1. Chemical Kinetics Rates of Reactions ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

2. What is a Reacation Rate? Deals with how the reactants concentrations change with time ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

3. Factors that Affect Reaction Rate Temperature Collision Theory: When two chemicals react, their molecules have to collide with each other with sufficient energy for the reaction to take place. Kinetic Theory: Increasing temperature means the molecules move faster. Concentrations of reactants More reactants mean more collisions if enough energy is present Catalysts Speed up reactions by lowering activation energy Surface area of a solid reactant Bread and Butter theory: more area for reactants to be in contact Pressure of gaseous reactants or products Increased number of collisions ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

4. How do reactions take place? Collision Theory Reactants must have… 1. Correct orientation to each other 2. Enough energy for the reaction to occur ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

5. Activation Energy Diagrams Exothermic vs Endothermic ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

6. A + B C + D Exothermic Reaction Endothermic Reaction The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. 13.4 $ What is 13.4? ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

7. 4PH 3 (g)  P 4 (g) + 6H 2 (g) If 0.0048 mol of PH 3 is consumed in a 2.0 L container during each second of the reaction, what are the rates of production of P 4 and H 2 ? $ The picture at the top show a solution reaction, not gaseous. Is the temperature of the desired reaction high enough that P4(g) is produced Instead of solid? Missing term inside [ ]. ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

8. Solution… ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

9. 2NO 2 (g)  2NO(g) + O 2 (g) at 300 o C Time (s)[NO 2 ][NO][O 2 ] 00.010000 500.00790.00210.0011 1000.00650.00350.0018 1500.00550.00450.0023 2000.00480.00520.0026 2500.00430.00570.0029 3000.00380.00620.0031 3500.00340.00660.0033 4000.00310.00690.0035 ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

10. Rate Laws Deal only with the concentration of the reactants Rate = k[NO 2 ] n k = proportionality constant (rate constant) n = order (may be a fraction) & must be determined experimentally $ Change k to italics. ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

11. The Rate Law 13.2 The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. aA + bB cC + dD Rate = k [A] x [B] y reaction is xth order in A reaction is yth order in B reaction is (x +y)th order overall $ Mention that x and y must be determined experimentally. What is the 13.2? ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

12. Types of Rate Laws Differential Rate Law (Rate Law): shows how rate depends on concentration Integrated Rate Law: shows how concentrations of a species changes depends on time ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

13. Finding the Form The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ] 1 $ Add and must be determined experimentally. ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

14. Method of Initial Rates Rate just after the reaction has begun but before there is a significant change in concentration ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

15. Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- (aq) + 3I - (aq) 2SO 4 2- (aq) + I 3 - (aq) Experiment [S 2 O 8 2- ][I - ] Initial Rate (M/s) 10.080.0342.2 x 10 -4 20.080.0171.1 x 10 -4 30.160.0172.2 x 10 -4 ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

16. Terms Molecularity – the number of molecules that must collide to produce the reaction indicated by that step Elementary Step – a reaction for which the rate law can be written from its molecularity Unimolecular – reactions involving one molecule Bimolecular – reactions involving two molecules ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

17. So what is a mechanism? A series of elementary steps that must add up to the overall balanced equation AND agree with the rate law Step 1 Step 2 Step 3 ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

18. Reaction Mechanisms Series of steps by which a chemical reaction occurs Balanced equation does not tell us HOW the reactants become the products It is a summary of the overall process Ex – 6CO 2 + 6H 2 O  C 6 H 12 O 6 + 6O 2 ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

19. Elementary Steps A  productsunimolecularRate = k[A] A + A  productsbimolecularRate = k[A] 2 A + B  productsbimolecularRate = k[A][B] A + A + B  productstermolecularRate = k[A] 2 [B] A + B + C  productstermolecularRate = k[A][B][C] Termolecular reactions are rare due to infrequent collisions of 3 molecules simultaneously. ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

20. Reactions Mechanisms Con’t The sum of the intermediate steps must agree with stoichiometry of the reaction Intermediates – are neither products or reactants and must cancel out ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

21. NO 2 (g) + CO(g)  NO(g) + CO 2 (g) NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) NO 2 (g) + CO(g)  NO(g) + CO 2 (g) Step 1 is the rate determining step (slow) and the rate law can be written Rate = k[NO] 2 Step 1 Step 2 ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

22. zerofirstsecond Rate Law Integrated Rate Law Straight line Graph Half-life Rate = kRate = k[A]Rate = k[A] 2 [A] = -kt + [A] 0 ln[A] = -kt + ln[A] 0 [A] vs t Slope = -k ln[A] vs t Slope = -k Summary of Rate Laws ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright