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International Baccalaureate Chemistry

International Baccalaureate Chemistry. Topic 6 - Kinetics. Review: Reaction Types. Sythesis : A + B → AB 2Na + Cl 2 → 2NaCl Decomposition: AB → A + B 2H 2 O → 2H 2 + O 2 S. Replacement: A + + BC → B + + AC Cu + + AgNO 3 → Ag + + Cu(NO 3 ) 2

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International Baccalaureate Chemistry

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  1. International Baccalaureate Chemistry Topic 6 - Kinetics

  2. Review: Reaction Types • Sythesis: A + B → AB 2Na + Cl2 → 2NaCl • Decomposition: AB → A + B 2H2O → 2H2 + O2 • S. Replacement: A+ + BC → B+ + AC Cu+ + AgNO3 → Ag+ + Cu(NO3)2 • D. Replacement: AB + CD → AD + BC KI + Pb(NO3)2 → KNO3 + PbI2 • Combustion: CxHy + O2 → CO2 + H2O CH4 + 2O2 → CO2 + 2H2O • Neutralization: HX + MOH → MX + H2O HCl + NaOH → NaCl + H2O • Redox: Still to come

  3. Kinetics: The Study of Reaction Rates • A Chemical reaction is the process in which chemical species react to form new substances. • Not all chemical reactions take place at the same rates. • Some reactions are very fast (explosions) and some are very slow (a car rusting). • Reaction Rate: A change in the concentration of reactants or products per unit of time R = Conc. of reactant used(D conc) Time interval (D time) = Conc. of Product Formed Time interval • Units are usually Mol/s = mol/L•s = mol•dm-3•s-1 • Other units are possible if properties other than concentration are used.

  4. Kinetics • Chemical reactions occur at different rates. • Chemical reaction that occur in nature: • Car rusting (Slow) • Engine Combustion (Fast) • Oxidation of copper on pennies (Slow) • Food spoiling at high temperatures (Fast) • Graphite → Diamond (Slow)

  5. Factors Affecting Reaction Rates 1. Nature of Reactants 2. Surface Area: Dust vs. solid 3. Concentration: 0.01 M vs 3 M Zn + HCl → 4. Temperature: Ice melting in hot vs cold water 5. Catalyst H2O2 decomposition normal vs with MnO2

  6. Example • If 3.0 mol dm-3 of HCl are used up after 15 s in a certain reaction with Zn metal, find the average rate of reaction. • 3.0M = 0.2 M 15 s 1 s

  7. Reaction Rates Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. [A] vs t 7

  8. Dependents • The rate of a chemical reaction is a measure of the rate at which reactants are consumed which is equal to the rate at which products are formed: Rate = -[R] = [P] t t • Where: • [R] and [P] are the concentration of the reactants and products respectively • t is a change in time

  9. Δ[B] Δ[A] 1 1 = - = - b a Δt Δt Δ[D] Δ[C] 1 1 = = d c Δt Δt General Rate of Reaction a A + b B → c C + d D Rate of reaction = rate of disappearance of reactants = rate of appearance of products

  10. Example • MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)Mn2+(aq) + 4H2O(l) + 5Fe3+(aq) Rate = -[MnO4-] = 1 [Fe3+]t 5 t • The rate of MnO4- consumption is equal to 1/5 the rate of Fe3+ production. • Any property that differs between reactants and products can be used to measure reaction rate. • Absorption of color or light • Volume, mass pressure of a gas • pH • Electrical conductivity • Temperature

  11. Δ[Fe2+] 0.0010 M Rate of formation of Fe2+= = = 2.6x10-5 M s-1 Δt 38.5 s Example • Rate of change of concentration with time. 2 Fe3+(aq) + Sn2+→ 2 Fe2+(aq) + Sn4+(aq) t = 38.5 s [Fe2+] = 0.0010 M Δt = 38.5 s Δ[Fe2+] = (0.0010 – 0) M

  12. Graphs • In all cases, a graphical plot of the measured property vs. time will allow determination of the slope of the graph, which will be equal to the reaction rate. • Example: Cu(s) + 4HNO3(aq)Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g) + heat • The rate of the above equation can be found by measuring several different properties:

  13. Possible Properties to Measure • Color change: Rate = Color Intensity Time • Temperature Change: Rate = Temperature Time • Pressure Change: Rate =  Pressure Time • Mass Change: Rate = Mass Time For the initial rate, draw a gradient (Tangent) at t = 0. These rates will be greater than those taken later on in the reaction (t=x)

  14. H2O2(aq) → H2O(l) + ½ O2(g) -Δ[H2O2] Rate = Δt Determining and Using an Initial Rate of Reaction. -(-2.32 M / 1360 s) = 1.7 x 10-3 M s-1 -(-1.7 M / 2600 s) = 6 x 10-4 M s-1

  15. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) [C4H9Cl] M In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times, t.

  16. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) Average Rate, M/s The average rate of the reaction over each interval is the change in concentration divided by the change in time:

  17. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • Note that the average rate decreases as the reaction proceeds. • This is because as the reaction goes forward, there are fewer collisions between reactant molecules (because there are fewer molecules).

  18. Collision Theory • Chemical reactions involve collisions of reactant particles. • Not all collisions lead to a chemical reaction. • For molecules to react (effective collisions), they must: • Collide with each other • Collide with correct orientation/geometry (Steric factor) • Collide with sufficient Kinetic Energy (meet the minimum activation energy (Ea) requirement) • Ineffective collisions involve particles that rebound essentially unchanged. • The rate of reaction depends upon the frequency of collisions and fractions of those collisions that are effective.

  19. Collision Theory • Kinetic-Molecular theory can be used to calculate the collision frequency. • In gases 1030 collisions per second. • If each collision produced a reaction, the rate would be about 106 M s-1. • Actual rates are on the order of 104 M s-1. • Still a very rapid rate. • Only a fraction of collisions yield a reaction.

  20. Collision Theory Furthermore, molecules must collide with the correct orientation and with enough energy to cause bonds to break and new bonds to form

  21. Factors Affecting Reaction Rate • Nature of Reactants • Surface Area Affect Collision Rate • Concentration • Temperature Affect proportion with required Ea • Catalyst

  22. Nature of Reactants • Generally, chemical reactions which involve a lot of old bond breaking and new bond forming are slow. • Example 1: 5Fe2+(aq) + MnO4-(aq) + 8H+(aq)  5Fe3+(aq) + Mn2+(aq) + 4H2O(l) • Example 2: 5C2O42+(aq) + 2MnO4-(aq) + 16H+(aq)  10CO2(g) + 2Mn2+(aq) + 8H2O(l) • Which reaction would occur faster? • In both reactions, MnO4-(aq) produces Mn2+(aq) • Change is accomplished by two different reactions. • Using Fe2+(aq), there are no bonds to break, thus the reaction would be faster. • In C2O42-(aq), there are lots of bonds to break, thus the reaction would be slower.

  23. Physical States • Another aspect to consider with respect to the nature of the reactants is the physical state the reactants are in (solid, liquid, gas, aqueous). • Example: • Solid + Gas → Slow • Liquid + gas → Slow • Gas + gas → Fast • Aqueous + Aqueous → Very fast

  24. Surface Area • The more surface area exposed to or in contact with the reactants, the faster the reaction will take place. • Smaller particles have more available surfacearea than larger particles.

  25. Example: • Surface Area only applies to heterogeneous reactions (ones involving different states) • Example 1: Wood + O2(g)→ Products + Heat • Example 2: Mg(s) + 2HCl(aq) → H2(aq) + MgCl2(aq) • Example 3: 2NaCl(aq) + Cu(OH)2(aq) → 2NaOH(aq)+ CuCl2(aq) • How would you increase the reaction rate of the above three reactions?

  26. Concentration • If the concentration of a reactant is increased, we can expect the reaction rate to increase because moreeffective collisions may occur due to the presence of more reactant particles in a given volume. • NOTE:Reducing volume has the same effect as adding more reactant. 2M H2 6M H2

  27. Temperature • A temperature increase always increases the rate of a reaction. (IB Question: State) • Reason: There will be more collisions between reacting particles at higher temperatures because they are moving faster. (IB Question: Explain) • Recall: For reactant particles to react and form products, they must collide with sufficient kinetic energy to break old bonds (i.e. collisions must be successful). • The minimum amount of energy required for a successful collision is called its activation energy (Ea).

  28. Maxwell–Boltzmann Distributions • Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. • At any temperature there is a widedistribution of kinetic energies. 28

  29. Maxwell–Boltzmann Distributions • As the temperature increases, the curve flattens and broadens. • Thus at higher temperatures, a larger population of molecules has higher energy. If the dotted line represents the activation energy, as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. As a result, the reaction rate increases. 29 29.

  30. Activation Energies • How does the change in Temperature affect the reaction rate? (possible exam question) • A temperature increase increases the number of particles with the necessary activation energy and therefore increases the reaction rate. (Full mark answer) • Activation Energy (Ea) is the minimum energy required for a successful collision, assuming optimum collision geometry.

  31. Catalysts • A catalyst is a substance that increases the reaction rate without apparently being consumed in the chemical reaction. • Example: The decomposition of hydrogen peroxide MnO2 2H2O2 2H2O + O2 • The rate of this reaction is increased by adding the catalyst MnO2. • How do Catalysts work? • They lower the Ea so that a greater percentage of moleculeswill have sufficient energy to react (E  Ea).

  32. Catalysts The catalyst is H+

  33. Potential Energy Change • To help explain Catalysts, we can also look at the Potential Energy Change in a chemical reaction. • All reactions take place in Three Steps: • Reactant particles approach each other. As they approach they begin to slow down because of electron repulsive forces. This means K.E. is decreasing and Potential Energy (P.E.) is increasing. If they have sufficient energy (E  Ea) they will: • Collide. This group of atoms formed on collisions is called the activated complex. If the molecules have enough activation energy they will react on collision. • Once product molecules form, they will separate because of repulsive forces. As they separate, KE will increase and PE will decrease.

  34. Reaction Mechanisms • The molecularity of a process tells how many molecules are involved in the process. • The rate law for an elementary step is written directly from that step.

  35. Reaction Mechanisms 4HBr(g) + O2(g) 2H2O(g) + 2Br2(g) • For this reaction to proceed as written, a five-particle collision would have to occur. • Chances of this happening are low and therefore we assume that this reaction, like most, occurs through a series of steps called a “Reaction Mechanism”. • A reaction mechanism is a sequence of simple 2-particle reactions (usually) by which a complex reaction takes place.

  36. The Steps Step 1 HBr + O2HOOBr (Slow) Intermediate Step 2 HOOBr + HBr 2 HOBr (Fast) Intermediate Step 3 HOBr + HBr H2O + Br2 (Fast) Step 4 HOBr + HBr H2O + Br2 (Fast) Net Reaction: 4 HBr(g) + O2(g) 2H2O(g) + 2Br2(g)

  37. Characteristics of a Reaction Mechanism • Each step in the mechanism is usually bimolecular (2 Particle) and never more than trimolecular (3 Particle). • The sum of the steps in the reaction mechanism must give the balanced equation for the chemical reaction. (Intermediates must be consumed) • It is always the slowest step in a series of steps which determine the rate of the overall reaction. (called the Rate Determining Step or R.D.S.) • The only way to increase the rate of the reaction is to speed up the slowest step. In the above reaction, this can be done by increasing the [HBr] and /or [O2] because these are the two reactants involved in the first step of the reaction sequence. • If a reactant is not in the Rate Determining Step, then changing its concentration will have no effect on the reaction rate.

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