Solving kinetics problems

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Theory and examples of solving seven types of chemical kinetics problems that may be found on standardized tests.

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Solving kinetics problems

  1. 1. Solving Kinetics Problems Writing Rate Expressions from Balanced Equations Finding Reaction Order from [A]-time Data Finding Reaction Order from Initial Rate Finding k from [A] & Rate Data Calculating ½ Life from k Drawing Reaction Profile from Data Creating Reaction Mechanisms from Rate Law & Finding the Slow Step
  2. 2. Writing Rate Expressions from Balanced Equations     The rate expression is not the rate law It tells us what we are looking for in a rate law experiment Example: H2 (g) + I2 (g) → 2 HI (g) Rate expression could be rate of disappearance of hydrogen It tells you in − Rate = ΔH2 = 1 ΔHI Δt 2 Δt the lab what to measure
  3. 3. Your turn  The equation is − −  Cu (s) + 2 Ag+ (aq) → Cu2+ (aq) + 2 Ag(s) Write at least two expressions by which we could measure the rate Answers: − Rate = ΔCu = 1 ΔAg+ = ΔCu2+ = 1ΔAg Δt 2 Δt Δt 2 Δt
  4. 4. Finding Reaction Order from [A]-time Data   Needed: a chart of concentration of a reactant we want to study vs. time Product: a fast graph of Ln[A] vs time or 1/[A] vs time  If the Ln[A] v time graph is linear, it's first order  If the Ln[A] v time graph is a curve, it's 2 nd order  If [A] v time goes down in a linear fashion, it's zero order (rare)
  5. 5. Your turn    Reactant A changes concentration with time. Here is the data: What is the order of the reaction in A? Hint: find the natural log of each [A] and graph on your graphing calculator
  6. 6. Solution  Plotting LN[A] v. time we get The straight line plot suggests a first order equation Rate = k[A]
  7. 7. Your Turn Butadiene changes concentration with time. Here is the data: Time (s) 0 [Butadiene] 0.01000 Mol/L 1000 1800 2800 3600 4400 0.00625 0.00476 0.00370 0,00313 0.00270 What is the order of the reaction in butadiene? Again, find the natural log of each [A] and graph on your graphing calculator.
  8. 8. Solution Plotting LN[butadiene] vs time we get This is not a straight line as shown by the straight line between the first and last points, so the reaction must be 2nd order.
  9. 9. Finding Reaction Order from Initial Rate     This is a favorite of test writers! You are given concentrations of all reactants and the rate of reaction for each set of conditions You must identify the controls and variables and determine how the rate is affected You are looking for doubling of rate when concentration doubles (1st), or quadrupling of rate when concentration doubles (2 nd)
  10. 10. Your Turn SO2 + O2 → SO3 Given the following data, determine the order of reaction in SO2 and O2 Look for doubling of concentrations with other concentration held constant!
  11. 11. Solution Given the following data, determine the order of reaction in SO2 and O2 In 2 and 1, oxygen concentration doubles while SO2 is held constant
  12. 12. Solution Given the following data, determine the order of reaction in SO2 and O2 In 2 and 1, rate of formation of the trioxide goes from 0.60 to 1.20, also a doubling
  13. 13. Solution Given the following data, determine the order of reaction in SO2 and O2 That means the rate is directly proportional to the concentration of the oxygen gas, so the reaction is first order in O2
  14. 14. Solution Given the following data, determine the order of reaction in SO2 and O2 When we look at Experiments 1 and 3, we see the oxygen concentration is held constant, and the SO 2 concentration is doubling.
  15. 15. Solution Given the following data, determine the order of reaction in SO2 and O2 But at the same time, the rate of trioxide formation goes from 1.2 to 4.8, which is a quadrupling. That is 2 2 times the initial rate, so the rate is going up faster than the concentration.
  16. 16. Solution Given the following data, determine the order of reaction in SO2 and O2 This means that Rate = k[SO2]2 The reaction is 2nd order in SO2
  17. 17. Predicting Concentrations, Rates Now that we have reaction order, let's see if we can fill in the table. The reaction is 1st order in O2 and 2nd order in SO2
  18. 18. Solutions To find the oxygen in Exp 4, we see that the rate is 17% lower than in Exp 1 The reaction is 1st order in O2 and 2nd order in SO2
  19. 19. Solutions And the [SO2] is 33% lower than in Exp 1 That is predictable since Rx is 2nd order The reaction is 1st order in O2 and 2nd order in SO2
  20. 20. Solutions The change in sulfur dioxide accounts for all the rate change, so oxygen is 0.20 M 0.20 M The reaction is 1st order in O2 and 2nd order in SO2
  21. 21. Solutions Now predict the rate of forming trioxide in experiment 5 0.20 M The reaction is 1st order in O2 and 2nd order in SO2
  22. 22. Solutions Both concentrations change. Let's do oxygen first 0.20 M The reaction is 1st order in O2 and 2nd order in SO2
  23. 23. Solutions The oxygen is 50% higher than in Exp 2, so rate should be 50% higher, or 9.0 x 10-3 0.20 M The reaction is 1st order in O2 and 2nd order in SO2
  24. 24. Solutions But SO2 is 17% higher than in Exp 2, so rate from that is 33% higher yet, 1.2 x 10 2 0.20 M 1.2 x 10-2 M/s The reaction is 1st order in O2 and 2nd order in SO2
  25. 25. Finding k from [A] & Rate Data Let's go with a first order reaction we have already looked at: Rate = -0.240[A] -2.49, from the equation of line And k = negative of slope, or 0.240 here
  26. 26. Finding k from [A] & Rate Data But suppose we have the data, but no equation, and know it's first order Look at Exp 1 and 2 and assure yourself that the reaction is first order in A
  27. 27. Finding k from [A] & Rate Data Look at Exp 1 and 3 and see that the reaction is 2nd order in B
  28. 28. Finding k from [A] & Rate Data Look at Exp 3 & 4 and see that the reaction is 2nd order in C
  29. 29. Finding k from [A] & Rate Data Thus the rate law is Rate = k[A][B]2[C]2
  30. 30. Finding k from [A] & Rate Data So solve for k and plug in the numbers from any of the fully known data lines K = Rate [A][B]2[C]2 Rate = k[A][B]2[C]2
  31. 31. Finding k from [A] & Rate Data So solve for k and plug in the numbers from any of the fully known data lines K = 2.85 x 1012 This is harder than anything on a test Rate = k[A][B]2[C]2
  32. 32. Your Turn Find the value of the rate constant: 2 NO (g) + Cl2 (g) → 2NOCl (g) Data: [NO]o mol/L [Cl2]o mol/L Initial Rate Mol/L-min 0.10 0.10 0.18 0.10 0.20 0.36 0.20 0.20 1.45 First you must find the order of the reaction in both reactants, and write the rate law, then plug in to solve for k.
  33. 33. Solution Find the value of the rate constant: The rate law is Rate = k [NO]2[Cl2] So k = (Rate)/[NO]2[Cl2] = 0.18 x 10-3 L2mol-2min-1 [NO]o mol/L [Cl2]o mol/L Initial Rate Mol/L-min 0.10 0.10 0.18 0.10 0.20 0.36 0.20 0.20 1.45 First you must find the order of the reaction in both reactants, and write the rate law, then plug in to solve for k.
  34. 34. Calculating ½ Life from k We'll stick with first order reactions here Remember that t ½ = 0.693 k Suppose the rate constant for a 1 st order reaction is 0.18 x 10-3s-1 What is the halflife? If the initial concentration is 2.0 M, what will it be after the reaction runs for 770 seconds?
  35. 35. Solution Suppose the rate constant for a 1st order reaction is 0.18 x 103 -1 s What is the half-life? If the initial concentration is 2.0 M, what will it be after the reaction runs for 770 seconds? T ½ = 0.693 = 0.693 = 385 s K 0.0018 s-1 Now 770/385 = 2.0 so that's 2 half-lives In the first half-life, the concentration goes to 1.0 M In the second half-life, the concentration goes to 0.50 M
  36. 36. Your Turn A certain first-order reaction is 45.0% complete in 65 s. What are the rate constant and half-life for this process? This is #37 on page 605
  37. 37. Your Turn A certain first-order reaction is 45.0% complete in 65 s. What are the rate constant and half-life for this process? If [A]o = 100.0, then after 65 s, [A] = 55.0. In 1 st order Rxn, LN([A]/[A]o = -kt, LN(55.0/100.0 = -k(65 s) You do the arithmetic, but K = 9.2 x 10-3 s-1 and t ½ = 0.693/k = 75 s This is #37 on page 605
  38. 38. Drawing Reaction Profile from Data This should be a little familiar to you if you recall the diagrams of endothermic and exothermic reaction We just add the activation energy and diagram it like a hill the reactants have to get over by colliding at the right energy and orientation.
  39. 39. Drawing Reaction Profile from Data Draw a reaction energy profile for an endothermic reaction with ΔH = +34 kJ/mol and a forward activation energy of 66 kJ/mol. Calculate the activation energy in the reverse direction.
  40. 40. Drawing Reaction Profile from Data Draw a reaction energy profile for an endothermic reaction with ΔH = +34 kJ/mol and a forward activation energy of 66 kJ/mol. Calculate the activation energy in the reverse direction.
  41. 41. Your turn Draw the reaction energy profile of an exothermic reaction with a forward activation energy of 116 kJ/mol and a ΔH of -225 kJ/mol. Calculate the activation energy in the reverse direction.
  42. 42. Your turn Draw the reaction energy profile of an exothermic reaction with a forward activation energy of 116 kJ/mol and a ΔH of -225 kJ/mol. Calculate the activation energy in the reverse direction. Ea for reverse direction is 341 kJ/mol From UC Davis
  43. 43. Creating Reaction Mechanisms from Rate Law & Finding the Slow Step The rate law gives us a mathematical picture of the initial and time-related concentrations or pressures at a given temperature. However, to control reactions we need to understand how they run. So we derive reaction mechanisms from the rate law, whenever possible.
  44. 44. Example For the reaction H2 (g) + 2 ICl → I2 + 2 HCl the rate law is found to be Rate = k [H2][ICl] What is the most rational two-step mechanism that fits all the information given? First, a termolecular collision is almost never seen. So a two-step mechanism is very reasonable. The rate law implies that both hydrogen and ICl are involved in the slow step.
  45. 45. Example For the reaction H2 (g) + 2 ICl → I2 + 2 HCl the rate law is found to be Rate = k [H2][ICl] The rate law implies that both hydrogen and ICl are involved in the slow step. So we can reasonably say • H2 + ICl → HI + HCl • HI + ICl → I2 + HCl Overall: (slow) (fast) H2 (g) + 2 ICl → I2 + 2 HCl
  46. 46. Another mechanism Here's one I found on YouTube: Organic mechanism Click on the link above

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