1. Suggested HW (Ch 17): 6, 12, 18, 22, 42, 47 50

2. About Kinetics Chemists ask three questions when studying chemical reactions: 1. What happens? 2. How fast? 3. To what extent? The first question is addressed by writing balanced chemical reactions, calculating yields, making qualitative observations, etc. The second question is addressed in an area of chemistry called KINETICS, which will be covered over the next two lectures The third question will be answered in chapter 19 (Equilibrium)

3. Reactions Rates Different reactions proceed with different rates The rate of a reaction is defined as the decrease in the concentration of reactant per unit time (or the increase in the concentration of product) The rate of a reaction depends on several factors, including: reactant concentration temperature catalysts surface area Today, we will focus exclusively on the relationship between reaction rates and reactant concentration

4. Intro Lets take the reaction: A ---> B. This reaction tells us that as A is consumed, B is formed at an equal rate. We can express this mathematically in terms of changing concentrations by: Imagine we have 10 moles of A in 1 L of solution. If we can freeze time for an instant, such that the reaction has not yet begin (t=0), the concentration of A is 10M.

5. = 1 mol of A [A] = 10 M t = 0 A B After 10 seconds, 3 moles of B have formed. [A] = 7 M [B] = 3 M t = 10 = 1 mol of B 10 more seconds [A] = 5 M [B] = 5 M t = 20 20 more seconds [A] = 4 M [B] = 6 M 40 more seconds [A] = 3 M [B] = 7 M t = 40 t = 80

6. Plotting the data from previous slide

7. Reactions Follow a Rate Law The graph in the previous slide shows that the disappearance of A (formation of B) is not linear. As the reactant concentration decreases, the reaction slows down. This dependence of rate on concentration suggests that reaction rates follow a rate law, a mathematical expression that ties concentration and rate together

8. Instantaneous Rates The rate of the reaction is constantly changing, but we can determine instantaneous rates (reaction rate at a specific time and concentration) Instantaneous rate at t=0 is the initial rate We can determine the instantaneous rate by taking the slope of the tangent at the point of interest Note: a tangent line is linear and ONLY touches the point in question. It does NOT cross the curve Instantaneous rate of disappearance of A at t=20 sec

9. Rates and Stochiometry In the previous example (A---->B), we had 1:1 stoichiometry. Thus, at any given time, the rate of disappearance of A equals the rate of formation of B. If the stoichiometry is NOT 1:1, we have a much different situation, as shown below: As you can see, 2 moles of HI are consumed for every 1 mole of H 2 and 1 mole of I 2 formed. Thus, the disappearance of HI is twice as fast as the appearance of the products.

10. Example: N 2 O 5 (g) ----> 2NO 2 (g) + ½ O 2 (g) Looking at average rates average rate of disappearance after 10 minutes average rate of disappearance after 100 minutes

11. N 2 O 5 (g) ----> 2NO 2 (g) + ½ O 2 (g) fast slow

12. Rate Laws We see that reducing reactant concentration lowers the reaction rate, but to what extent? What is the mathematical correlation? The equation that relates the concentration of the reactants to the rate of reaction is called the rate law of the reaction. We can derive the rate law of a reaction by seeing HOW THE REACTION RATE CHANGES WITH REACTANT CONCENTRATION. For any reaction aA + bB……. ----> cC + dD….. In this expression, k is the rate constant, m and n are reaction orders.

13. Lets go back to the previous reaction: Below is a table of data, showing the initial reaction rate as a function of the starting concentration of N 2 O 5 (g). We perform multiple experiments to collect enough data to determine our rate law. Our rate law will be in the form: Rate = k [N 2 O 5 ] m. Never include products in a rate law !!!!! Let’s use the data in the table to obtain the value of m. We must pay attention to how the rate is affected by [N 2 O 5 ] Reaction Orders and the Method of Initial Rates N 2 O 5 (g) 2NO 2 (g) + ½ O 2 (g) Experiment[N 2 O 5 ] o (M)Rate, M/s 10.010.018 20.020.036 30.040.072

14. Let’s solve for m. We can choose any two sets of data values to set up a system of equations. We’ll choose data from experiments 1 & 2. Now, we have two equations and two variables (k and m), but the k values can cancel, allowing us to solve for m as shown below: Rate Laws

15. Reaction Orders (example continued) This means that the reaction is FIRST ORDER WITH RESPECT TO [N 2 O 5 ] When we double [N 2 O 5 ] o, the rate also doubles. When we quadruple [N 2 O 5 ] o, the rate quadruples. Physically speaking, a reaction order of 1 means that one molecule of the reactant is involved in the formation of products. Thus, the rate is directly proportional to [N 2 O 5 ] o by the rate constant, k. We can write the rate law as: The overall reaction order is the sum of the individual reaction orders. (1 st order reaction). We can easily solve for k by plugging in any corresponding rate and concentration (k = 1.8 s -1 )

16. Rate Laws/Reaction Orders Reaction orders of non-elementary reactions must be determined experimentally. You can not assume based on the stoichiometry. When you have multiple reactants, you must determine the reaction order of each one. To do this, you must vary the concentration of only one reactant at a time while holding the others fixed. Let’s attempt to determine the rate law for the reaction below: 2NO(g) + O 2 (g) ---> 2NO 2

17. Example: 2NO(g) + O 2 (g) ---> 2NO 2 Using the data below, determine the rate law of this reaction in the form: Experiment[NO] o (M)[O 2 ] o (M)Rate (M/s) 1.0126.01252.82 x 10 -2 2.0252.02501.13 x 10 -1 3.0252.01255.64 x 10 -2 This time, we have two reactants. Lets start by determining the value of ‘m’. To do so, we hold [O 2 ] o fixed and vary [NO] o. This will show how the rate depends on [NO] o. In experiments #1 and #3, [O 2 ] o is fixed, so we will use these experiments to find ‘m’.

18. Experiment[NO] o (M)[O 2 ] o (M)Rate (M/s) 1.0126.01252.82 x 10 -2 2.0252.02501.13 x 10 -1 3.0252.01255.64 x 10 -2 Remember, rate is proportional to [NO] by the power m. The factor of change in the rate is equal to the factor of change of [NO] to the m th power: factor of rate change factor of change in [NO] order m = 1 The reaction is 1 st order with respect to [NO]

19. Run[NO] o (M)[O 2 ] o (M)Rate (M/s) 1.0126.01252.82 x 10 -2 2.0252.02501.13 x 10 -1 3.0252.01255.64 x 10 -2 factor of rate change factor of change in [O 2 ] order n = 1 The reaction is 1 st order with respect to [NO], 1 st order with respect to [O 2 ] and 2 nd order overall. Now we can find ‘n’ by varying [O 2 ] o and holding [NO] o fixed. We can use experiments #2 and #3 for this. This will show how the rate depends on [O 2 ] o. k = 179 M -1 s -1

20. Pay Attention to the Units of k, As They Change with Overall Reaction Order The rate constant, k, is the constant of proportionality between rate and concentration. Higher values of k = faster reactions It is important to note that the units of k depend on the overall reaction order. Ex: Rate is always in units of molarity per unit time (sec, hr, etc). Concentrations are always expressed as molarities (M or mol/L). Thus, we have: Recall for a 1 st order reaction: For z th order (z = any integer) Units of k for a 2 nd order reaction Units of k for a 1 st order reaction

21. Group Example Determine the relative (m & n) and overall (m+n) reaction order of the reaction below. Then, derive the rate law and determine the value of k. Experiment[NO 2 ] o (M)[CO] o (M)Rate (M/s) 1.0300.2001 x 10 5 2.0900.2009 x 10 5 3.300.04001 x 10 7 4.300.08001 x 10 7 Tripling [NO 2 ] causes the rate to increase nine-fold. This means that the rate is squarely proportional to [NO 2 ], so the reaction is second order with respect to NO 2 (m=2). Doubling [CO] does nothing. Thus, the rate does not depend on [CO], and is zero order with respect to CO (m=0). Overall 2 nd order. k = 1.11 x 10 8 M -1 s -1

22. Determining the Overall Rate Order of A Reaction Graphically As we have shown, a first-order reaction depends on the concentration of a single reactant to the 1 st power. For the reaction: A----> products Using calculus, we can convert this to: This equation is in y = mx + b form. Therefore, for any 1 st order reaction, the plot of the natural log of [A] t vs time will be linear. The slope of the line will be –k. natural log of concentration at time t natural log of starting concentration rate constant y axis m (slope) b time x axis

23. Plotting 1 st Order Reactions b time values on x-axis slope = -k units: s -1 natural log of [A] t on y-axis

24. Determining the Overall Rate Order of A Reaction graphically A second-order reaction depends on the concentration of [A] to the 2 nd power. For the reaction: A ----> B Therefore, for any 2 nd order reaction, the plot of the inverse of [A] t vs time will be linear. The slope of the line will be k. y mxmx b

25. Important!!! Natural logs (ln) are exponential terms! To solve for the value of a term within a natural log function, use the exponential function on your calculator (e x )

26. Plotting a 2 nd Order Reaction b slope = k units = M -1 s -1 time values on x-axis1/[A] t on y-axis

27. Determining Overall Rate Order From Plotting Time- Dependent Data We can determine if a process is first or second order by plotting the data against both equations. Which ever fitting method yields a linear plot gives the overall order. not linear: NOT 1 st order linear! 2 nd order

28. Example From the given data, write the rate law of the reaction, including k. What is the rate of the reaction at t=4 seconds? t=5 seconds? What concentration of A remains at t=10 seconds?

29. Suggested HW: Ch 18: 38, 41, 44, 47 Ch 18: Kinetics Pt. 2 Temperature Dependence of Rate Constants

30. Temperature and Rate The rates of most chemical reactions increase with temperature. How is this temperature dependence reflected in the rate expression? Rates increase with temperature because rate constants increase with temperature. An example is the 1 st order reaction: CH 3 NC ---> CH 3 CN Variation in k with temperature.

31. Collision Model The collision model makes sense of this. This model is based on kinetic theory We’ve seen that the thermal energy of a molecule is converted to kinetic energy in order to facilitate motion, and we’ve seen that the velocity of a molecule increases with T. The central idea of the collision model is that molecules must collide to react. The more collisions per second, the faster the reaction goes. This model also rationalizes the concentration dependence. The more molecules present, the more collisions you have.

32. Activation Energy Of course, there is more to a chemical reaction than just collisions of molecules. Molecules must have some minimum energy in order to overcome the energy barrier of reaction Upon colliding, the kinetic energy of molecules is used to stretch, bend, and break bonds in order to cause a reaction. But if the molecules don’t have enough kinetic energy, they simply bounce off one another. This minimum energy that molecules must have is called the activation energy,(E a ) E too low

33. Activation Energy Needed to Reach Transition State Lets go back to the reaction CH 3 NC ---> CH 3 CN The molecule passes through a transition state, a high-energy intermediate in which the CN bond gets rotated 90 o. This is an unstable configuration and requires an activation energy to be reached. Once this transition state is reached, the process is energetically downhill. The transition state is always the highest energy point in the reaction pathway. As drawn, this reaction is exothermic because the energy of the products is less than the energy of the reactants. Transition state nitrogen carbon ΔHΔH Enthalpy

34. Arrhenius noted that reaction-rate data depended on three aspects (1) the fraction of molecules possessing an energy E a or greater (2) the number of collisions per second (3) the fraction of molecules oriented in the right way for a reaction to occur These factors are incorporated into the Arrhenius Equation This equation relates k to temperature. A is the frequency factor, which is related to criteria 2) and 3) above, and the fraction of particles possessing an energy of E a or greater is given by

35. Rearranging the Arrhenius Equation Taking the natural log of both sides, the Arrhenius equation becomes: As you see, we once again have an equation in y=mx+b form. Plotting ln k vs 1/T yields a linear plot with slope –E a /R and a y- intercept of ln A

36. Showing Changes in k Graphically From the data below, determine the activation energy of the 1 st order reaction Temperature, o CK (s -1 ) 189.72.52 x 10 -5 198.95.25 x 10 -5 230.36.30 x 10 -4 251.23.16 x 10 -3 Plot ln k vs 1/T to get the slope. We know R, so we can calculate E a. T must be in Kelvin. 1/T ( o K -1 )ln K 0.002161-10.5887 0.002119-9.8547 0.001987-7.36979 0.001908-5.75718

37. Calculating Changes in k Mathematically The Arrhenius equation can be rearranged once again to compare k values at different temperatures Remember:

38. Example The reaction above has an activation energy of 43.5 kJ/mol. The reaction occurs at 298 K with a rate constant of 110 s -1. What will the rate constant be if we increase the temperature to 308 K?

39. Reaction Mechanisms Consider the following reaction: In looking at this equation, and applying your understanding of collision theory, you would assume that this reaction proceeds when NO 2 and CO collide The activation energy is exceeded The atoms rearrange to form NO and CO 2 Thus, we would expect the rate of the reaction to depend on both [NO 2 ] and [CO]

40. Reaction Mechanisms However, according to experimental data, the reaction is 2 nd order with respect to NO 2 and zero order with respect to CO. The rate law is: This means that the rate of the reaction does not depend on [CO] at all. How can this be? What does this tell us about the reaction mechanism? This particular reaction must proceed through multiple steps, and the rate-determining step must only depend on NO 2

41. Reaction Mechanisms For this particular reaction, there are two steps: This step-by-step description of the molecular pathway is the reaction mechanism. From reaction (1), we can physically understand why the reaction is 2 nd order with respect to [NO 2 ]. The 1 st step is substantially slower than the 2 nd. Therefore, step (1) acts as the “reaction bottleneck”, and the overall reaction speed can only be as fast as its slowest step. Thus, step (1) is the rate determining step.

42. Each step in a reaction mechanism is called an elementary reaction. An elementary reaction is any reaction that proceeds in a single step. Mechanisms consist of sums of elementary reactions. Going back to rate laws, each elementary reaction must have a corresponding rate constant *** For elementary reactions only, you can assume that the rate law depends directly on the number of species present. Thus, the rate law of an elementary reaction follows the stoichiometry. k1k1 k2k2 rate of overall reaction intermediate (transition step)

43. Examples Write the rate law for the following reaction given the elementary reaction steps of the mechanism: k1k1 k2k2

44. Catalysis The rates of reactions can be increased by using catalysts. Catalysts work by providing a different reaction path (mechanism) between reactants and products with a lower activation energy The catalyst itself is not consumed in the reaction.

45. Example: Oxidation of Sulfur Dioxide Uncatalyzed 2SO 2 (g) + O 2 (g)  2SO 3 (g) (1) SO 2 (g) ---> S(s) + O 2 (g) ΔH o rxn (1) = 296.8 kJ/mol (slow step) (2) 2S(s) + 3O 2 (g) ---> 2SO 3 (g) ΔH o rxn (2) = -791. 4 kJ/mol 2SO 2 Formation of S(s) combustion of S to SO 3 ΔH o rxn (E a ) ΔH o 1 ΔH o 2

46. Example: Oxidation of Sulfur Dioxide Pt-Catalyzed 2SO 2 (g) + O 2 (g)  2SO 3 (g) (1) O 2 (g) ---> 2O ΔH o rxn (1) = 93.4 kJ/mol (slow step) (2) SO 2 (s) + O(g) ---> SO 3 (g) ΔH o rxn (2) = -791. 4 kJ/mol Pt 2SO 2 Formation of O ΔHΔH EaEa E a,cat

47. Example: Oxidation of Sulfur Dioxide Pt-Catalyzed