1. Dr. Harris Lecture 18 HW: Ch 17: 5, 11, 18, 23, 41, 50 Ch 17: Kinetics Pt 1

2. Reactions Rates Chemical kinetics is the area of chemistry that investigates how fast reactions occur Different reactions proceed with different rates 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

3. 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.

4. = 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

5. Plotting the data from previous slide

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

7. Instantaneous Rates Although the rate of the reaction is constantly changing with reactant concentration, we can determine the instantaneous rate (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 Tangent at t = 20s, [A] = 5M

8. 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.

9. 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

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

11. 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.

12. Reaction Orders and the Method of Initial Rates 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. We see that when we double [N 2 O 5 ] o, the rate also doubles. When we quadruple [N 2 O 5 ] o, the rate quadruples. Thus, the rate is directly proportional to [N 2 O 5 ] o by the rate constant, k. This means that the reaction is FIRST ORDER WITH RESPECT TO [N 2 O 5 ] (m=1). We can write the rate law as: 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

13. Reaction Orders The overall reaction order is the sum of the individual reaction orders. In our previous example, there was only one reactant, so the overall order is 1 (1 st order reaction). We can easily solve for k by plugging in any corresponding rate and concentration. Lets plug in the values from run # 1 Run[N 2 O 5 ] o (M)Rate, M/s 10.010.018 20.020.036 30.040.072

14. Rate Laws/Reaction Orders Reaction orders 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

15. 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 to find ‘m’.

16. 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]

17. 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 [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.

18. 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 molarity per unit time (sec, hr, etc). Concentration is always M (mol/L). Thus, we have: Recall for a 1 st order reaction: Units of k for a 2 nd order reaction Units of k for a 1 st order reaction

19. 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 (with correct units) 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 proportional to the square of [NO 2 ], so the reaction is second order with respect to NO 2 (n=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

20. Example: 2NO(g) + Br 2 (L) ---> 2NOBr (g) Experiment[NO] o [Br 2 ] o Rate (M/s) 10.100.2024 20.250.20150 30.100.5060 Using the information below, determine the rate law of this reaction in the form: 1.) Find m. We can use runs 1 & 2: 2.) Find n. We can use runs 1 & 3 m = 2 n = 1 The reaction is 2 nd order with respect to [NO], 1 st order with respect to Br 2, and the reaction is overall 3 rd order.

21. 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

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

23. 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

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

25. 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