CH111 SOS Study Session

1. CH111 SOS Study Session CH111 SOS SessionBy: Matt Nichols + Mike Golian

2. Goals of chapter 12 • Describe and predict changes in concentration with time • Describe and predict how the rate of the reaction is affected by concentration. • Use reaction rates to figure out mechanism: which bonds break and form in what order in a reaction • Understand the temperature dependence of reactions. • Understand how catalysts speed reactions.

3. Reactions • Molecules only undergo a reaction if: • They have enough energy • The geometry of the collision is conducive to the reaction • All reactions undergo elementary reactions (the most basic reactions) • Usually involve only one or two molecules • Reaction mechanisms are multi-step reactions • Many elementary reactions simplified into one chemical equation • The slowest elementary reaction in the reaction mechanism is called the rate determining step

4. Reactions • Atoms, ions, and molecules recombine during chemical reactions • Reactions do not occur all at once, but are made up of sequential molecular events making a reaction mechanism • A reaction mechanism is a step by step process, where each step is referred to as an elementary reaction • Elementary reactions makes up each step in the sequential molecular events ,and when put together make up the reaction mechanism • Elementary reactions are generally comprised 1,2 or 3 reactants • Overall reaction describes the starting materials and final products of a reaction • Overall reactions are not generally elementary as they do not show the individual steps of the reaction

5. Elementary reactions • A bimolecular reaction is the most common elementary reaction • Occurs with the collision of 2 atoms/molecules/ions • Characteristics feature is the collision of 2 species creating a collision complex that results in rearrangement of chemical bonds

6. Elementary reactions continued • A unimolecular reactions occurs when a single molecule fragments into 2 species or rearranges to form a new isomer

7. Elementary reactions cont’d • A termolecular reaction occurs when 3 or more species collide and react • Less common as the probability of 3 species colliding at the same time with the correct reaction geometry is small • 3 species reaction commonly occur in 2 steps • with the collision of 2 species forming a complex in the first step • And the collision of the 3rd species with the complex from the first step • This means that it can be written as a product of bimolecular and unimolecular elementary reaction

8. Reaction intermediates • A species that is produced in the first step and then consumed in a later step • Will not appear in the overall reaction • Intermediate represented by the collision

9. Rate Law Relates the rate of reaction with the concentration of reactants through a rate constant (k)  experimentally determined Rate law can contain concentrations of chemical species that do not appear in the overall chemical reaction - occurs when an intermediate is a rate determining step aA + bB + cC...  products Rate = k[A]m[B]n[C]p... • Exponents in rate law do not depend on stoichiometric coefficients • Exponent is the order of reaction with respect to that species • Exponent dependant on the reaction mechanism • Sum of the exponents is the overall reaction order • M+N+P =overall order of reaction

10. Rate constants • Every reaction has characteristic rate constant • Has units of concentration/time • Remains constant over the reaction even with change concentrations of reactants/products • Make sure units of K give a rate unit that makes sense • A rate law has the form R= k[A]^2[B], when [A]=1.12M and [B]=0.87M and the rate of reaction is 4.78x10^-2 M s-1 what is the value of K including units?

11. First Order Reaction The general rate law is Rate = k[A] • Integrated rate law  • Plotting on a graph  • Plot as vs. time Gives rise to a linear graph

12. First order example • Decomposition of hydrogen peroxide is a first order reaction • 2H202 2H20 + 02 • Initial [H202] = 2.32 M • K= 7.30 x 10^-4 S^-1 • What is the rate law? • What will the concentration of hydrogen peroxide be after 1200 s • For appearance of product

13. Half life of first order reaction • The time it takes to reduce initial product concentration to 50% of original value • t1/2 isThe point where [A] = 0.5[Ao] • Half life is only dependant upon the rate constant • What is the half life of hydrogen peroxide from the previous slide (K= 7.30 x 10^-4 S^-1)

14. Example – first order • At what time will the 30.6% of the initial hydrogen peroxide have decomposed? • K= 7.30 x 10^-4 S^-1 • Do not need initial concentration as they cancel out!!! • Assuming 100% initially

15. Zero Order Reaction Rate = k[A]^0 • Rate constant is independent of the initial reactant concentrations • Will take form of y=mx+b, where [At]= -kt + [Ao] • Rate of reaction remains constant throughout the reaction • Don’t confuse with 1st order decomposition reaction • Has K units of concentration over time ( ie. M s^-1) as there are no units to cancel • Plotting on a graph • Plot as [A] vs. time • Will be a straight line with slope equal to the negative value of K

16. Second Order Reaction Rate = k[A]2 • Integrated rate law  • Plotting on a graph  • Plot vs. time • Half-life 

17. Second order example • The data in the table below is from the decomposition of A  products • At this point it is very useful to graph your data • Data analyzed by each method to save time • What is the value of k with units? • Derive the half life equation!! ( simple substitution) • What is the half life time?

18. Experimental determination of reaction rate • A + B  2C • What is the order of the reaction • What is the K value with units?

19. Pseudo first order reactions • Can be possible to simplify some second order reactions where one reagent is in great excess of the other to a pseudo first order reaction • Suppose we follow a 1L hydrolysis reaction of 0.01M ethyl acetate to completion ( creates ethanol and acetic acid) • Initially there is ~ 1000g water 55.5 M, when completed there is 55.5-0.01= ~55.5M • As Water concentration essentially remains constant, reaction rate can be treated as independent of water concentration ( exponent 0), which simplifies this to a first order reaction • Can apply first order reaction kinetics

20. Summary of rate parameters

21. What if the first step isn’t rate determining?? • All examples this far have had the first step being the rate determining step, and as no step can go faster than the slowest elementary reaction, all steps after this have no effect on the reaction rate • For many reactions the rate determining step occurs in the 2nd or 3rd elementary reaction, thus the overall reaction rate is dependant upon this step and must be related to the initial reactants • Further, some 2nd and 3rd elementary reactions are comprised of intermediates that do not appear in the overall reaction • Doesn’t make sense to have a rate law with an intermediate…must relate to the initial reactants

22. Rate determining step not first problem • H2 + Br2  2 HBr • Appears as simple 1:1 stoichiometry, however if 1st step were to be rate determining a ½ Br2 would be necessary, which doesn’t exist • Reaction mechanism: Br2 2Br (fast, reversible) Br + H2  HBr + H (slow rate determining) H + Br2  HBr (fast)

23. Br2 2Br (K1) 2Br  Br2 ( K -1) Br + H2 HBr + H ( K2)*** H + Br2  HBr (K3) --------------------------------------------------- Net reaction H2 + Br2 2 HBr rate = k2[Br][H2] although this rate law accurately describes the reaction it cannot be used as it is very hard to measure the concentration of the intermediates in reactions as they are very short lived and hard to detect

24. Relate [Br] to [Br2] • Assume forward and reverse rates are equal • Br2 2Br • Relate equations • K1 [Br2]=K-1 [Br]2 …………..rearrange • [Br]2= (K1/K-1) [Br2] ……square root both sides • [Br] = (K1/K-1) ^(1/2) [Br2] ^(1/2)…substitute back for [Br] • Rate = k[H2] [Br2] ^(1/2)

25. Activation Energy • The minimum amount of energy needed above the average kinetic energy for a molecule to undergo a chemical reaction • The activation energy directly affects the rate of a reaction • Activation arises from the energy requirement to distort and/or break chemical bonds of the reactants • Activated complex is the highest point on an activation energy diagram, and is not considered an intermediate as these are stable for a short period

26. Temperature and Reaction rates and the Boltzman distribution • At any T, a larger fraction of molecules can react if Ea is small compared to a larger Ea • For any reaction a larger fraction of molecules can react if T is high rather than low • Vertical line represent the activation energy requirement of a reaction • The greater the number of molecules to the right of the vertical line the faster the reaction will proceed

27. Arrhenius Equation -Rate constant is dependant upon activation energy, temperature of the system, and collision frequency, and collision orientation -Arrhenius equation relates these variables • Energy, orientation and collisions taken into account in the equation

28. The Arrhenius equation can be rewritten to remove the exponent: • When there are two different temperatures and the activation energy is being calculated, the Arrhenius equation can be modified into:

29. Example 1 • If T2 has a value k = 9.63x10^-5 s^-1, and • T1 = 298K and has a value K1 = 3.46x10^-5 s^-1 • Ea = 106 kj/mol • What is the value of T2?

30. Example 2 • In a reaction that was done at 0 °C, the reaction coefficient was found to be 4.92x10-12 sec-1. In a second reaction at room temperature, the reaction coefficient was found to be 4.15x10-12 sec-1. What is the activation energy of the reaction?

31. Catalysts • Molecules that are involved in a reaction but are regenerated at the end • Help to speed up a reaction by providing a different reaction mechanism • Reaction mechanism has a lower activation energy • Substrate – the species that is catalysed • Two types of catalysts: • Homogenous • Heterogenous

32. Heterogenous Catalysis • A catalyst that is not in the same phase as the reaction • Example is a Catalytic converted in car • Four steps are involved in the catalysis of a substrate • Adsorption of substrate to catalyst • Migration to active sites • Reaction • Desorption of products

33. Homogenous catalysts • A catalyst that is in the same phase as the reaction • Example. Chlorine gas produced from CFC’s that react with ozone • CF2Cl2  CF2Cl + Cl • Cl + O3 ClO + O2 • O3 + O2O2 • Chlorine acts as the catalyst in ozone decomposition

34. Biological Catalysts • Organisms have developed enzymes to catalyze reactions • Enzymes are highly substrate-specific • Follow the same mechanism as heterogenous catalysis • Derive the rate law for an enzyme K1 K2 K-1

35. Dynamic Equilibrium • State of a reaction in which there is no net change of the amount of products and reactants • Formation of products = formation of reactants • Reaction is assumed to be reversible • Equilibrium is shown as: aA + bB ↔ cC + dD

36. Equilibrium Constant • Describes the reversibility of a reaction • Equation denoted as: • aA + bB ↔ cC + dD • Many different forms are present: • Kp rate of production (forward) • Kd rate of decomposition ( reverse) • Reaction has to be at equilibrium to use the equilibrium constant! • Eq’m constant ( Keq) = (Kp/ Kd )

37. Properties of Equilibrium Constants • Keq is related to the stoichiometry of the balanced net reaction • Keq applies only under equilibrium conditions • Keq is independent of the initial conditions • Is constant for any particular reaction at a given temperature, initial conditions can be any mixture of reactants and products and will reach a state of equilibrium determined by value of Keq

38. Homogenous Rules All components in the same phase (usually aqueous or gas) All components are used in the equilibrium equation N2 (g) + H2 (g) ↔ NH3 (g) 1 2 3

39. Heterogeneous equilibrium • Concentration of a pure solid or liquid does not vary significantly • The amount of a solid liquid can change but he number of moles per unit volume essentially remains fixed • Means that their concentrations are always equal to their standard concentration • If the concentrations are the same when they are entered into the equilibrium equation will results in a value of 1 • Allows us to omit pure solids and liquids from the equilibrium constant

40. Heterogeneous • CaCO3 (s) ↔ CaO (s) + CO2 (g) • By definition the equilibrium constant should be • Pure liquids and solids are excluded from the equilibrium equation, results in…

41. Water in aqueous equilibrium • Water is often a reagent in aqueous equilibrium • H2O (l) + CO2 (g)  H2CO3 (aq) • The concentration of carbon dioxide is in bars and concentration of carbonic acid is in moles • What are the appropriate units for water as it is neither a pure liquid or a solute? • Is expressed in its mole fraction Xsolvent = nsolvent / ntotal

42. Water in aqueous equilibrium • The mole fraction of water varies only slightly as solutes are added • 1L of pure water contains 55.5 moles, if a solute is dissolved to 0.5M the mole fraction of water is 0.99 ( very close to 1) • Equilibrium calculations are rarely more accurate than 5%, thus water can be treated as a pure liquid as its concentration is essentially constant

43. Relationships involving chemical equilibrium • Forward and reverse reactions are balanced exactly, the equilibrium constant can be expressed in the reverse order • Reverse equilibrium constant in the reciprocal of the forward eq’m constant • Multiplying reaction stoichiometry by a constant rasies Keq to that power • If 2 reactions are combined their Keq are multiplied

44. Example problem • N2(g) + O2(g) 2 NO(g)   Kc1 = 2.3 x 1019 • 2 NO(g) + O2(g) 2 NO2(g)   Kc2 = 3 x 106 • What is the net reaction? • If the reactions are combined to show the relationship between N2(g), O2(g) and NO2(g) what is the equilibrium constant?

45. Magnitudes of chemical equilibria • Equilibrium constants vary tremendously in value • If Keq is large (eg. 10^15) majority will be products • If Keq is small (eg. 10^-15) majority will be reactants • If Keq is neither small or large ( eg. 3) there will be a mix of both reactants and products N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)

46. Reaction Quotient • Denoted as Q • Uses the same formula as K • Used to determine whether an equilibrium reaction is at equilibrium or not • If not, can also be used to determine which way the reaction will proceed to get to equilibrium • If Q < Keq, reaction proceeds toward reactants • If Q = Keq, reaction is at equilibrium • If Q > Keq ,reaction proceeds towards products

47. Gibbs’ Free Energy and Equilibrium • Can elucidate the spontaneity of a reaction • Negative value indicates a reaction will proceed forward in the direction written • ΔG = Δ Go + RT ln Q • Equation is used to relate ΔG with the concentration of reactants through the reaction quotient Q • ΔG = 0 when Q= Keq , when the system in in equilibrium

48. Example problem • N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) • The production of ammonia gas has a Δ Goformation= -16.4 kJ/mol • Δ Goformation = 0 for N2 (g) and H2 (g) • What is Δ Goreaction ? • What is the Keq ?

49. Temperature and Equilibrium • Temperature affects equilibrium just like it affects the rate of a reaction • Will cause the Keq to change • Exothermic reaction has a negative Δ Ho Which makes the first term positive, increasing temperature will lower the value of the first term, which lowers Keq

50. Example problem • CH4 (g) + H2O (g)  CO(g) + 3H2(g) • Hydrogen gas is produced by the above reaction • This is an endothermic reaction and for it to proceed forward reaction temperatures are often carried out above 1000K • Δ Ho = +206 kJ/mol, Keq =1.3x10^-25 at 298 k • What is Keq at 1500 K? what does value imply about hydrogen gas production compared to 298K?