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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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**The effect of surface area on reaction rate.**

Figure 16.3 The effect of surface area on reaction rate.

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**Collision energy and reaction rate.**

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-1s-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 O3 at Various Time in its **

Reaction with C2H4 at 303K Time (s) Concentration of O3 (mol/L) 0.0 3.20x10-5 (conc A) - t 10.0 2.42x10-5 20.0 1.95x10-5 30.0 1.63x10-5 40.0 1.40x10-5 50.0 1.23x10-5 60.0 1.10x10-5

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**The concentrations of O3 vs. time during its reaction with C2H4**

Figure 16.5 The concentrations of O3 vs. time during its reaction with C2H4 rate = - [C2H4] t = - [O3] t

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**Plots of [C2H4] and [O2] vs. time.**

Figure 16.6 Plots of [C2H4] and [O2] vs. time. Tools of the Laboratory

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Problem 1 P.1.2H2 (g) +O2 (g)→ 2H2O (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 [H2O] 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]- 1st order Rate = k[A]2- 2nd 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.NO2(g) + CO (g) → NO (g) + CO2(g) rate=k[NO2]m [CO]n

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Sample Problem 16.3 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[NO2]m[CO]n Use the following data to determine the individual and overall reaction orders. Experiment Initial Rate(mol/L*s) Initial [NO2] (mol/L) Initial [CO] (mol/L) 1 2 3 0.0050 0.080 0.10 0.40 0.20 PLAN: Solve for each reactant using the general rate law using the method described previously. SOLUTION: rate = k [NO2]m[CO]n First, choose two experiments in which [CO] remains constant and the [NO2] varies.

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Sample Problem 16.3 Determining Reaction Order from Initial Rate Data continued [NO2] 2 [NO2] 1 m = rate 2 rate 1 k [NO2]m2[CO]n2 k [NO2]m1 [CO]n1 = The reaction is 2nd order in NO2. 0.40 0.10 = m 0.080 0.0050 ; 16 = 4m and m = 2 [CO] 3 [CO] 1 n = rate 3 rate 1 = k [NO2]m3[CO]n3 k [NO2]m1 [CO]n1 The reaction is zero order in CO. 0.20 0.10 n 0.0050 = ; 1 = 2n and n = 0 rate = k [NO2]2[CO]0 = k [NO2]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 Order First Order Second Order Rate law rate = k rate = k [A] rate = k [A]2 Units for k mol/L*s 1/s L/mol*s [A]t = k t + [A]0 Integrated rate law in straight-line form ln[A]t = -k t + ln[A]0 1/[A]t = k t + 1/[A]0 Plot for straight line [A]t vs. t ln[A]t vs. t 1/[A]t = t Slope, y-intercept k, [A]0 k, 1/[A]0 -k, ln[A]0 Half-life [A]0/2k ln 2/k 1/k [A]0

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**Integrated rate laws and reaction order**

Figure 16.7 Integrated rate laws and reaction order 1/[A]t = kt + 1/[A]0 [A]t = -kt + [A]0 ln[A]t = -kt + ln[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-1st 1/[reactant] vs time-2nd [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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**A plot of [N2O5] vs. time for three half-lives.**

Figure 16.9 A plot of [N2O5] vs. time for three half-lives. t1/2 = for a first-order process ln 2 k 0.693 =

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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 Effect of Temperature on Reaction Rate**

The Arrhenius Equation where k is the kinetic rate constant at T Ea is the activation energy R is the energy gas constant T is the Kelvin temperature ln k = ln A - Ea/RT A is the collision frequency factor ln k2 k1 = Ea RT - 1 T2 T1

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Figure 16.11 Graphical determination of the activation energy ln k = -Ea/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 16.12 Information sequence to determine the kinetic parameters of a reaction. Series of plots of concentra-tion vs. time Initial rates Determine slope of tangent at t0 for each plot Compare initial rates when [A] changes and [B] is held constant and vice versa Reaction orders Rate constant (k) and actual rate law Substitute initial rates, orders, and concentrations into general rate law: rate = k [A]m[B]n Find k at varied T Activation energy, Ea Rearrange to linear form and graph Rate constant and reaction order Find k at varied T Use direct, ln or inverse plot to find order Integrated rate law (half-life, t1/2) Plots of concentration vs. time

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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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**The dependence of possible collisions on the product **

Figure 16.13 The dependence of possible collisions on the product of reactant concentrations. A B 4 collisions A B A B Add another molecule of A 6 collisions A B A A B Add another molecule of B A B A B

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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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**The effect of temperature on the distribution of collision energies**

Figure 16.14 The effect of temperature on the distribution of collision energies

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**Energy-level diagram for a reaction**

Figure 16.15 Energy-level diagram for a reaction Collision Energy Collision Energy ACTIVATED STATE Ea (forward) Ea (reverse) REACTANTS PRODUCTS The forward reaction is exothermic because the reactants have more energy than the products.

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Figure 16.17 The importance of molecular orientation to an effective collision. NO + NO NO2 A is the frequency factor Z is the collision frequency p is the orientation probability factor A = pZ where

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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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**Nature of the transition state in the reaction between CH3Br and OH-.**

Figure 16.18 Nature of the transition state in the reaction between CH3Br and OH-. CH3Br + OH CH3OH + Br - transition state or activated complex

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**Reaction energy diagram for the reaction of CH3Br and OH-.**

Figure 16.19 Reaction energy diagram for the reaction of CH3Br and OH-.

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**Reaction energy diagrams and possible transition states.**

Figure 16.20 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**

REACTION MECHANISMS Table Rate Laws for General Elementary Steps Elementary Step Molecularity Rate Law 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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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. NO2Cl(g) → NO2(g) + Cl(g) NO2Cl(g) + Cl(g)→ NO2(g) + Cl(g)

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**Problem 8 (a) Write the overall balanced equation.**

Determine the molecularity of each step. 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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Kinetics Chemistry—Introduction

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