University:
Kentucky State University
Course:
CHE 101 | General Chemistry I
Academic year:

2024
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266

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56
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duckgorcxj

Chemical Kinetics CHEM-102 1 The Rate of Reaction Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Chemical kinetics: ➢ Study of reaction rates, ➢ how reaction rates change with varying conditions and which molecular events occur during the overall reaction. CHEM-102 2 The Rate of Reaction Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B D[A] rate = Dt D[A] = change in concentration of A over time period Dt D[B] rate = Dt D[B] = change in concentration of B over time period Dt Because [A] decreases with time, D[A] is negative. Molar concentrations are denoted by enclosing the substance in square brackets. CHEM-102 3 A B D[A] rate = Dt D[B] rate = Dt CHEM-102 4 Reaction Rates and Stoichiometry 2A B Two moles of A disappear for each mole of B that is formed. 1 D[A] rate = 2 Dt aA + bB D[B] rate = Dt cC + dD 1 D[A] 1 D[B] 1 D[C] 1 D[D] rate = == = a Dt b Dt c Dt d Dt CHEM-102 5 Example 1 2N2O5 (g ) → 4NO2 (g )+ O2 (g ) Rate of formation of O2 = Δ[O 2 ] Δt Δ[NO 2 ] Rate of formation of NO2 = …. Δt Rate of decomposition of N2O5 = …. CHEM-102 Δ[N2 O 5 ] − Δt 6 Example 2 Write the rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of the products: • (a) • (b) 3O2(g) → 2O3(g) • (c) CHEM-102 7 Example 3 Consider the reaction Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of 0.024 M/s. (a)At what rate is N2O5 being formed? (b)At what rate is NO2 reacting? CHEM-102 8 Solution (a) (b) CHEM-102 9 Example 4 Calculate the average rate of decomposition of N2O5, – Δ[N2O5]/Δt, by the reaction 2N2O5 (g ) → 4NO2 (g )+ O2 (g ) During the time interval from t = 600s to t = 1200s (regard all time figures as significant). Use the following data: Time 600s 1200s CHEM-102 [N2O5] 1.24×10− 2 M 0.93×10− 2 M 10 Average rate of decomosition of N2O 5 = − Δ N2O5  Δt 0.93 − 1.24 ) ×10 − 2 M ( =− (1200 − 600 ) s − 0.31× 10 − 2 M = 600 s = 5.2×10 − 6 M / s Note that this rate is twice the rate of formation of O2 in the same time interval. CHEM-102 11 Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g) slope of tangent slope of tangent slope of tangent [Br2]final – [Br2]initial D[Br2] average rate = =Dt tfinal - tinitial instantaneous rate = rate for specific instance in time CHEM-102 12 2H2O2 (aq) 2H2O (l) + O2 (g) PV = nRT n P= RT = [O2]RT V 1 [O2] = P RT D[O2] 1 DP rate = = RT Dt Dt measure DP over time CHEM-102 13 The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. aA + bB cC + dD Rate = k [A]x[B]y Reaction is xth order in A Reaction is yth order in B Reaction is (x +y)th order overall CHEM-102 14 Example 5 F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2]x[ClO2]y Double [F2] with [ClO2] constant Rate doubles x=1 Quadruple [ClO2] with [F2] constant rate = k [F2][ClO2] Rate quadruples y=1 CHEM-102 15 Rate Laws • Rate laws are always determined experimentally. • Reaction order is always defined in terms of reactant (not product) concentrations. • The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2][ClO2] 1 CHEM-102 16 Example 6 The reaction of nitric oxide with hydrogen at 1280°C is Determine: (a) the rate law (b) the rate constant (c) the rate of the reaction when [NO] = 12.0 × 10−3 M and [H2] = 6.0 × 10 −3 M Solution CHEM-102 17 (a) the rate law or x = 2, that is, the reaction is second order in NO. or y = 1, that is, the reaction is first order in H2. (2 + 1) or third-order reaction overall CHEM-102 18 (b) The rate constant k : (c) CHEM-102 19 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions Order 0 Rate Law rate = k 1 rate = k [A] 2 [A]2 rate = k Concentration-Time Equation [A] = [A]0 - kt ln[A] = ln[A]0 - kt 1 1 = + kt [A] [A]0 Half-Life t½ = [A]0 2k t½ = ln 2 k 1 t½ = k[A]0 Integrated Rate Law CHEM-102 20 First-Order Reactions A k= product rate M/s = = 1/s or s-1 M [A] [A] = [A]0e−kt CHEM-102 D[A] rate = Dt rate = k [A] D[A] = k [A] Dt [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 ln[A] = ln[A]0 - kt 21 Graphical Determination of k 2N2O5 CHEM-102 4NO2 (g) + O2 (g) 22 First-Order Reactions The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t½ = t when [A] = [A]0/2 ln t½ = CHEM-102 [A]0 [A]0/2 k ln 2 0.693 = = k k 23 First-order reaction A product # of half-lives [A] = [A]0/n 1 2 CHEM-102 2 4 3 8 4 16 24 Second-Order Reactions 2A product D[A] rate = Dt rate M/s = = 1/M•s k= 2 2 M [A] 1 1 = + kt [A] [A]0 rate = k [A]2 D[A] = k [A]2 Dt [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 t½ = t when [A] = [A]0/2 1 t½ = k[A]0 CHEM-102 25 Zero-Order Reactions A product D[A] rate = Dt D[A] =k Dt rate = M/s k= 0 [A] [A] = [A]0 - kt rate = k [A]0 = k [A] is the concentration of A at any time t [A]0 is the concentration of A at time t = 0 t½ = t when [A] = [A]0/2 [A]0 t½ = 2k CHEM-102 26 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions Order 0 Rate Law rate = k 1 rate = k [A] 2 [A]2 rate = k Concentration-Time Equation [A] = [A]0 - kt ln[A] = ln[A]0 - kt 1 1 = + kt [A] [A]0 Half-Life t½ = [A]0 2k t½ = ln 2 k 1 t½ = k[A]0 Integrated Rate Law CHEM-102 27 Example 7 The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate constant of 6.7 × 10−4 s−1 at 500°C. (a)If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 min? (b)How long (in minutes) will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M? (c)How long (in minutes) will it take to convert 74 percent of the starting material? CHEM-102 28 Solution we note that because k is given in units of s−1, we must first convert 8.8 min to seconds: CHEM-102 29 (b) (c) CHEM-102 30 Example 8 The decomposition of ethane (C2H6) to methyl radicals is a first-order reaction with a rate constant of 5.36 × 10−4 s−1 at 700°C: Calculate the half-life of the reaction in minutes. Solution CHEM-102 31 Example 9 Iodine atoms combine to form molecular iodine in the gas phase This reaction follows second-order kinetics and has the high-rate constant 7.0 × 109/M · s at 23°C. (a)If the initial concentration of I was 0.086 M, calculate the concentration after 2.0 min. (b)Calculate the half-life of the reaction if the initial concentration of I is 0.60 M and if it is 0.42 M. CHEM-102 32 Solution (a) (b) For [I]0 = 0.60 M CHEM-102 For [I]0 = 0.42 M 33 Activation Energy The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction. A+B + AB+ C+D Ea 1 ln k = + lnA R T CHEM-102 34 Activation Energy A+B Exothermic Reaction + AB+ C+D Endothermic Reaction The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction. CHEM-102 35 Activation Energy and Temperature Dependence of the Rate Constant k = A  e( −Ea / RT ) (Arrhenius equation) Ea is the activation energy (J/mol) R is the gas constant (8.314 J/K•mol) T is the absolute temperature A is the frequency factor Alternate format: Ea 1 ln k = + lnA R T CHEM-102 Y =m X + b 36 NO + Cl2 → NOCl2‡ → NOCl + Cl Example 10 The rate constants for the decomposition of acetaldehyde were measured at five different temperatures. The data are shown in the table. Plot ln k versus 1/T, and determine the activation energy (in kJ/mol) for the reaction. Note that the reaction is “3/2” order in CH3CHO, so k has the units of 1/M½ ·s. CHEM-102 39 Solution: CHEM-102 40 Solution: CHEM-102 41 Alternate Form of the Arrhenius Equation At two temperatures, T1 and T2 or CHEM-102 42 Example 12 The rate constant of a first-order reaction is 3.46 × 10−2 s−1 at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol? Solution CHEM-102 43 Example 11 The rate constant for the formation of hydrogen iodide from the elements H2 (g )+I2 (g ) → 2HI(g ) is 2.27×10–4 L/(mol·s) at 600 K and 3.5×10–3 L/(mol·s) at 650 K. a. Find the activation energy Ea. b. Calculate the rate constant at 700 K. Solution (a) Ea 3.5×10 −3 1   1 In = −   −4 2.7×10 8.31J / ( mol  K )  600 K 650 K  Ea In 1.30×10 = 2.56 = × (1.28×10 −4 ) 8.31J / mol 1 2.56×8.31 J / mol Ea = 1.28×10 −4 Ea = 1.66×105 J / mol Solution (b) Substitute Ea = 1.66×105 J/mol. k1 = 2.7×10–4 L/(mol·s) k2 = unknown as yet (T1 = 600 K) (T2 = 700 K) k2 1.66×105 J / mol  1 1  In = × −   = 4.77 −4 2.7×10 L / ( mol  s ) 8.31J / (mol  K )  600 K 700 K  k2 4.77 2 = e = 1.2×10 2.7×10 −4 L / ( mol  s ) Taking antilogarithms: k2 = (1.2×102 ) × ( 2.7×10−4 ) L / (mol  s ) k2 = 3.2×10−2 L / (mol  s ) Importance of Molecular Orientation effective collision ineffective collision CHEM-102 47 Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. 2NO (g) + O2 (g) 2NO2 (g) N2O2 is detected during the reaction! CHEM-102 Elementary step: NO + NO N2O2 + Elementary step: N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 48 2NO (g) + O2 (g) 2NO2 (g) Mechanism: CHEM-102 49 Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step. Elementary step: NO + NO N 2O 2 + Elementary step: N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 The molecularity of a reaction is the number of molecules reacting in an elementary step. • Unimolecular reaction – elementary step with 1 molecule • Bimolecular reaction – elementary step with 2 molecules • Termolecular reaction – elementary step with 3 molecules CHEM-102 50 Rate Laws and Elementary Steps Unimolecular reaction A products rate = k [A] Bimolecular reaction A+B products rate = k [A][B] Bimolecular reaction A+A products rate = k [A]2 Writing plausible reaction mechanisms: • The sum of the elementary steps must give the overall balanced equation for the reaction. • The rate-determining step should predict the same rate law that is determined experimentally. The rate-determining step is the slowest step in the sequence of steps leading to product formation. CHEM-102 51 Sequence of Steps in Studying a Reaction Mechanism CHEM-102 52 Example 13 The gas-phase decomposition of nitrous oxide (N2O) is believed to occur via two elementary steps: Experimentally the rate law is found to be rate = k[N2O]. (a)Write the equation for the overall reaction. (b)Identify the intermediate. (c)What can you say about the relative rates of steps 1 and 2? CHEM-102 53 Solution (a) Adding the equations for steps 1 and 2 gives the overall reaction (b) Because the O atom is produced in the first elementary step and it does not appear in the overall balanced equation, it is an intermediate. (c) If we assume that step 1 is the rate-determining step (that is, if k2 >> k1), then the rate of the overall reaction is given by and k = k1. CHEM-102 54 Catalyst A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A  e( −Ea / RT ) Ea Uncatalyzed k Catalyzed ratecatalyzed > rateuncatalyzed CHEM-102 Ea′ < Ea 55 In heterogeneous catalysis, the reactants and the catalysts are in different phases. • Haber synthesis of ammonia • Ostwald process for the production of nitric acid • Catalytic converters In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid. CHEM-102 • Acid catalysis • Base catalysis 56

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