY RECOMMEND that you go through the Sample Exercise 14.4, then belowSHOW how to do the practice exercise: Using these data, determine (a) the rate law for the reaction Exp. 1 vs. Exp. 2, we doubled the concentration of B and there was NO CHANGE in the rate. This means that B is zero order [B]0

  37. STRONGLY RECOMMEND that you go through the Sample Exercise 14.4, then belowSHOW how to do the practice exercise: Using these data, determine (a) the rate law for the reaction Exp. 1 vs. Exp. 2, we doubled the concentration of B and there was NO CHANGE in the rate. This means that B is zero order [B]0 Exp. 2 vs. Exp. 3, we doubled the concentration of A and the rate quadrupled or 22 . This means that A is 2nd order [A]2 a) Rate = k[A]2[B]0

  38. STRONGLY RECOMMEND that you go through the Sample Exercise 14.4, then belowSHOW how to do the practice exercise: Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [A] = 0.050 M and [B] = 0.100 M. Rate = k[A]2[B]0 or k[A]2

  39. STRONGLY RECOMMEND that you go through the Sample Exercise 14.4, then belowSHOW how to do the practice exercise: Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [A] = 0.050 M and [B] = 0.100 M. Rate = k[A]2 k = 4.0  103M1 s1 Rate = (4.0  103M1 s1) (0.050 M)2 c) Rate = 1.0  105M/s

  40. STRONGLY RECOMMEND that you go through the Sample Exercise 14.4, then belowSHOW how to do the practice exercise: A particular reaction was found to depend on the concentration of hydrogen ions [H+]. The initial rates varied as a function of [H+] as follows: (a)What is the order of the reaction in [H+]? (b)Predict the initial reaction rate when [H+] is 0.400 M

  41. STRONGLY RECOMMEND that you go through the Sample Exercise 14.4, then belowSHOW how to do the practice exercise: A particular reaction was found to depend on the concentration of hydrogen ions [H+]. The initial rates varied as a function of [H+] as follows: (a)What is the order of the reaction in [H+]? (b)Predict the initial reaction rate when [H+] is 0.400 M As the [H+] increases the rate of reaction decreases Doubling the [H+] the rate drops by a factor of 2. a) [H+]-1

  42. STRONGLY RECOMMEND that you go through the Sample Exercise 14.4, then belowSHOW how to do the practice exercise: A particular reaction was found to depend on the concentration of hydrogen ions [H+]. The initial rates varied as a function of [H+] as follows: (a)What is the order of the reaction in [H+]? (b) Predict the initial reaction rate when [H+] is 0.400 M At 0.200 M with a rate of 1.6 x 10-7. The next trend is 0.400 so cut the rate in half. b) 0.80 x 10-7M/s

  43. 14.17 Consider the following reaction: CH3Br (aq) + OH- (aq)  CH3OH (aq) + Br- (aq) The rate law for this reaction is first order in CH3Br and first order in OH-. When [CH3Br] is 5.0 x 10-3 M and [OH-] is 0.050 M, the reaction rate at 298K is 0.0432 M/s. Provide the reaction law: Rate = k[CH3Br]1[OH-]1

  44. 14.17 Consider the following reaction: CH3Br (aq) + OH- (aq)  CH3OH (aq) + Br- (aq) The rate law for this reaction is first order in CH3Br and first order in OH-. When [CH3Br] is 5.0 x 10-3 M and [OH-] is 0.050 M, the reaction rate at 298K is 0.0432 M/s. a) What is the value of the rate constant? Rate = k[CH3Br]1[OH-]1 Rate 0.0432 M/s = 1. x 102M -1s-1 k= = [CH3Br]1[OH-]1 [5.0 x 10-3]1[0.050]1

  45. 14.17 Consider the following reaction: CH3Br (aq) + OH- (aq)  CH3OH (aq) + Br- (aq) The rate law for this reaction is first order in CH3Br and first order in OH-. When [CH3Br] is 5.0 x 10-3 M and [OH-] is 0.050 M, the reaction rate at 298K is 0.0432 M/s. b) What would happen to the rate if the concentration of OH- were tripled? If you triple [OH-], you’ll triple the rate of reaction

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