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Ch #16: Kinetics Kinetics: Rates and Mechanisms of Chemical Reactions

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Chemical kinetics: Study of reaction rates, changes in concentrations of reactants or products as a function of time.

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Factors that influence reaction rate: 1.Concentration - molecules must collide to react; Rate α collision frequency α conc 2.Physical state - molecules must mix to collide; Smaller the particle size, greater the surface area and more the collisions. 3.Temperature - molecules must collide with enough energy to react; Rate α collision frequency α conc.

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Figure 16.3 The effect of surface area on reaction rate.

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Figure 16.4 Collision energy and reaction rate.

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Factors that influence reaction rate: 4.The use of a catalyst.

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Expressing the Reaction Rate reaction rate - changes in the concentrations of reactants or products per unit time reactant concentrations decrease while product concentrations increase For a reaction A → B Rate of reaction = -∆[A]/∆t [A]= conc of A in mol/L ∆=change

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Expressing the Reaction Rate Units= moles per liter per second. Mol L -1 s -1 or mol/ L.s Change I product conc is positive so the rate is Rate of reaction = ∆[B]/∆t Rate decreases during course of reaction. Instantaneous rate: Rate at a particular instant: considering closer values.

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Expressing the Reaction Rate Use initial rate as soon as the reactants are present( no products at this time) For a reaction aA + bB →cC+ dD, Rate =-1/a∆[A]/∆t=- 1/b∆[B]/∆t==1/c∆[C]/∆t==1/d∆[A]/∆t

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Table 16.1 Concentration of O 3 at Various Time in its Reaction with C 2 H 4 at 303K Time (s) Concentration of O 3 (mol/L) x x x x x x x10 -5 (conc A) - tt

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Figure 16.5 The concentrations of O 3 vs. time during its reaction with C 2 H 4 - [C 2 H 4 ] tt rate = - [O 3 ] tt =

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Figure 16.6 Plots of [C 2 H 4 ] and [O 2 ] vs. time. Tools of the Laboratory

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Problem 1 P.1.2H 2 (g) +O 2 (g)→ 2H 2 O (g) Write the rate equation in terms of reactants and products. If [O2] is decreasing at 0.23 mol/L.s at what rate is [H 2 O] increasing?

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The Rate Law and its components: Rate = k[A] m [B] n K= rate constant(does not change as reaction proceeds) M and n are reaction orders. Coefficients are not related to reaction orders. Rate constant and orders can only be found by experimental data.

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Determining the Initial Rate: By performing expt and collecting data.

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Reaction order: Rate=k[A]- 1 st order Rate = k[A] nd order Rate = k[A] 0 - Zero order OR Rate=k(1)=k (not dependent on conc of A)

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Problem 2 P.2.For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law. 2NO(g) + O2(g) → 2NO2(g); rate = k[NO] 2 [O2] CH3CHO(g) → CH4(g) + CO(g); rate = k[CH3CHO] 3/2

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Problem 2 H2O2(aq) + 3I-(aq) + 2H+(aq) → I3- (aq) + 2H2O(l); rate = k[H2O2][I-]

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Determining Reaction Orders: P.3.NO 2 (g) + CO (g) → NO (g) + CO 2 (g) rate=k[NO 2 ] m [CO] n

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Sample Problem 16.3 PLAN: SOLUTION: Determining Reaction Order from Initial Rate Data PROBLEM:Many gaseous reactions occur in a car engine and exhaust system. One of these is rate = k[NO 2 ] m [CO] n Use the following data to determine the individual and overall reaction orders. ExperimentInitial Rate(mol/L*s)Initial [NO 2 ] (mol/L)Initial [CO] (mol/L) Solve for each reactant using the general rate law using the method described previously. rate = k [NO 2 ] m [CO] n First, choose two experiments in which [CO] remains constant and the [NO 2 ] varies.

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Sample Problem 16.3Determining Reaction Order from Initial Rate Data continued rate 2 rate 1 [NO 2 ] 2 [NO 2 ] 1 m = k [NO 2 ] m 2 [CO] n 2 k [NO 2 ] m 1 [CO] n 1 = = m ;16 = 4 m and m = 2 k [NO 2 ] m 3 [CO] n 3 k [NO 2 ] m 1 [CO] n 1 [CO] 3 [CO] 1 n = rate 3 rate 1 = = n ;1 = 2 n and n = 0 The reaction is 2nd order in NO 2. The reaction is zero order in CO. rate = k [NO 2 ] 2 [CO] 0 = k [NO 2 ] 2

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Determining the Rate constant: Calculated from collected data. Units of Rate constant depend on order.

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Table 16.4 An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions Zero OrderFirst OrderSecond Order Plot for straight line Slope, y-intercept Half-life Rate law rate = k rate = k [A]rate = k [A] 2 Units for k mol/L*s 1/sL/mol*s Integrated rate law in straight-line form [A] t = k t + [A] 0 ln[A] t = -k t + ln[A] 0 1/[A] t = k t + 1/[A] 0 [A] t vs. t ln[A] t vs. t1/[A] t = t k, [A] 0 -k, ln[A] 0 k, 1/[A] 0 [A] 0 /2kln 2/k1/k [A] 0

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Figure 16.7 Integrated rate laws and reaction order ln[A] t = -kt + ln[A] 0 1/[A] t = kt + 1/[A] 0 [A] t = -kt + [A] 0

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Problem 5 P.5At 1000 o C, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 87s-1, to two molecules of ethylene (C2H4). a) If the initial C4H8 concentration is 2.00M, what is the concentration after s? (b) What fraction of C4H8 has decomposed in this time?

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Rules for orders through graphs If a straight line is produced with the foll: Ln[reactant] vs time-1 st 1/[reactant] vs time-2 nd [reactant]vs time -0 order.

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Half Life Half Life: Time required for the reactant concentration to reach half of its initial value.

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Figure 16.9 A plot of [N 2 O 5 ] vs. time for three half-lives. t 1/2 = for a first-order process ln 2 k k =

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Problem 6 P.6 Cyclopropane is the smallest cyclic hydrocarbon. Because its 600 bond angles allow poor orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 10000C via a first-order reaction: The rate constant is 9.2s-1; (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value?

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Effect of Temperature: Temperature affects rate by affecting rate constant. Arrhenius equ: k=Ae -Ea/RT K=rate const, e=base of natural logs, T=absolute temp, R= Universal gas const, Ea=activation energy (minimum energy that molecules must have to react)

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Effect of Temperature: As T increases, k increases, rate increases. Ln k =ln A- Ea/R(1/T) A plot of ln k vs 1/T gives a straight line Slope=-Ea/R Y-int= ln A Ln K2/k1=-Ea/R(1/T2-1/T1)

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The Arrhenius Equation The Effect of Temperature on Reaction Rate ln k = ln A - E a /RT ln k2k2 k1k1 = EaEa RT - 1 T2T2 1 T1T1 - where k is the kinetic rate constant at T E a is the activation energy R is the energy gas constant T is the Kelvin temperature A is the collision frequency factor

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Figure Graphical determination of the activation energy ln k = -E a /R (1/T) + ln A

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Problem 7 P.7.The decomposition of hydrogen iodide 2HI(g) → H2(g) + I2(g) has rate constants of 9.51x10-9L/mol*s at 500. K and 1.10x10-5 L/mol*s at 600. K. Find Ea.

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Figure Information sequence to determine the kinetic parameters of a reaction. Series of plots of concentra- tion vs. time Initial rates Reaction orders Rate constant (k) and actual rate law Integrated rate law (half-life, t 1/2 ) Rate constant and reaction order Activation energy, E a Plots of concentration vs. time Find k at varied T Determine slope of tangent at t 0 for each plot Compare initial rates when [A] changes and [B] is held constant and vice versa Substitute initial rates, orders, and concentrations into general rate law: rate = k [A] m [B] n Use direct, ln or inverse plot to find order Rearrange to linear form and graph Find k at varied T

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Effects of concentration and temperature: Collision Theory: Reactant particles must collide with each other. Concs are multiplied in rate law. Ea: Energy required to activate the molecules into a state from which reactant bonds can change into product bonds. Only those collisions with enough energy to exceed Ea can lead to reaction.

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Figure The dependence of possible collisions on the product of reactant concentrations. AA AA BB BB AA AA BB BB AA 4 collisions Add another molecule of A 6 collisions Add another molecule of B AA AA BB BB AABB

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Effects of concentration and temperature: Rise in temp enlarges the fraction of collisions with enough energy to exceed Ea. F= e -Ea/RT e-base of natural log, T=temp, R=gas const. In exothermic reaction: Eaf > Ear In endothermic reaction: Eaf < Ear : k=Ae -Ea/RT A is the frequency factor, A=pZ, Z=collision frequency and p=orientation probability factor.

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Figure The effect of temperature on the distribution of collision energies

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Figure Energy-level diagram for a reaction REACTANTS PRODUCTS ACTIVATED STATE Collision Energy E a (forward) E a (reverse) The forward reaction is exothermic because the reactants have more energy than the products.

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Figure The importance of molecular orientation to an effective collision. NO + NO 3 2 NO 2 A is the frequency factor A = pZ where Z is the collision frequency p is the orientation probability factor

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Transition state theory: Transition state/activated complex: neither reactant nor product present, a transitional species with partial bonds is present. Ea is the energy required to stretch and deform bonds in order to reach transition state. Reaction energy diagram : TB. Pg.699

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Figure Nature of the transition state in the reaction between CH 3 Br and OH -. CH 3 Br + OH - CH 3 OH + Br - transition state or activated complex

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Figure Reaction energy diagram for the reaction of CH 3 Br and OH -.

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Figure Reaction energy diagrams and possible transition states.

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Molecularity of a reaction: Elementary steps make up the reaction mechanism. An elementary step is not made up of simpler steps. Molecularity means the number of reactant particles involved in the step.

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Molecularity of a reaction: In an elementary step equation coeff= order. Reaction order= molecularity A→ product Unimolecular Rate = k[A] 2A→ product Bimolecular Rate = k[A] 2 A + B→ product Bimolecular Rate = k[A][B] 2A + B→ product Termolecular Rate =k [A] 2 [B]

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Table 16.6 Rate Laws for General Elementary Steps Elementary StepMolecularityRate Law A product 2A product A + B product 2A + B product Unimolecular Bimolecular Termolecular Rate = k [A] Rate = k [A] 2 Rate = k [A][B] Rate = k [A] 2 [B] REACTION MECHANISMS

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Problem 8 P.8. The following two reactions are proposed as elementary steps in the mechanism of an overall reaction: NO2Cl(g) → NO2(g) + Cl(g) NO2Cl(g) + Cl(g)→ NO2(g) + Cl(g) (a) Write the overall balanced equation. NO 2 Cl(g) → NO 2 (g) + Cl(g) NO 2 Cl(g) + Cl(g)→ NO 2 (g) + Cl(g)

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Problem 8 (a) Write the overall balanced equation. (b)Determine the molecularity of each step. (c)Write the rate law for each step

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Rate-Determining step: Also called as rate limiting step. It represents rate law for overall reaction. Slow reaction is a rate determining reaction. The elementary steps must add up to the overall equation. The elementary steps must be physically reasonable. The mechanism must correlated with the rate law.

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Catalysis: Each catalyst has its own specific way of functioning. In general a catalyst lowers the energy of activation. Lowering the Ea increases the rate constant, k, and thereby increases the rate of the reaction A catalyst increases the rate of the forward AND the reverse reactions.

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Catalysis: A catalyzed reaction yields the products more quickly, but does not yield more product than the uncatalyzed reaction. A catalyst lowers Ea by providing a different mechanism, for the reaction through a new, lower energy pathway

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