Advanced Placement Presenter Copy Chemistry Rates and Mechanisms 2014
47.90 91.22 178.49 (261) 50.94 92.91 180.95 (262) 52.00 93.94 183.85 (263) 54.938 (98) 186.21 (262) 55.85 101.1 190.2 (265) 58.93 102.91 192.2 (266) H Li Na K Rb Cs Fr He Ne Ar Kr Xe Rn Ce Th Pr Pa Nd U Pm Np Sm Pu Eu Am Gd Cm Tb Bk Dy Cf Ho Es Er Fm Tm Md Yb No Lu Lr 1 3 11 19 37 55 87 2 10 18 36 54 86 Be Mg Ca Sr Ba Ra B Al Ga In Tl C Si Ge Sn Pb N P As Sb Bi O S Se Te Po F Cl Br I At Sc Y La Ac Ti Zr Hf Rf V Nb Ta Db Cr Mo W Sg Mn Tc Re Bh Fe Ru Os Hs Co Rh Ir Mt Ni Pd Pt Cu Ag Au Zn Cd Hg 1.0079 6.941 22.99 39.10 85.47 132.91 (223) 4.0026 20.179 39.948 83.80 131.29 (222) 140.12 232.04 140.91 231.04 144.24 238.03 (145) 237.05 150.4 (244) 151.97 (243) 157.25 (247) 158.93 (247) 162.50 (251) 164.93 (252) 167.26 (257) 168.93 (258) 173.04 (259) 174.97 (260) 9.012 24.30 40.08 87.62 137.33 226.02 10.811 26.98 69.72 114.82 204.38 12.011 28.09 72.59 118.71 207.2 14.007 30.974 74.92 121.75 208.98 16.00 32.06 78.96 127.60 (209) 19.00 35.453 79.90 126.91 (210) 44.96 88.91 138.91 227.03 58.69 106.42 195.08 (269) 63.55 107.87 196.97 (272) 65.39 112.41 200.59 (277) 4 12 20 38 56 88 5 13 31 49 81 6 14 32 50 82 7 15 33 51 83 8 16 34 52 84 9 17 35 53 85 21 39 57 89 58 90 *Lanthanide Series: Actinide Series: * 59 91 60 92 61 93 62 94 63 95 64 96 65 97 66 98 67 99 68 100 69 101 70 102 71 103 22 40 72 104 23 41 73 105 24 42 74 106 25 43 75 107 26 44 76 108 27 45 77 109 28 46 78 110 29 47 79 111 30 48 80 112 Periodic Table of the Elements Not yet named
Throughout the test the following symbols have the definitions specified unless otherwise noted. L, ml liter(s), milliliter(s) mm Hg millimeters of mercury g gram(s) J, kj joule(s), kilojoule(s) nm nanometer(s) V volt(s) atm atmosphere(s) mol mole(s) ATOMIC STRUCTURE E hν c λν E energy ν frequency λ wavelength Planck s constant, h 6.626 10 34 J s Speed of light, c 2.998 10 8 ms 1 Avogadro s number 6.022 10 23 mol 1 Electron charge, e 1.602 10 19 coulomb EQUILIBRIUM c d [C] [D] K c a b, where a A + b B c C + d D [A] [B] c d ( P K p C)( PD ) a b ( PA) ( PB) + - [H ][A ] K a [HA] - + [OH ][HB ] K b [B] K w [H + ][OH ] 1.0 10 14 at 25 C Equilibrium Constants K c (molar concentrations) K p (gas pressures) K a (weak acid) K b (weak base) K w (water) K a K b ph log[h + ],, poh log[oh ] 14 ph + poh ph pk a + log [A - ] [HA] pk a logk a, pk b logk b KINETICS ln[a] t ln[a] 0 kt 1-1 kt [ A] [ A] 0 t t ½ 0.693 k k rate constant t time t ½ half-life
GASES, LIQUIDS, AND SOLUTIONS PV nrt P A P total X A, where X A P total P A + P B + P C +... n m M K C + 273 D m V KE per molecule 1 2 mv2 moles A total moles Molarity, M moles of solute per liter of solution A abc P V T n m M D KE Ã A a b c pressure volume temperature number of moles mass molar mass density kinetic energy velocity absorbance molarabsorptivity path length concentration -1-1 Gas constant, R 8.314 J mol K 0.08206 L atm mol K -1-1 62.36 L torr mol K 1 atm 760 mm Hg 760 torr STP 0.00 C and 1.000 atm -1-1 THERMOCHEMISTRY/ ELECTROCHEMISTRY q mcdt DS ÂS products -ÂS reactants DH ÂDH products -ÂDH reactants f DG ÂDG products -ÂDG reactants DG DH - TDS -RT ln K -nfe q I t f f f q heat m mass c specific heat capacity T temperature S standard entropy H standard enthalpy G standard free energy n number of moles E I q t Faraday s constant, F standard reduction potential current (amperes) charge (coulombs) time (seconds) 96,485 coulombs per mole of electrons 1 joule 1volt 1coulomb
KINETICS Collisions, Differential Rates, and Mechanisms What I Absolutely Have to Know to Survive the AP Exam The following might indicate the question deals with kinetics: rate; reactant concentration; order; rate constant; mechanisms; rate determining step; intermediate; catalyst; half-life; instantaneous rate; relative rate; activation energy; integrated rate law; rate expression; rate law Temperature Nature of the Reactants Concentration of the Reactants Catalyst Surface Area Inert Gas FACTORS THAT AFFECT REACTION RATE Reaction rate increases with increasing temperature heat `em up, speed `em up The higher the temp the more likely there are more collisions that are able to overcome the activation energy and form products Phases of matter play a role, chemical identity (ions of opposite charge typically react faster; molecules with more bonds to break and form and substances with stronger bonds react more slowly) Reaction rate typically increases with increasing concentration of reactants (pressure changes have the same effect on gaseous reactions) more reactants the more effective collisions, the faster the reactions proceeds Reaction rate increases with the addition of a catalyst; not a part of the reaction and is not used up but typically participates in the rate determining step of the reaction (slowest step) it LOWERS the activation energy! Reaction rate increases with increased surface area of the reactant; reactions occur at the exposed surface of the two substances (except in g or aq) thus the greater the surface area the more exposure the greater chance of effective collisions, the greater the rate. Aqueous solutions offer the ultimate exposure. DON T FALL FOR THIS adding an inert gas will NOT effect the reaction rate; it is not part of the reaction mechanism thus is has NO EFFECT Collision Theory of Reactions For a reaction to happen two things must occur: 1. The molecules must collide with the appropriate orientation for a reaction to occur 2. Molecules must collide with enough kinetic energy to react i.e. they must meet or exceed the energy of activation, E a In other words the collisions MUST BE EFFECTIVE. If this happens the product of an effective collision is at the peak of the energy hump (activated complex or the transition state. This complex can either fall back to reactants or produce products E a the activation energy Is a measure of the energy barrier colliding molecules must overcome if they are to react rather than recoil from one another. It is assumed that reacting species with energy less than E a will not react and those with energy greater than E a and the proper orientation will typically react. Copyright 2013 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 1
Kinetics 1 Collision Theory of Reactions Con t When the temperature is increased, the molecules speed up and collide more frequently with more energetic collisions. Only those particles represented by the area to the right of the activation energy mark will react when they collide. The majority don't have enough energy, and will simply bounce apart. If we increase the temperature from T 1 to T 2, more molecules are energetic enough to achieve the E a, thus more molecules react and the overall reaction rate increases. Notice how many more molecules are to the right of the activation energy and thus will react when they collide. A general rule of thumb is that reaction rate doubles for each 10 C increase in temperature! EXPRESSING REACTION RATE Reaction rate is expressed in terms of how fast the concentration of a substance changes. It does not matter whether you measure the rate the products are formed or the rate at which the reactants are consumed they are stoichiometrically linked just focus on either the disappearance of the reactants ( ) or the appearance of the (+). For example: The relative rate can be expressed as: rate 1 Δ[N 2 O 5 ] 1 Δ[NO 2 ] 2 Δt 4 Δt Rate of reaction Δ[O 2 ] Δt Δ[product] time or Rate of reaction 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) Δ[reactant] time Translated The rate of consumption (disappearance) of N 2 O 5 is equal to half the rate of the production (appearance) of NO 2 and equal to twice the production (appearance) of O 2 Or better yet NO 2 is produced at twice the rate at which N 2 O 5 is consumed O 2 is produced at half the rate at which N 2 O 5 is consumed Instantaneous Rate Copyright 2013 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 2
Kinetics 1 Instantaneous rate is the rate at any one point in time during the experiment. To find instantaneous reaction rate you find the slope of the curve at the time in question (for those of you in calculus aka derivative) i.e. the slope of the tangent line to that point in time. Differential Rate Laws Concentration and Rate The rate law is a mathematical equation relating the instantaneous rate at a particular point in the progress of a chemical reaction to the concentration of the reacting species. Rate generally refers to the initial rate. The initial rate is the fastest rate of the reaction and occurs at the very beginning of the reaction. At this point there are few competing reactions, little chance for the reverse reaction (think equilibrium) to get involved, etc. It should be noted when using the initial rate the concentration of the reactants are initial concentrations. For the following reaction 2 X + Y Z the general form of the rate law is rate k[x] m [Y] n Where k is the rate constant The exponent m represents the order of the reaction with respect to reactant X The exponent n represents the order of the reaction with respect to reactant Y The sum of m + n represents the overall order of the reaction. Reactant orders must be determined experimentally; they cannot be written from the overall balanced equation Differential Rate Laws Concentration and Rate Con t Using the experimental data shown below you can determine the order of each reactant, the overall reaction order, and Copyright 2013 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 3
Kinetics 1 calculate the rate constant, k. Examine the data with an analytical eye. Look for two trials where the concentration of ONE reactant was held constant. Then compare how the other reactant changed and its effect on the initial rate. Look at Experiments 1 & 4 [A] was held constant and [B] doubled the initial rate of the reaction doubled, therefore the reaction is FIRST ORDER with respect to reactant B. Look at Experiments 1 & 3 [B] was held constant and [A] doubled the initial rate of the reaction was unchanged, therefore the reaction is ZERO ORDER with respect to reactant A. rate k[a] 0 [B] 1 or rate k[b] Use experiment 5 (can use any of them) to solve for the rate constant, k. rate k[b] rate 1.50 10 2 k 0.025 hr 1 [B] 0.60 Reaction Mechanisms Reaction mechanisms attempt to describe the stepwise sequence of elementary reactions that take reactants to products. The mechanism describes in detail the bonds that are broken and formed as the reaction proceeds. Each elementary step of a mechanism typically involves one or two reactants forming products. Every mechanism consists of a series of stepwise reactions. Each reaction in the mechanism has a rate associated with it. The overall speed of the reaction depends upon the slowest step of the mechanism. The rate law of this step is identical or equivalent to the experimental rate law. The slow step of the mechanism is also called the rate-determining step of the mechanism. The sum of all the steps of the mechanism must equal the overall balanced chemical equation. The coefficients of the reactants in the rate-determining step of the mechanism must correspond to the exponents or order of the reactants in the experimental rate law. Catalysts and Intermediates A catalyst is used up early in a reaction (reactant) and is regenerated (product) in a subsequent step. A catalyst is a substance that acts to increase the rate of a chemical reaction by providing an alternate path for the reaction to occur. This means that there will be a change in the magnitude of the rate constant and possibly a change in the order of the reaction. An intermediate is produced early in the reaction (product) and used up (reactant) in a subsequent step. Copyright 2013 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 4
Kinetics 1 Kinetics Cheat Sheet Differential Rate Law (concentration vs. rate data): Relationships Rate k [A] x [B] y Be able to explain with algebraic equations or words how an order is determined. It is important to state which concentration(s) is/are held constant and which concentration is varied as well the effect that has on the rate of the reaction if you choose not to justify with algebraic equations. Discuss number of effective collisions in relation to increasing or decreasing rates rate in terms of is code for relative rates use stoichiometry ratios on rate value Mechanisms must agree with the stoichiometry of the reaction and the summary rate law must agree with the slow step; identify intermediates and catalysts and clearly state that the correct mechanism agrees with the experimentally determined rate law. E a predicts speed but the relationship is an inverse one; high slow rate; low fast rate Instantaneous rate slope of the line tangent to the time point in question 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) rate 1 Δ[N 2 O 5 ] 1 Δ[NO 2 ] 2 Δt 4 Δt Δ[O 2 ] Δt Catalyst lowers Activation Energy and provides an alternate Distinguish between catalyst and intermediate pathway/mechanism especially in the steps of a mechanism! Temperature increases of approx.. 10 C Connections Stoichiometry using up one component of the system Electrochemistry if reaction is redox in nature, rate might indicate a limiting reactant in effect problems could come into play Thermochemistry E a and ΔH rxn and reaction diagrams Potential Pitfalls Units on k! Make sure you can solve for units for k Copyright 2013 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 5
Kinetics 1 NMSI SUPER PROBLEM X(g) + 2Y (g) Z(g) The reaction represented above was studied at 25 C. The data collected are shown in the table below. Experiment [X] [Y] Initial rate of formation of Z (mol L 1 sec 1 ) 1 0.200 0.200 1.20 10 5 2 0.200 0.400 2.40 10 5 3 0.100 0.200 6.00 10 6 (a) Calculate the initial rate of disappearance of substance Y in Experiment 1. Since 2 mol Y is reacted for every 1 mol of Z formed the rate of disappearance of Y must be TWICE the rate of formation of Z 2.4 10 4 mol L 1 s 1 1 point is earned for the correct value of the rate of disappearance of substance Y. or rate 1 Δ[Y] Δ[Z] 2 Δt Δt Δ[Y] 2 Δ[Z] Δt Δt Δ[Y] 2 1.2 10 5 Δt ( ) 2.4 10 5 (b) Determine the order of the reaction with respect to each reactant. Show your work. i. X 0.200 0.100 2 x 2 x 0.200 0.200 y 1.20 10 5 6.00 10 6 x 1 FIRST order with respect to X Alternatively, in experiments 1 and 3, [Y] was held constant while [X] was doubled which doubled the rate. Thus the reaction is FIRST order in [X]. 1 point is earned for the correct order with work or justification Copyright 2013 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 6
Kinetics Rates and Mechanisms ii. Y 0.200 0.200 2 y 2 x 0.400 0.200 y 2.40x10 5 1.20x10 5 y 1 1st order with respect to Y Alternatively when experiment 2 is compared to 1, [X] remains constant while [Y] doubles. This caused the reaction rate to to double, thus that the reaction is 1 st order in [Y] 1 point is earned for the correct order with work or justification (c) Write the rate law for the reaction consistent with part B. Rate k[x][y] 1 point is earned for the correct rate law consistent with part (a) (d) Calculate the value of the rate constant, k. Be sure to include proper units. rate k[x][y] rate [X][Y] 6 6.00 10 k [0.100][0.200] 3.00 10 4 L mol 1 sec 1 1 point is earned for the correct value of k 1 point is earned for the correct units for k (e) In a closed 2.50 L reaction vessel at 22 C, 0.0254 mole of substance X was reacted with 0.0495 mol Y i. Determine the limiting reactant. Justify your answer mathematically. X(g) + 2Y (g) Z(g) 0.0254 mol 0.0495 mol 0.0248 mol 1 2 ( 0.0495 mol Y ) Therefore X is in excess by 0.0006 mol and Y is limiting 1 point is earned for the correct limiting reactant with justification ii. Calculate the number of moles of Z formed. Since Y is limiting: X(g) + 2Y (g) Z(g) 0.0254 mol 0.0495 mol ( 0.0495 mol Y) 1 mol Z 2 mol Y 0.0248 mol Z 1 point is earned for the correct number of moles of Z consistent with the answer in part (e) i. Copyright 2013 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 7
Kinetics Rates and Mechanisms iii. Calculate the total pressure in the flask at the completion of the reaction. Since Y is limiting: X(g) + 2Y (g) Z(g) 0.0254 mol 0.0495 mol 0 mol 0.0248 mol 0.0495 mol +0.0248 mol At completion of the reaction: 0.0006 mol X and 0.0248 mol Z present P Total V n Total RT P n Total RT V (0.0006 + 0.0248)(0.0821)(295) 0.246 atm (2.50) 1 point is earned for the total mol of gas present consistent with the answers given in part (e) i and ii. 1 point is earned for the total pressure consistent with the answers given in part (e) i and ii (f) Three possible mechanisms for this reaction are shown below. Y + Y D X + D Z Mechanism 1 Mechanism 2 Mechanism 3 (fast) (slow) X D D + Y G G + Y Z (slow) (fast) (fast) X + Y D Y + D Z (slow) (fast) i. Select the one most consistent with the experimental data. Justify your choice by writing a rate law for each of the three mechanisms. Mechanism 3 is most consistent with the experimental data. Its rate law is the same as that determined from the experimental data, rate k [X][Y] The rate law for Mechanism 1 would be rate k [X][D] which would be rate k [X][Y] 2 solving for the intermediate D which is not consistent with the experimental data. The rate law for Mechanism 2 would be rate k [X] which is not consistent with the experimental data. 1 point is earned for the correct mechanism. 1 point is earned for justification, with the rate laws, for each mechanism. ii. Identify substance D in the mechanisms shown above as an intermediate or a catalyst. Justify your answer. Substance D is an intermediate because it is a product of one step of the reaction that is used as a reactant in a later step. 1 point is earned for identifying and justifying D as an intermediate. Copyright 2013 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 8
Kinetics 1 (g) The following diagram shows the energy of the reaction as the reaction progresses. i. Clearly label the activation energy for the forward reaction. ii. Clearly label the enthalpy change for the reaction. iii. On the diagram draw a second energy curve showing the effect of a catalyst on the reaction. 1 point is earned for correctly labeling the activation energy, E a E a 1 point is earned for correctly labeling the enthalpy change, ΔH. ΔH 1 point is earned for sketching the plot of the catalyzed reaction must start and end at the same energy level and exhibit a lower activation energy, E a (h) The collision between X and Y occur with enough energy to overcome the activation energy barrier, E a, however no products are formed. Identify and explain one other factor that affects whether the collision will result in a reaction. For a collision to be successful, the molecules must have the correct orientation. Only collisions with the correct orientation will be able to begin to form and begin to break the bonds as the transition state is approached. In other words the molecules must contact each other at very specific locations on their surfaces for the transition state to be possible. 1 point is earned for the identifying that the correct orientation for the collision is required. 1 point is earned for the explanation that refers to specific locations of the molecules must collide. Copyright 2013 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 9