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AP Chemistry

AP Chemistry. Chemical Kinetics: The Steps and Speed of Reactions. Reaction Rate. The speed at which a reaction occurs measured by the change in concentration of a reactant or product over time. Rate = Δ [A] or = - Δ [A] Δ t Δ t

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AP Chemistry

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  1. AP Chemistry Chemical Kinetics: The Steps and Speed of Reactions

  2. Reaction Rate • The speed at which a reaction occurs measured by the change in concentration of a reactant or product over time. • Rate = Δ[A] or = -Δ[A] • Δt Δt • Can be either related to the appearance of products or disappearance of reactants.

  3. We study reaction rates to begin to learn how reactants turn into products. • C3H8 + 5 O2 3 CO2 + 4 H2O • Impossible for 6 molecules to collide at one time with proper orientation and speed. Must be accomplished by series of steps. These steps called a REACTION MECHANISM. More about these later.

  4. Factors Affecting Reaction Rate. • Nature of reactants – • Differences in reactivity • Ease of loss or gain electrons • Ability of reactants to meet- • Liquid or gas, more movement • Vaporized liquid • Vaporized with air

  5. Homogeneous rxn- same phase reactants can mix easily • Heterogeneous rxn- different phases interface is only place where reaction can occur. • Surface area determines reaction rate • Can be explosive with small particle size. • LICOPODIUM POWDER DEMO • LYSOL DEMO

  6. Factors con’d • Initial Concentration of reactants • Higher concentration, faster rate. • Temperature of system • Almost all reactions • Metabolism slows with decreased temp • Catalysts- increasing reaction rate by lowering activation energy without being consumed in reaction.

  7. Measuring Reaction rates • Ratio with time as denominator • Wealth/time pay (+) • debt (-) • Reactants decrease, products increase over time • Rate = Δ[ ] = mol L-1 s-1 or mol/L/ s • Δ time

  8. Relative rates of reactants and products related through coefficients in balanced eq. • Combustion of butane by following equation: • 2 C4H10 + 13 O2 8 CO2 + 10 H2O • Because all products and reactants are related, we need only study one species to determine rate of a reaction.

  9. If butane reacts at -0.20 mol/l/s, at what rate is O2consumed? At what rates are CO2 and H2O formed? • -0.20 mol C4H10x 13 mol O2= -1.3 mol/L/s • L • s 2 mol C4H10 • For products use positive rate • 0.20 mol C4H10x 8 molCO2 = 0.80 mol/L/s • L • s 2 mol C4H10 • 0.20 mol C4H10x 10 mol H2O = 1.0 mol/L/s • L • s 2 mol C4H10 • Kin 1a

  10. As reactants are consumed, what should happen to the rate of the reaction? • Decreases due to decreased concentration • Graphically, concentration vs. time gives parabolic curve. • To find rate at any given time, INSTANTANEOUS RATE, find slope of tangent line at that time.

  11. The Differentiated Rate LawConcentration and Rate • Rate proportional to concentration of reactants each raised to some power. • 2A + 3B  products • Rate = k [A]m [B]n • We must determine the exponents experimentally, they have nothing to do with coefficients in the equation. k is a proportionality constant, dependent upon Temp and the reaction.

  12. Rate = k [A]m [B]n The rate law for the reaction of A and B Once we know values of m, n , and k, we can calculate rate if concentrations are known. H2SeO3 + 6 I-1 + 4 H+ Se + 2I3-1 + 3H2O Rate = k[H2SeO3]x[I-1]y[H+]z Experimentally it is found that x = 1, y = 3 and z = 2. It is also found that at 273 K. k = 5.0 x10-5 L5mol-5s-1 Rate = 5.0 x10-5 L5mol-5s-1[H2SeO3]1 [I-1]3 [H+]2

  13. We can calculate the initial rate of reaction with any combination of concentrations using k and the exponents. • Find the initial rate of reaction when : • [H2SeO3] = 0.020 M, [I-1] = 0.002 M, and [H+] = 0.001 M Rate = 5.0 x10-5 L5mol-5s-1[H2SeO3]1 [I-1]3 [H+]2 • Rate = 5.0 x10-5 L5mol-5s-1 • (0.02)(0.002)3(0.001)2 • Rate = 8.0 x 10-21 mol L-1 s-1

  14. Units L5___ x mol6 = mol L-1 s-1 Mol5 s1 L6

  15. The exponents in the rate law are referred to as the order of the reaction WRT to that particular reactant Rate = 5.0 x10-5 L5mol-5s-1[H2SeO3]1 [I-1]3 [H+]2 • 1st order H2SeO3 3rd order I-1 and • 2nd order H+. Overall the reaction is 6th order. (The sum of all the individual orders.) • The order of the reaction will help us determine the mathematics used to analyze data for the reactions.

  16. BrO3-1 + 3 SO3-2 Br-1 + 3 SO4-2 • Rate = k [BrO3-1][SO3-2] • 1st order WRT bromate and 1st order WRT sulfite, 2nd order overall. • Must be experimentally determined.

  17. To determine the exponents in the rate law, we most often use the RATIO OF INITIAL RATES METHOD. We use several trials of the experiment, varying concentration of one reactant at a time and then measuring rate. By comparing these different trials, we can calculate the exponents. • Rate 2 = [A]2m[B]2n Rate 1 [A]1m [B]1n

  18. Rate 2 = [A]2m[B]2n • Rate 1 [A]1m [B]1n • By using trials where [B] is constant, then the [B]/[B] term is one and the rate can be attributed to only Δ[A]. • Simplified for one reactant systems, as any two trials will do.

  19. The initial rate of decomposition of acetaldehyde, CH3CHO, at 600 ºC was measured with the following results. • CH3CHO (g)  CH4 (g) + CO (g) • Trial [CH3CHO] Rate (mol L-1s-1) • 1 0.10 0.085 • 2 0.20 0.34 • 3 0.30 0.76 • 4 0.40 1.40 • Determine the order of the reaction.

  20. Rate = k[CH3CHO]m Rate 2 = [A]2m Rate 3 = [A]3m Rate 1 [A]1m Rate 1 [A]1m 0.34 = 0.20m0.76 = 0.30m 0.085 0.10m 0.085 0.10m 4 = 2m 8.941 = 3m 2 = m 2 = m The reaction is 2nd order WRT CH3CHO and 2nd order overall

  21. Once the orders are known, the value of the specific rate constant, k can be calculated by plugging in concentrations and the rate from any trial. • Choose trial 2: • Rate = k[CH3CHO]2 • 0.34 mol L-1s-1 = k (0.20 mol L-1)2 • 0.34 mol L-1s-1 = k • 0.04 mol2 L-2 • 8.5 L mol-1 s-1 = k

  22. A + B  Products • Rate = k [A]x[B]y • Trial [A] [B] rate (mol L-1s-1) • 1 0.10 0.10 0.20 • 2 0.20 0.10 0.40 • 3 0.30 0.10 0.60 • 4 0.30 0.20 2.40 • 5 0.30 0.30 5.40 • Find rate law and k

  23. Rate2 = [A]x[B]yRate4 = [A]x[B]y • Rate1 [A]x[B]y Rate3 [A]x[B]y • 0.40 = 0.20x)(0.10y2.40 = 0.30x)(0.20y • 0.20 0.10x)(0.10y 0.60 0.30x)(0.10y • 2 = 2x)(1y 4 = 1x)(2y • 1 = x and 2 = y • Rate = k [A]1[B]2 • Pick a trial and find value of k. 0.20 mol L-1s-1= k(0.10mol L-1)(0.10mol L-1)2 200 L2 mol-2s-1 = k

  24. A certain reaction follows: • A + B  C + D Trial [A] [B] rate(mol L-1s-1) 1 0.40 0.30 1.0 x 10-4 2 0.80 0.30 4.0 x 10-4 3 0.80 0.60 1.6 x 10-3 Determine the order of the reaction WRT to each reactant and the rate law for the reaction. Then determine the value of the rate constant, k with units.

  25. Rate2 = [A]2x[B]2yrate3 = [A]3x[B]3y • Rate1 [A]1x[B]1y rate2 [A]2x[B]2y • 4 x 10-4 = 0.80x)(0.30y • 1 x 10-4 0.40x)(0.30y • 4 = 2x • 2 = x • 1.6 x 10-3= 0.80x)(0.60y • 4 x 10-4 = 0.80x)(0.30y • 4 = 2y • 2 = Y

  26. Rate = k [A]2[B]2 • Rate = k [A]2[B]2 • 1 x 10-4mol L-1s-1= k (0.40)2(0.30)2mol4L-4 • 6.9 x10-3 L3 mol-3s-1 = k

  27. Integrated Rate Lawsconcentration and time related • Once the rate of a reaction is known or determined, the importance of our studies will shift to monitoring or finding concentrations at various times. To do this we must use a different form of the rate law, the integrated rate law. Fortunately, all similar reactions behave according to the same integrated rate law.

  28. Integrated rate laws involve k, t, the initial concentration, [A]0, and [A]t,a later concentration. • For all first order reactions the integrated rate law has the form: • Rate = k[A] • = ln([A]0/[A]t) = kt (by integrating) • ln[A]0 – ln[A]t = kt (law of ln) • ln[A]t = -kt + ln[A]0 (eq of line m=-k) • Graph of ln[A] vs. t yields straight line with rate constant - k = slope. • Also for two reactant systems where one order is 0 r = k[A]1[B]0

  29. Useful to graph ln [A] vs time to see if a reaction is first order or not. If no straight line, then not first order. • The rate law itself can be used to find later concentrations or time to get a certain concentration. • For decomposition of N2O5 • Rate = k[N2O5] k = 0.35 • min • Find: [N2O5] after 4.0 min if initial is 0.160 • Find time for [N2O5] to reach 0.100 m • Find time required for [N2O5] to decrease by ½

  30. Find: [N2O5] after 4.0 min if initial is 0.160 ln[A]0 – ln[A]t = kt ln 0.160– ln [N2O5]4 = (0.35/min)(4.00 min) -1.83 – 1.4 = ln[N2O5]4 -3.23 = ln[N2O5]4 e-3.23 = [N2O5]4 = 0.040 • Find time for [N2O5] to reach 0.100 m ln[A]0 – ln[A]t = kt ln 0.160 – ln 0.100 = 0.35t 1.3 min = t

  31. Find time required for [N2O5] to decrease by ½ ln[A]0 – ln[A]t = kt ln 0.160 – ln 0.080 = 0.35 t 0.693 = t = 2.0 min 0.35/min Half life for first order process is independent of [A]0. t1/2 = 0.693 k ln 2 = 0.693

  32. Half life is the time required for half a sample to react or decompose, or decay. • Sucrose, C12H22O11, decomposes to glucose and fructose in a first order process. R = k[suc] k = 0.208 h-1 at 25C • Find the half life of sucrose at 25 C. • t1/2 = 0.693/0.208 h-1 • t1/2 = 3.33 hr

  33. Zero and Second order rxn’s • Zero order rate = k[A]0 • [A] = -kt + [A]0 (integrated rate law) • Plot of [A] vs time is linear with slope = -k • Half life t1/2 = [A]0 • 2k

  34. 2nd order • Rate = k[A]2 • 1_ = kt + _1_ [A] [A]0 • Plot of 1/[A] vs t is linear. • t1/2 = _1_ k[A]0

  35. Summary • rate law integrated rate law • Zero: rate = k[A]0 [A] = -kt + [A]0 • [A] vs t linear slope = -k • 1st : Rate = k[A] ln[A]t = -kt + ln[A]0 • ln[A] vs t linear slope = -k • 2nd : Rate = k[A]2 _1_ = kt + _1_ • [A] [A]0 • 1/[A] vs t linear slope = k

  36. Time(h) 0 2 4 6 • [HI] 1.00 0.50 0.33 0.25 • Determine the order of this reaction.

  37. (CH3)3CBr + OH-1 (CH3)3COH + Br-1 • Trial [(CH3)3CBr] [OH-1] rate • 1 0.5 0.05 0.005 • 2 1.0 0.05 0.010 • 3 1.5 0.05 0.015 • 4 1.0 0.10 0.010 • 5 1.0 0.10 0.010 • Find the rate law for the reaction, • Determine k with units • Find the time for [(CH3)3CBr] to drop from 0.85M to 0.275 M • Find the ½ life of (CH3)3CBr

  38. rate = k [(CH3)3CBr]1 • k = 0.01 s-1 • Time from 0.85  0.275 • 112.8 s • t1/2= 69.3 s

  39. For half life, the fraction remaining after n half lives is : • (½)n or 1/2n • I-131 decays by first order process with a half life of 8 days. What fraction would be left after 24 days? • 24 days/8 days = 3 half lives • (½)3 = 1/8 left or 12.5 %

  40. Collision Theory : rate effects from the micro-scale. • For a reaction to occur, the particles must collide with proper orientation at the proper location and with enough energy to disrupt existing bonds and form new bonds. (enough energy to over come activation energy). Productive collision. • Most collisions are not productive.

  41. If we increase the number of overall collisions, we will increase the number of productive collisions and thus the rate. • Nature of reactants • Ability to meet (phases) • Initial concentration • Catalysts • Temperature- Increasing T gives more molecules enough energy to react, but has no effect on the different PE of the reactants and products. T measure of KE

  42. Reaction Mechanisms • Proposed pathway for a reaction to occur. • Derived from the rate law and intuition. • Never exact, rather plausible. • To be possible, 2 criteria must be met. • Sum of steps must add to give reaction equation. • Kinetics of rate determining step must equal rate law

  43. Simple mechanisms- with two reactants colliding, a one step mechanism • Complex – several steps with intermediates formed and consumed. • Ex. CO + NO2 NO + CO2 all gaseous • High temp 1 step mech • Low temp • 1. NO2 + NO2 NO3 + NO • 2. NO3 + CO  NO2 + CO2 • Net NO2 + CO  NO + CO2

  44. Rate expressions: • High temp r = k [NO2][CO] • Low temp r = k [NO2]2 • Same reaction with different expressions for different conditions. • Determining rate expressions for mechanism steps. Can use coefficients • A  B + C r = k[A] • A + A  B + C r = k[A]2 ([A][A]) • A + B + C  D + E r = k [A][B][C]

  45. Often one step will be slower than the others. Slowest step is rate determining step. • Overall rate can be taken to be the rate of the slowest step. Nothing else can happen until the slowest step occurs. • The rate expression often mirrors the rate expression of the slowest step.

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