Advanced Placement Chemistry Integrated Rates 204
47.90 9.22 78.49 (26) 50.94 92.9 80.95 (262) 52.00 93.94 83.85 (263) 54.938 (98) 86.2 (262) 55.85 0. 90.2 (265) 58.93 02.9 92.2 (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 3 9 37 55 87 2 0 8 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.0079 6.94 22.99 39.0 85.47 32.9 (223) 4.0026 20.79 39.948 83.80 3.29 (222) 40.2 232.04 40.9 23.04 44.24 238.03 (45) 237.05 50.4 (244) 5.97 (243) 57.25 (247) 58.93 (247) 62.50 (25) 64.93 (252) 67.26 (257) 68.93 (258) 73.04 (259) 74.97 (260) 9.02 24.30 40.08 87.62 37.33 226.02 0.8 26.98 69.72 4.82 204.38 2.0 28.09 72.59 8.7 207.2 4.007 30.974 74.92 2.75 208.98 6.00 32.06 78.96 27.60 (209) 9.00 35.453 79.90 26.9 (20) 44.96 88.9 38.9 227.03 58.69 06.42 95.08 (269) 63.55 07.87 96.97 (272) 65.39 2.4 200.59 (277) 4 2 20 38 56 88 5 3 3 49 8 6 4 32 50 82 7 5 33 5 83 8 6 34 52 84 9 7 35 53 85 2 39 57 89 58 90 *Lanthanide Series: Actinide Series: * 59 9 60 92 6 93 62 94 63 95 64 96 65 97 66 98 67 99 68 00 69 0 70 02 7 03 22 40 72 04 23 4 73 05 24 42 74 06 25 43 75 07 26 44 76 08 27 45 77 09 28 46 78 0 29 47 79 30 48 80 2 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 0 34 J s Speed of light, c 2.998 0 8 ms Avogadro s number 6.022 0 23 mol Electron charge, e.602 0 9 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 ].0 0 4 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 ] 4 ph + poh ph pk a + log [A - ] [HA] pk a logk a, pk b logk b KINETICS ln[a] t ln[a] 0 kt - kt [ A] [ A] 0 t t ½ 0.693 k k rate constant t t ½ half-life Page 43
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 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 - - Gas constant, R 8.34 J mol K 0.08206 L atm mol K - - 62.36 L torr mol K atm 760 mm Hg 760 torr STP 0.00 C and.000 atm - - 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 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 standard reduction potential current (amperes) charge (coulombs) (seconds) q I t Page 44 Faraday s constant, F 96,485 coulombs per mole of electrons joule volt coulomb
KINETICS Integrated Rates Putting it All Together 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 Instantaneous Rate Instantaneous rate is the rate at any one point in during the experiment. To find instantaneous reaction rate you find the slope of the curve at the in question (for those of you in calculus aka derivative) i.e. the slope of the tangent line to that point in. Copyright 204 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page
Integrated Rate Concentration and Time Where the differential rate law expresses rate as a function of reactant concentration(s) at an instant in (hence instantaneous rate), integrated rates express the reactant concentrations as a function of. To solve integrated rate problems, construct a graph with on the x-axis and then make 3 plots where the y-axis is Concentration of A [A] vs. t Natural log of the concentration of A ln [A] vs. t Reciprocal of [A] [A] vs. t LINEAR IS THE WINNER Zero Order reaction are linear for [A] vs. t First Order reactions are linear for ln [A] vs. t Second Order reactions are linear for [A] vs. t Think y mx + b It is imperative that you can determine reaction order simply by analyzing a graph. What is important? What is plotted on each axis? What does the slope of the line indicate? If you know this, the order and rate constant can easily be determined. For the reaction Zero Order Reactions A(g) products Δ [ A] rate k[a] 0 Δt When this relationship is integrated from t to t t then [A] t kt + [A] 0 y mx + b Where [A] t is the concentration at t [A] t represents what is plotted on the y-axis [A] 0 is the initial concentration t is the represent what is plotted on the x-axis k is the rate constant - which is the SLOPE of the graph! Notice that in this equation k has a negative sign; thus the SLOPE of the graph is negative not the rate constant If the reaction is zero order then the graph will be linear with a negative slope. [A] slope k Copyright 204 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 2
For the reaction First Order Reactions A(g) products Δ [ A] rate k[a] Δt When this relationship is integrated from t to t t then ln [A] t kt + ln [A] 0 y mx + b Where [A] t is the concentration at t ln [A] t represents what is plotted on the y-axis [A] 0 is the initial concentration t is the represent what is plotted on the x-axis k is the rate constant which is the SLOPE of the graph! Notice that in this equation k has a negative sign; thus the SLOPE of the graph is negative not the rate constant If the reaction is first order then the graph will be linear with a negative slope. slope k ln[a] Half life the required for the initial concentration, [A] 0, to decrease to ½ of its value. ln [A] t kt + ln [A] 0 ln kt + ln 2 ln 2 kt 0.693 kt 0.693 t k Copyright 204 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 3
For the reaction Second Order Reactions A(g) products Δ [ A] rate k[a] 2 Δt When this relationship is integrated from t to t t then Where kt + [ A] [ A] y mx + b [A] t is the concentration at t represents what is plotted on the y-axis [A] [A] 0 is the initial concentration t is the represent what is plotted on the x-axis k is the rate constant which is the SLOPE of the graph! Notice that in this equation k is positive; thus the SLOPE of the graph is positive If the reaction is 2 nd order then the graph will be linear with a positive slope. 0 [A] slope +k Copyright 204 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 4
Kinetics Cheat Sheet Differential Rate Law (concentration vs. rate data): Relationships Rate k [A] x [B] y Integrated Rate Laws (concentration vs. data): Zero order [A] ( )kt + [A] 0 First order ln [A] ( )kt + ln[a] 0 Second order k t + [A] [A] 0.693 t/2 for first order reactions and all nuclear decay k 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 0 Copyright 204 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 5
NMSI SUPER PROBLEM The decomposition of substance X was experimentally observed at 25 C and shown to be first order with respect to X. Data from the experiment are shown below. [X] M Time (min) 0.00 0 0.088 2 0.069 6 0.054 0 0.043 4 0.030??? [X] ln [X] (a) For each of the graphs above i. Sketch the expected curve based on the labeled axes. You do not need to plot the exact data. ii. Write the rate law for the decomposition of substance X. iii. Explain how one of the two graphs above can be used to determine the rate constant, k. Be sure to specify which graph. (b) Based on the above data i. Calculate the rate constant for this reaction. Be sure to include units. ii. How many minutes will it take for [X] to become 0.030 M? Copyright 204 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 6
In a different experiment, the decomposition of substance Y at 50 C was determined to have the following rate law. Rate k [Y] 2 (c) On the axes above i. Sketch the graph that is expected to provide a linear relationship when plotted against. Be sure to label the y-axis. ii. What does the slope of this line represent? (d) The temperature of this reaction was increased from 50 C to 00 C. Predict the effect this would have on each of the following. i. Rate of the reaction ii. Rate constant, k (e) Sketch the graph of the reaction at 00 C on the plot in part (c) Copyright 204 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 7