1. Chemical Kinetics By: Ms. Buroker

2. Chemical Kinetics Spontaneity is important in determining if a reaction occurs- but it doesn’t tell us much about the speed of the reaction. Kinetics is the area of chemistry that deals with reaction rates.

3. Factors Affecting Reaction Rates 1.) Concentration of reactants 2.) Temperature (Increasing temperature increases the reaction rates- in what direction is a different question.) 3.) Surface Area 4.) Catalysis (Substances that participate but are not changed by the reaction. It is not included in the balanced equation. Inhibitors slow down reactions.)

4. 2N 2 O 5 → 4NO 2 + O 2 Let’s say that at t=0, the concentration of N 2 O 5 is 0.0100mol/L and the concentrations of NO 2 and O 2 are 0.00mol/L. Over time, the concentration of N 2 O 5 will decrease and the concentrations of NO 2 and O 2 will increase. Kinetics deals with the speed at which this takes place!! Reaction Rate = change in concentration of a reactant or product per unit time

5. Reaction Rates Rate =  [A]  t Reaction Rate = concentration of A @ t 2 - concentration of A @t 1 t 2 - t 1 Reaction Rate = change in concentration of a reactant or product per unit time Change  positive means increase & Negative means decrease These brackets mean “concentratio n of”

6. Rate =  [A]  t Reaction Rates Once the rate of one component is determined, others can be determined (i.e. the rate of appearance and disappearance) using stoichiometric ratios. (More on this later)

7. Plotting the concentration of a chemical in a reaction as a function of time is called a Kinetic Curve or Concentration vs. Time Curve. Rates are not constant by change over time; however it is possible to calculate the slope at a given time (Instantaneous rate) by drawing a tangent at that point and calculating the slope (derivative) of the line. Always use a positive value for rate Kinetic Curves or Concentration vs. Time Curves Rate =-  [A]  t

8. 2N 2 O 5 → 4NO 2 + O 2

9. Finding the instantaneous rates at several times during the reaction.

10. 2N 2 O 5 → 4NO 2 + O 2 As 2mol of N 2 O 5 is consumed 4 moles of NO 2 is being formed 1 mole of O2 is being formed → +

11. Rate Laws Chemical reactions are reversible! 2N 2 O 5 → 4NO 2 + O 2 * When enough NO 2 and O 2 have formed, they will react to form N 2 O 5 so the  [N 2 O 5 ] depends on the rate of the forward and reverse reaction. We aren’t going to deal with anything that complicated in this class … so we’ll study on the rate of the FORWARD reaction!! In other words, we’ll study the rate shortly after the reactants are mixed, before the products have significant time to build up.

12. Rate Laws Rate = k[A] n Proportionality Constant Concentration of reactant A Order of the reactant Remember that the products are not being considered- only the rate of the reactants disappearing! “n,” the order of the reaction must be determined experimentally!!

13. Rate Laws Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction. Integrated rate laws express (reveal) the relationship between concentration of reactants and time *The differential rate law is usually just called “the rate law.”

14. Writing a (differential) Rate Law 2 NO(g) + Cl 2 (g)  2 NOCl(g) Experiment[NO] Mol/L [Cl 2 ] Mol/L Rate Mol/L·s 10.250 1.43 x 10 -6 20.5000.2505.72 x 10 -6 30.2500.5002.86 x 10 -6 40.500 11.4 x 10 -6 Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction:

15. Step 1 – Determine the values for the exponents in the rate law: Rate= k[NO] x [Cl 2 ] y Writing a (differential) Rate Law Step 2 – We want to know how the reaction is affected according to each reactant, so we look at the data and choose a situation where the reactants do not change. Then we plug and chug using the Rate law from above to determine the order of the reaction … x and y!!

16. Experiment[NO] Mol/L [Cl 2 ] Mol/L Rate Mol/L·s 10.250 1.43 x 10 -6 20.5000.2505.72 x 10 -6 30.2500.5002.86 x 10 -6 40.500 11.4 x 10 -6 Writing a (differential) Rate Law In Experiments 1 and 3, the concentration of NO is held constant, so we can see how the rate is affected by changing the concentration of Cl 2.

17. Writing a (differential) Rate Law We solve by dividing the two rates where the data is the same: Rate 3/Rate 1 2.86 x 10 -6 = k(0.250) x (0.500) y 1.43 x 10 -6 = k (0.250) x (0.250) y SO… AND … 2 = 2 y therefore y=1 2.86 x 10 -6 = (0.500) y 1.43 x 10 -6 = (0.250) y So what this means is that the reaction is first order with respect to Cl 2  so whatever you do to change [Cl 2 ], the same thing will happen to the rate.

18. Experiment[NO] Mol/L [Cl 2 ] Mol/L Rate Mol/L·s 10.250 1.43 x 10 -6 20.5000.2505.72 x 10 -6 30.2500.5002.86 x 10 -6 40.500 11.4 x 10 -6 Writing a (differential) Rate Law Now, we’re going to look at experiments 3 and 4, where the concentration of Cl 2 is held constant, so we can see how the rate is affected by changing the concentration of NO.

19. Writing a (differential) Rate Law We solve by dividing the two rates where the data is the same: Rate 3/Rate 4 2.86 x 10 -6 = k(0.250) x (0.500) y 11.4 x 10 -6 = k (0.500) x (0.500) y SO… AND ….25 =.5 x therefore y=2 2.86 x 10 -6 = (0.250) x 11.4 x 10 -6 = (0.500) x So what this means is that the reaction is second order with respect to NO  so whatever you do to change [NO], the rate will double.

20. Writing a Rate Law Part 2 – Determine the value for k, the rate constant, by using any set of experimental data:Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s10.2500.250 1.43 x 10 -6 R = k[NO] 2 [Cl 2 ]

21. Writing a Rate Law Part 3 – Determine the overall order for the reaction. R = k[NO] 2 [Cl 2 ] Overall order is the sum of the exponents, or orders, of the reactants 2 + 1= 3  The reaction is 3 rd order

22. Determining Order with Concentration vs. Time data The Integrated Rate Law Zero Order: graphing time vs. concentration produces a linear graph First Order: graphing time vs. (ln concentration) produces a linear graph Second Order: graphing time vs. 1/concentration produces a linear graph

23. Solving an Integrated Rate Law Time (s)[H 2 O 2 ] (mol/L) 01.00 1200.91 3000.78 6000.59 12000.37 18000.22 24000.13 30000.082 36000.050 Problem: Find the integrated rate law and the value for the rate constant, k A graphing calculator with linear regression analysis greatly simplifies this process!!

24. Let’s Graph!! In your graphing calculator, put the data in this way: 1.) time vs. [H 2 O 2 ] 2.) time vs. ln[H 2 O 2 ] 3.) time vs. 1/[H 2 O 2 ] Time (s)[H 2 O 2 ] (mol/L) 01.00 1200.91 3000.78 6000.59 12000.37 18000.22 24000.13 30000.082 36000.050 The graph that gives you the straight line … is our winner!!

25. And the winner is… Time vs. ln[H 2 O 2 ] 1. As a result, the reaction is 1 st order 2. The (differential) rate law is: 3. The integrated rate law is: 4. But…what is the rate constant, k ?

26. Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why?

27. Other Factors Affecting Reaction Rate Temperature affects rate – Higher temp = more collisions – More collisions = faster rate However (there’s always a “however” isn’t there?) – Measured rate is much smaller than the predicted number of collisions – Why? Not all collisions have enough “oomph” to make a reaction go

28. Collision Model 1.) Collisions must have sufficient energy to produce the reaction (must equal or exceed the activation energy). 2.) Colliding particles must be correctly oriented to one another in order to produce a reaction.

29. Activation Energy Arrhenious proposed the concept of an activation energy needed to react Not all collisions occur with the necessary energy to cause bonds to break and then reform time Energy EaEa

30. Endothermic Reactions

31. Exothermic Reactions

32. Activation Energy Arrhenious found that the number of collisions with the correct activation energy increase exponentially with temp (collisions with E a ) = (total collisions)e -Ea/RT However ( ) the observed reaction rate is still smaller than the rate predicted by the above equation This is due to molecular orientation – The molecules must not only hit with the right energy, but the right orientation as well

33. Arrhenious Equation Taking the orientation into account, Arrhenious proposed this equation: k=Ae -E a /RT where A = frequency factor (a measure of number of collisions with correct orientation) Taking the natural log of the equation gives: ln(k) = -(E a /RT) + ln(A)

34. The Arrhenius Equation, Rearranged  Simplifies solving for E a  -E a / R is the slope when ( 1/T ) is plotted against ln(k)  ln(A) is the y-intercept  Linear regression analysis of a table of (1/T) vs. ln(k) can quickly yield a slope  E a = -R(slope)

35. Catalysis Catalyst: A substance that speeds up a reaction without being consumedCatalyst: A substance that speeds up a reaction without being consumed Enzyme: A large molecule (usually a protein) that catalyzes biological reactions.Enzyme: A large molecule (usually a protein) that catalyzes biological reactions. Homogeneous catalyst: Present in the same phase as the reacting molecules.Homogeneous catalyst: Present in the same phase as the reacting molecules. Heterogeneous catalyst: Present in a different phase than the reacting molecules.Heterogeneous catalyst: Present in a different phase than the reacting molecules.

36. Lowering of Activation Energy by a Catalyst

37. Catalysts Increase the Number of Effective Collisions

38. Reaction Mechanism The reaction mechanism is the series of elementary steps by which a chemical reaction occurs.  The sum of the elementary steps must give the overall balanced equation for the reaction  The mechanism must agree with the experimentally determined rate law

39. Rate-Determining Step In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction. The experimental rate law must agree with the rate-determining step

40. Overall Reaction: 2NO 2(g) + F 2(g) → 2FNO 2(g) Experimental rate law: Rate = k[NO 2 ][F 2 ]

41. Identifying the Rate-Determining Step For the reaction: 2H 2 (g) + 2NO(g)  N 2 (g) + 2H 2 O(g) The experimental rate law is: R = k[NO] 2 [H 2 ] Which step in the reaction mechanism is the rate-determining (slowest) step? Step #1 H 2 (g) + 2NO(g)  N 2 O(g) + H 2 O(g) Step #2 N 2 O(g) + H 2 (g)  N 2 (g) + H 2 O(g) Step #1 agrees with the experimental rate law

42. Identifying Intermediates For the reaction: 2H 2 (g) + 2NO(g)  N 2 (g) + 2H 2 O(g) Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?) Step #1 H 2 (g) + 2NO(g)  N 2 O(g) + H 2 O(g) Step #2 N 2 O(g) + H 2 (g)  N 2 (g) + H 2 O(g) 2H 2 (g) + 2NO(g)  N 2 (g) + 2H 2 O(g)  N 2 O(g) is an intermediate

43. The Raschig reaction produces hydrazine, N 2 H 4, an industrially important reducing agent, from NH 3 and OCl - in basic, aqueous solution. A proposed mechanism is: Step 1: (fast) NH 3(aq) + OCl - (aq) → NH 2 Cl (aq) + OH - (aq) Step 2: (slow) NH 2 Cl (aq) + NH 3(aq) → N 2 H 5(aq) + Cl - (aq) Step 3: (fast) N 2 H 5 + (aq) + OH - (aq) → N 2 H 4(aq) + H 2 O (l) a. What is the overall stoichiometric equation? b. Which step of the three is rate-determining? c. Write the rate equation for the rate-determining step. d. What reaction intermediates are involved? An Example …

44. Rate Laws Summary Zero Order First Order Second Order Rate Law Rate = kRate = k[A]Rate = k[A] 2 Integrated Rate Law [A] = -kt + [A] 0 ln[A] = -kt + ln[A] 0 Plot the produces a straight line [A] versus tln[A] versus t Relationship of rate constant to slope of straight line Slope = -k Slope = k Half-Life