1. Chemical Kinetics Sorry not all reactions are instantaneous!

2. What is Chemical Kinetics? Kinetics examines the rates at which chemical reactions occur. Consider the decomposition of: 2NI 3(s)  N 2(g) and 3I 2(s) This reaction is SOOO rapid that we say it occurs instantaneously. NI 3 is very unstable with respect to its elements, it is sensitive to light and vibration.

3. What about a non-instantaneous reaction? Na 2 S 2 O 3(aq) + 2HCl  S (s) + H 2 SO 3(aq) + 2NaCl (aq) We saw that the rate which S (s) was formed, so the rate of the reaction could be altered by adjusting the concentration of S 2 O 3 2- or by varying the temperature.

4. What do we mean by rate of reaction? In general terms rate means how many of something can be done per unit of time. (ie km/h, m/s etc) In chemistry though we refer to rates of reaction based on how much reactant was consumed or how much product was produced per unit of time (usually seconds). Consider our experiment... Na 2 S 2 O 3(aq) + 2HCl  S (s) + H 2 SO 3(aq) + 2NaCl (aq) We agreed that the reaction was complete when 0.25mmol of sulfur was present. (Amount of product formed) Rate of reaction = ([A] final – [A] initial ) / (t final – t initial )

5. The 5 Factors That Can Affect the Rate of Reaction 1)The rate of reaction can (generally) be increased by increasing the temperature. 2)Increasing reactant concentrations typically increases reaction rate. 3)An increase in surface are will increase the reaction rate. 4)The nature of the reactants (ie metal react quickly with strong acids) 5)The use of a catalyst will increase the rate of reaction (if the catalyst can interact with the reactants).

6. What causes molecules to interact? Kinetic Molecular Theory (KMT) or Collision Theory are often used to describe this process. KMT maintains that: 1)all particles of matter are in constant motion (they possess kinetic energy) 2)there are spaces between the particles, how big the space and the speed determine the state of matter (solid, liquid or gas) 3)as we increase heat, the particles move faster increasing they’re kinetic energy

7. Collision Theory This theory is literally based around the notion that in order for two molecules, atoms or ions to react they must be able to successfully “bump into” each other. These collisions are increased at higher temperatures, with greater concentrations, if the reactants are ionic rather that molecular, with greater surface area. A catalysts helps bring together the two reacting pieces thereby facilitating a collision. Read pages 469 – 473 for general information. 481 – 482 Section Review Questions 1, 2, 7, 11 (484)

8. So what do we measure? Consider the following reaction 2H 2 O 2(aq)  2H 2 O + O 2(g) We can say the rate of this reaction is equal to the disappearance of H 2 O 2 or the appearance of O 2. Rate = -  [H 2 O 2 ] Rate =  [O 2 ].  t  t But for every H 2 O 2 that decomposes only 0.5 O 2 is formed, therefore Rate = -0.5  [H 2 O 2 ] =  [O 2 ].  t  t Units = mol/L/s or M/s or M s -1 We call this the average rate of reaction.

9. What about the Instantaneous Rate? 2H 2 O 2(aq)  2H 2 O + O 2(g) Time (s) Accumulated Mass O 2 (g) [H 2 O 2 ] (M) 000.882 602.9600.697 1205.0560.566 1806.7840.458 2408.1600.372 3009.3440.298 36010.3360.236 42011.1040.188 48011.6800.152 54012.1920.120 60012.6080.094 In order to get the rate of reaction at any given instant we need to find the negative slope of the tangent line... Huh? The derivative for those of you in calculus. Initial rate = 3.21 x 10 -3 M s -1

10. Ok well what about a different set of conditions? We know from our experiments with the 5 factors that affect the rate of reaction that generally if we increase the concentration of the reactants the reaction will occur faster. The problem with our experimental determination of rates on the previous slide is that it works only for an initial concentration of H 2 O 2 = 0.882 M. Enter the Rate Law. The rate law for a chemical reaction relates the rate of reaction to concentration of reactants so it can be used to describe any concentration. Consider: a A + b B  c C + d D Rate = -(1/a)(  [A]/t) = -(1/b)(  [B]/t) (Negative b/c disappearance of reactants) Rate = k[A] m [B] n (Rate Law) (m and n are determined experimentally)

11. Rate Law eh... ? a A + b B  c C + d D Rate = k[A] m [B] n k is a proportionality constant called the rate constant. The numerical values of k depends on the reaction, the temperature and the catalyst (if any). The units of k depend on m and n. The values of the exponents determine the order of the reaction. If m = 1, the reaction is first order in A. If n = 2 the reaction is second order in B and the reaction is third order overall. In general, the higher the order the more complex the mathematical expression describing the reaction. Therefore if you have a choice to treat a reaction using either reactant, choose the one with the lower order.

12. Back to H 2 O 2 The reaction is 1 st order in H 2 O 2, therefore the rate law is as follows: Rate = k[H 2 O 2 ] 1 = k[H 2 O 2 ] Remember the exponents are determined experimentally NOT from the balanced chemical equation. If we have determined the rates experimentally we can determine k mathematically. k = Rate/[H 2 O 2 ] = 3.21 x 10 -3 M s -1 / 0.882M = 3.64 x 10 -3 s -1 Once we have the value of k, we can use it to determine the initial rate for any initial concentration of H 2 O 2 (at the given temperature). 2H 2 O 2(aq)  2H 2 O + O 2(g) If I tell you that a reaction is zero order with respect to A, what is the rate law... ?

13. Determining the Rate Law The Method of Initial Rates for Determining m and n 2NO (g) + Cl 2(g)  2NOCl (g) ExperimentInitial [NO] Initial [Cl 2 ] Initial Rate, M s -1 10.01250.02552.27E-05 20.01250.05104.55E-05 30.0250.02559.08E-05 (Initial rate) 3 = k[NO] m [Cl 2 ] n = (2 x 0.0125) m = 2 m x (0.0125) m = 2 m (Initial rate) 1 k[NO] m [Cl 2 ] n (0.0125) m (0.0125) m (Initial rate) 3 = 9.08 x 10 -5 M s -1 = 4 (Initial rate) 1 2.27 x 10 -5 M s -1 4 = 2 m = 2 2, therefore the reaction is second order in NO If the reaction is first order, when you double the concentration of a reactant, the rate doubles.

14. Applications of the Rate Law Determining Concentrations at a Later Time First Order Integrated Rate Law ln [A] t [A] 0 = -kt or ln[A] t – ln[A] 0 = -kt ln[A] t = (-k)t + ln[A] 0 y = m x + b Time (s) [H 2 O 2 ] (M)ln[H 2 O 2 ] 00.882-0.12556 600.697-0.36097 1200.566-0.56916 1800.458-0.78089 By plotting these points we get a straight line, this confirms that the reaction is first order. Given the table you could solve for k.

15. Applications of the Rate Law Half Life What is half life?  the amount of time it takes for the concentration of something to decrease to ½ its initial amount ln ½ [A] 0 [A] 0 = -kt = ln (½) = -kt ½ t ½ = 0.693 / k If you have the need for any diagnostic imaging you may be required to ingest some radioactive dye. This dye traces your insides so the image can be enhanced by X-ray. Biological dyes usually have a very short half-life and are nearly completely excreted within a few days.