Kinetics and Mechanism

1. KineticsandMechanism

2. 16.2 Reaction mechanism 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.

3. 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.

4. Sometimes the molecularity of a reaction can be used to guess if an elementary reaction is a rate–determining step. Molecularity is the count of reacting particles in an elementary reaction. Termolecular reaction are almost always rate limiting because it is a low probability event to get three particles to hit in the same location at the same time

5. 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.

6. Step 1 Rate = k1[I2] Step 2 Rate = k–1[I]2 Step 3 Rate = k2[H2][I]2

7. Reaction Mechanisms • The rate-determining step is the crucial step in establishing the rate of the overall reaction.( it is the slowest step and also the one with the highest AE) • 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.

8. A Mechanism With A Slow Step Followed By A Fast Step- A Plausible Mechanism Slow step: H2O2 + I- H2O + OI- Fast step: H2O2 + OI- H2O + O2 + I- ___________________________________________________________ Net equation: 2 H2O2 2 H2O + O2 I- - catalyst; OI- - intermediate The slow step is the rate-determining step. Rate = rate of slow step = k[H2O2][I-]

9. Mechanisms • Intermediate • A species that is created in one step and consumed in the other • Catalyst • 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.

11. Since the energy of activation for the first step is so much higher than that for the second step, the first step of the an SN1 mechanism is the rate-limiting or rate-determining step.

12. A Mechanism With A Slow Step Followed By A Fast Step Slow step: H2O2 + I- H2O + OI- Fast step: H2O2 + OI- H2O + O2 + I- Overall: 2 H2O2 (aq)  2 H2O (l) + O2 (g) Facts: (1) The rate of decomposition of H2O2 is first order in both H2O2 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.

13. A Mechanism With A Fast Reversible Step Followed By A Slow Step 2 NO (g) + O2 (g)  2 NO2 (g) Experimentally found: Rate = k[NO]2[O2]

14. A Mechanism With A Fast Reversible Step Followed By A Slow Step – A Plausible Mechanism k1 Fast step: 2 NO < -- > N2O2 k-1 k2 Slow step: N2O2 + O2 2 NO2 Net equation: 2 NO + O2 2 NO2

15. Molecularity • The number of molecules that participate as reactants in anelementary step • Unimolecular: a single molecule is involved. • Ex: CH3NC (can be rearranged) • Radioactive decay • Its rate law is 1st order with respect to that reactant

17. Bimolecular: Involves the collision of two molecules (that form a transition state that can not be isolated) • Ex: NO + O3 NO2 + O2 • It’s rate law is 1st order with respect to each reactant and therefore is 2nd order overall. • Rate =k[NO][O3]

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

21. Rate Laws for Elementary steps • Can use the equation coefficients as the reaction orders in the rate law for an elementary step

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

23. 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.

24. Exothermic • activation energy (the energy needed so that the reactants bonds will break and reform to make product)

25. 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

26. 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.

28. Fast and slow reactions • The smaller the activation energy, the faster the reaction will occur regardless if exothermic or endothermic.

29. Practice: • 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. • What type of reaction is this? • Calculate the activation energy for the reverse reaction. • Calculate the ΔH for the reverse reaction.

30. 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?

31. 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.

32. 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-Ea/RT Ea – activation energy R – gas constant, 8.3145 J mol-1K-1 T - Kelvin temperature A – Arrhenius constant (depends on collision rate and shape of molecule)

33. k = A e-Ea/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

34. ln k and 1/T is linear • With R known, we can find Ea graphically from a series of k values at different temperatures. • lnk2 = - Ea ( 1/T2 – 1/T1) k1 R • Ea = -R (ln k2 / k1) ( 1/T2 – 1/T1) -1

35. 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.

36. Solution Ea = -R (ln k2 / k1) ( 1/T2 – 1/T1) -1 Ea = - (8.314)( ln 1.10 x10-5/ 9.51 x109)(1/600.00 – 1/500.0) = 1.76 x 105 J/mol

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

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

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