1. 16.2.1 Explain that reactions can occur by more than one step and that the slowest step determines the rate of the reaction (rate- determining step) 16.2.2 Describe the relationship between reaction mechanism, order of reaction and rate- determining step. 16.2 Reaction mechanism

2. Chemical reactions The chance of more than two particles colliding simultaneously with correct geometry and minimum energy required is very small. If there are more than 2 reactants, the reaction must occur by a number of simpler steps

3. Reaction Mechanisms A reaction mechanism is a series of simple steps that ultimately lead from the initial reactants to the final products of a reaction. The mechanism must account for the experimentally determined rate law. The mechanism must be consistent with the stoichiometry of the overall or net reaction.

4. Reaction Mechanisms The rate-determining step is the crucial step in establishing the rate of the overall reaction. Some two-step mechanisms have a slow first step followed by a fast second step, while others have a fast reversible first step followed by a slow second step.

5. A Mechanism With A Slow Step Followed By A Fast Step - A Plausible Mechanism Slow step:H 2 O 2 + I -  H 2 O + OI - Fast step:H 2 O 2 + OI -  H 2 O + O 2 + I - ___________________________________________________________ Net equation:2 H 2 O 2  2 H 2 O + O 2 I - - catalyst;OI - - intermediate The slow step is the rate-determining step. Rate = rate of slow step = k[H 2 O 2 ][I - ]

6. Mechanisms Intermediate A species that is created in one step and consumed in the other A species that is created in one step and consumed in the otherCatalyst A species that is present originally then reforms later on during the reaction A species that is present originally then reforms later on during the reaction It is not written in the overall equation, but you may see it noted above the reaction arrow. It is not written in the overall equation, but you may see it noted above the reaction arrow.

7. A Mechanism With A Slow Step Followed By A Fast Step Slow step:H 2 O 2 + I -  H 2 O + OI - Fast step:H 2 O 2 + OI -  H 2 O + O 2 + I - Overall:2 H 2 O 2 (aq)  2 H 2 O (l) + O 2 (g) Facts: (1) The rate of decomposition of H 2 O 2 is first order in both H 2 O 2 and I -, or second order overall. (2) The reactant I - is unchanged in the reaction and hence does not appear in the equation for the net reaction. (2) The reactant I - is unchanged in the reaction and hence does not appear in the equation for the net reaction.

8. A Mechanism With A Fast Reversible Step Followed By A Slow Step 2 NO (g) + O 2 (g)  2 NO 2 (g) Experimentally found: Rate = k[NO] 2 [O 2 ] Rate = k[NO] 2 [O 2 ]

9. A Mechanism With A Fast Reversible Step Followed By A Slow Step – A Plausible Mechanism k 1 k 1 Fast step:2 NO N 2 O 2 k -1 k -1 k 2 k 2 Slow step:N 2 O 2 + O 2  2 NO 2 Net equation:2 NO + O 2  2 NO 2

10. Molecularity The number of molecules that participate as reactants in an elementary step Unimolecular: a single molecule is involved. Ex: CH 3 NC (can be rearranged) Ex: CH 3 NC (can be rearranged) Radioactive decay Radioactive decay Its rate law is 1 st order with respect to that reactant Its rate law is 1 st order with respect to that reactant

11. Bimolecular: Involves the collision of two molecules (that form a transition state that can not be isolated) Ex: NO + O 3  NO 2 + O 2 Ex: NO + O 3  NO 2 + O 2 It’s rate law is 1 st order with respect to each reactant and therefore is 2 nd order overall. It’s rate law is 1 st order with respect to each reactant and therefore is 2 nd order overall. Rate =k[NO][O 3 ] Rate =k[NO][O 3 ]

12. Termolecular: simultaneous collision of three molecules. Far less probable. Some possible mechanisms for the reaction; 2A + B  C + D

15. Rate Laws for Elementary steps Can use the equation coefficients as the reaction orders in the rate law for an elementary step Elementary step Molecularity Rate law A  product Uni Rate = k[A] 2A  product Bi Rate = k[A] 2 A + B  product Bi Rate = k[A][B] 2A + B  product Ter Rate = k[A] 2 [B]

16. Differences between intermediates and transition states NOTE: transition state and activated complex are the same thing!

17. 16.3 Activation energy 16.3.1 Describe qualitatively the relationship between the rate constant (k) and temperature (T). 16.3.2 Determine activation energy (Ea) values from the Arrhenius equation by a graphical method.

18. Review of Exothermic Reactants Ep is higher than Products Ep. Now, we must consider the activation energy (the energy needed so that the reactants bonds will break and reform to make product)

19. Review of Endothermic Reactants Ep is lower than Products Ep. Need to add more energy to the system for the forward reaction to take place. Still need to consider activation energy

20. Activated Complex Is the short-lived, unstable structure formed during a successful collision between reactant particles. Old bonds of the reactants are in the process of breaking, and new products are forming Ea is the minimum energy required for the activation complex to form and for a successful reaction to occur.

21. Fast and slow reactions The smaller the activation energy, the faster the reaction will occur regardless if exothermic or endothermic. If there is a large activation energy needed, that means that more energy (and therefore, time) is being used up for the successful collisions to take place.

22. Sample Problem The following reaction has an activation energy of 120kJ and a ΔH of 113kJ. 2NO 2  2NO + O 2 Draw and label a potential energy (activation energy) diagram for this forward reaction. Draw and label a potential energy (activation energy) diagram for this forward reaction. Calculate the activation energy for the reverse reaction (if the reaction went backwards) Calculate the activation energy for the reverse reaction (if the reaction went backwards)

23. Another Problem The following hypothetical reaction has an activation energy of 70kJ and a ΔH of -130kJ A + B  C + D Draw and label a potential energy diagram for the reaction Draw and label a potential energy diagram for the reaction Calculate the activation energy for the reverse reaction. Calculate the activation energy for the reverse reaction.

24. Watch the following Flash Review of what is occurring during a chemical reaction for both endothermic and exothermic. KNOW THIS!! http://mhhe.com/physsci/chemistry/essenti alchemistry/flash/activa2.swf http://mhhe.com/physsci/chemistry/essenti alchemistry/flash/activa2.swf

25. Practice: 1.The following hypothetical reaction has an Ea of 120kJ and a ΔH of 80kJ 2a + B  2C + D Draw and label a potential energy diagram for this reaction. Draw and label a potential energy diagram for this reaction. What type of reaction is this? What type of reaction is this? Calculate the activation energy for the reverse reaction. Calculate the activation energy for the reverse reaction. Calculate the ΔH for the reverse reaction. Calculate the ΔH for the reverse reaction.

26. 2.Analyze the activation energy diagram below. What is the Ea for the forward reaction? For the reverse reaction? What is the ΔH for the forward reaction? For the reverse reaction? What is the energy of the activated complex?

27. At higher temperatures there is a greater proportion of molecules that would have enough Ea for the reaction to proceed. This is the major reason why high temps increase rate.

28. Effect of Temperature on the Reaction Rate The Arrhenius equation show the effect of temperature on the rate constant, k It indicates that k depends exponentially on temperature Arrhenius equation : k = A e - E a /RT E a – activation energy R – gas constant, 8.3145 J mol -1 K -1 T - Kelvin temperature A – Arrhenius constant (depends on collision rate and shape of molecule)

29. k = A e - E a /RT As T increases, the negative exponent becomes smaller, so that value of k becomes larger, which means that the rate increases. Higher T  Larger k  Increased rate

30. ln k and 1/T is linear With R known, we can find Ea graphically from a series of k values at different temperatures. ln k 2 = - Ea ( 1/T 2 – 1/T 1 ) k 1 R k 1 R Ea = -R (ln k 2 / k 1 ) ( 1/T 2 – 1/T 1 ) -1

31. Problem The decomposition of hydrogen iodide has rate constants of 9.51 x10 -9 L/mol.s at 500.0 K and 1.10 x 10 -5 L/mol.s at 600.0 K. Find Ea. Ea = - (8.314)( ln 1.10 x10 -5 / 9.51 x10 9 )(1/600.00 – 1/500.0) = 1.76 x 10 5 J/mol

32. If rearrange the equation and convert it to: lnk = - E a. 1 + lnA R T R T A graph of ln k against 1/T will be linear with a slope/gradient of –Ea/R and an intercept on the y-axis of lnA lnk = - E a. 1 + lnA R T R T y = m. x + b y = m. x + b

33. Plot lnk vs. 1/T = straight line

34. Readings Section 16.5, 16.6, 16.7, 16.8 Effect of temperature, concentration and catalysts on rate Effect of temperature, concentration and catalysts on rate Reaction mechanisms and rate law Reaction mechanisms and rate law Pg 705-722