Kinetics Chapter 14

Kinetics Study of rates of chemical reactions -factors that affect these rates -reaction mechanisms -models Goals of Kinetics -measure and predict rates of chemical reactions -understand the steps in which a reaction takes place -understanding the mechanisms allows us to find ways to facilitate the reaction

Important Concepts Spontaneity - tendency of a reaction to occur -NOT related to speed! 2. Collision Theory - reacting particles collide to break bonds and form new ones -the particles need proper orientation and energy in order for a reaction to occur 3. Activation energy - minimum energy required for a reaction to occur

Reaction Rates Speed at which reactants form products Factors influencing reaction rates: Temperature - as temperature increases, the energy and frequency of particle collisions increases Concentration - a greater concentration of particles causes more frequent collisions surface area - an increases in surface area results in more collisions Catalysts - a substance that increases reaction rate without being used up provides alternate pathway that requires a lower activation energy sometimes you hear “lowers activation energy”

Calculating Reaction Rates A + B → C + D Rate = - Δ[A] = rate of disappearance of A (reactant) Δt Rate = Δ[C] = rate of appearance of C (product)

Change of Rate with Time What happens to the concentration of A as it continues to react? How will this affect the rate of reaction? Calculate the average rate at which A disappears of the time interval from 20s to 40s. Calculate the average rate of appearance of B over the time interval from 0s to 40s Sample exercise 14.1 p. 560 answer: #1. 1.2x10-2M/s; #2. 1.8x10-2M/s

Change of Rate with Time

Relative Rate Reaction Rates and Stoichiometry For 2HI(g) → H2(g) + I2(g) Rate = -1 Δ[HI] = Δ[H2] = Δ[I2] 2 Δt Δt Δt Rate of disappearance of HI is twice that of the rate of appearance of the products, so to make them equal, divide disappearance of HI by 2

Relative Rate Examples How is the rate of disappearance of ozone related to the rate of appearance of O2? 2O3(g) → 3O2(g) Rate = -½ Δ[O3] = ⅓ [O2] Δt Δt If the rate of appearance of O2 is 6.0x10-5M/s, what is the rate of disappearance of O3? 4.0x10-5M/s

Relative Rate Practice N2O5(g) → NO2 (g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2x10-7 mol/Ls, what is the rate of appearance of NO2? of O2? NO2 = 8.4x10-7; O2 = 2.1x10-7

Instantaneous Rate To determine the rate, can use a table to obtain average rates (from experimental data) -compute the slope of a line tangent to the curve at that point on a graph of concentration as a function of time Practice: What is the instantaneous rate of disappearance of C4H9Cl at t=0s? What is the inst. rate of disap. of C4H9Cl at t=300s? sample exercise 14.2 p. 562. #1. 1.9x10-4M/s; #2. 1.1x10-4M/s

Differential Rates Rate Laws - an expression which shows how the rate of a reaction depends on the concentration of reactants Example: for aA + bB → P the rate law is: Rate = k[A]x[B]y k is the rate constant specific for this reaction x is the order with respect to species A y is the order with respect to species B The overall order for the reaction is x+y -The rate constant and order must be determined experimentally (coefficients are NOT used) -only the reactants are used in the rate law

-the rate constant is a proportionality constant that relates the concentration of reactants to the rate of the reaction -the units change for different reactions (important) -the reaction order indicates the effect the concentration of the reactant has on the reaction rate example: an order of one indicates that as the concentration of the reactants doubles, the rate doubles -an order of 2 indicates that as the reactant concentration doubles, the reaction rate increases 4 times -an order of 2 indicates that as the reactant concentration triples, the reaction rate increases 9 times -an order of 0 indicates that concentration has no effect on reaction rate

First Order Reaction

Second Order Reaction

Calculating Reaction Rates to determine the differential rate law, the method of initial rates is used -initial rate - instantaneous rate of reaction at t=0 2. The initial rate is determined in several experiments using different initial concentrations of reactants 3. The results are compared to see how the initial rate depends on the initial concentrations 4. data is in a table

Differential Rate Law Example BrO3- + 5Br- + 6H+ → 3Br2 + 3H2O Rate = k[BrO3-]n[Br-]m[H+]p Exp. Initial Concentrations Inital Rate [BrO3-] [Br-] [H+] mol/Ls 1 0.10 8.0x10-4 2 0.20 1.6x10-3 3 3.2x10-3 4

Magnitudes & Units of Rate Constants -in general, a large value of k (109 or greater) means a fast reaction -a small value of k (10 or lower) means a slow reaction Calculate k for the prior example Rate of reaction is determined by concentration. The rate constant depends also on temperature and catalysts (this will also change the rate of the reaction).

Integrated Rate Law First Order Reactions Rate = - Δ[A] = k[A] Δt with the magic of calculus you can relate the initial concentration of A,[A]0, to its concentration at any other time, [A]t Integrated Rate Law: ln[A]t = -kt + ln[A]0 If you think of this formula with the format y = mx + b, then the slope is -k and it gives you the line from slide 14; the y intercept is ln[A]0 formula isin reference tables

Integrated Rate Law Variations of 1st order reaction equation: ln[A]t - ln[A]0 = -kt (this is in your reference table) ln([A]t/[A]0) = -kt ln([A]0/[A]t) = kt

Integrated Rate Law Second Order Reactions Rate = - Δ[A] = k[A]2 Δt again, using the magic of calculus… 1 - 1 = kt or 1 = kt + 1 [A]t [A]0 [A]t [A]0 with y = mx + b, the slope is k as seen in slide 15; the y-intercept is 1/[A]0

Integrated Rate Law Zero Order Reactions Rate = k[A]0 = k more calculus magic... [A]t = -kt + [A]0 slope = -k y-intercept = A0

Integrated Rate Law What can you conclude from the fact that the plot of lnP versus t is linear? We can use pressure instead of concentration because from the ideal-gas law, pressure is directly proportional to the number of moles per unit volume

Integrated Rate Law and Half Life Half-life - the time required for the concentration of a reactant to drop to ½ of the initial concentration 1st order: t1/2 = 0.693 (in your ref. table) k 2nd order: t1/2 = 1 k[A]0 zero order: t1/2 = [A]0 2k

Half-Life If a solution containing 10.0g of a substance reacts by first-order kinetics, how many grams remain after 3 half-lives?

Half-Life The reaction of C4H9Cl with water is a first order reaction. Use the graph (p. 561) to estimate the half-life for this reaction. Use the half-life to calculate the rate constant.

Integrated Rate Law and Half Life The decomposition of a certain insecticide in water at 12C follows first-order kinetics with a rate constant of 1.45yr-1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0x10-7g/cm3. Assume that the average temperature of the lake is 12C. (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the insecticide concentration to decrease to 3.0x10-7g/cm3? (c) Calculate the half-life for the insecticide. (d) How long does it take for the concentration of the insecticide to reach one-quarter of the initial value?

Temperature and Reaction Rate reaction rate increases at higher temperatures due to the increase in the rate constant

Orientation and Reaction Rates

Activation Energy Minimum energy required to initiate a chemical reaction. -The reaction rate depends on the magnitude of Ea not ΔE -lower Ea = faster reaction -the activated complex (aka transition state) is not stable and not part of the reactants or products point out the bond bends and forms an intermediate state - the activated complex - also called the transition state the reverse reaction is endothermic - the activation energy is equal to Ea + dE

Activation Energy Maxwell-Boltzmann distribution curve - shows a wide range of energy values -increasing temperature results in a larger number of molecules having sufficient activation energy to react p.578

Activation Energy 1.Suppose we have two reactions, A-->B and B-->C. You can isolate B, and it is stable. Is B the transition state for the reaction A-->C? 2. In a chemical reaction, why does not every collision between reactant molecules result in formation of a product molecule? no - transition states are not stable not enough energy or incorrect orientation

Activation Energy/Arrhenius Equation -increase in reaction rate due to increasing temperature is not linear -depends on: -fraction of molecules with sufficient activation energy (Ea) -the number of collisions per second -the fraction of collisions with appropriate orientation k = Ae-Ea/RT k = rate constant A = frequency factor (constant as temp varies) Ea = activation energy R = gas constant (8.314 J/molK) T = absolute temperature As Ea increases, the fraction of molecules with sufficient energy decreases, so k will be smaller and reaction rate decreases absolute temperature is rankine or kelvin (rankine is based on fahrenheit temp scale)

Activation Energy/Arrhenius Equation Rank the reactions from slowest to fastest assuming that they have nearly the same value for frequency factor (A) Rank the reverse reactions slowest to fastest the lower the activation energy, the faster the reaction. dE does not affect the rate. 2<3<1 2<1<3

Reaction Mechanisms A balanced equation tells us the reactants and products, but does not tell us how the reactants became products. Vocabulary: Reaction mechanism: a series of elementary steps by which a reaction occurs elementary step: a reaction whose rate law can be written from its molecularity molecularity: the number of species that must collide to produce the reaction indicated by that step (unimolecular, bimolecular, termolecular) rate determining step: one step in the reaction mechanism that is much slower than all the other steps. This it determines the rate of reaction

Reaction Mechanisms 5. intermediate: a species that is formed in one step and then is a reactant in the next step. It is never seen as a product in the overall reaction 6. Catalyst: a species that is seen in the reactant and product of the overall reaction 7. Elementary Reaction: single step reaction -reactant rearranges: H3C-N=C → H3C-C=N -if there is only one reactant, what collides? -NO(g) + O3(g) → NO2(g) + O2(g) -what is the molecularity of this reaction? molecules collide with each other; don’t react with each other, just provide enough energy for the molecule to rearrange bimolecular (2 reactants)

Molecularity Why are termolecular (or higher) very rare?

Reaction Mechanisms Goal: choose a plausible reaction mechanism from 2 or 3 possibilities -the correct reaction mechanism will satisfy 2 requirements: the sum of the elementary steps must give the overall balanced equation for the reaction the mechanism must agree with the experimentally determined rate law the rate determining step (the slow step) will have the molecularity of the rate law to check for the slow step, assume one is slow and then check to see if it matches the rate law

Rate Laws for Elementary Reactions -rate laws are not simply coefficients -may be multiple steps; each step with it’s own rate law and relative speed -can’t tell just by looking at a reaction if it has one or more elementary steps -if a reaction is elementary, it’s rate law is based on molecularity example: A → products Rate = k[A] unimolecular -as the number of A molecules increases, the rate increases A + B → products Rate = k[A][B] bimolecular -rate is first order in both [A] and [B], second order overall

Rate Laws for Elementary Reactions

Rate Laws for Elementary Reactions If the following reaction occurs in a single elementary reaction, predict its rate law: H2(g) + Br2(g) → 2HBr(g) 2. Given: 2NO(g) + Br2(g) → 2NOBr(g) write the rate law (assuming single elementary reaction) Is a single-step mechanism likely? sample 14.13 p. 584 Rate = k[H2][Br2] (experiments show that it is actually k[H2][Br2]½ a. Rate = k[NO]2[Br2] b. no, termolecular reactions are rare

Multistep Mechanism sequence of elementary steps Reaction: NO2(g) + CO(g) → NO(g) + CO2(g) elementary steps: NO2(g) + NO2(g) → NO3(g) + NO(g) NO3(g) + CO(g) → NO2(g) + CO2(g) NO3 is an intermediate = formed in one elementary step and consumed in another cancel NO3 in the elementary steps - point out how the elementary steps combine to the original reaction

Multistep Mechanisms intermediates can be stable (isolated and identified) unlike transition state point out that the transition states have more PE and are therefore less stable than the intermediates

Multistep Mechanisms Proposed conversion of ozone into O2: O3(g) → O2(g) + O(g) O3(g) + O(g) → 2O2(g) describe the molecularity of each elementary reaction write the equation for the overall reaction identify the intermediates p. 582 sample 14.12 unimolecular; bimolecular 2O3 → 3O2 O is the intermediate

Multistep Mechanisms For the reaction: Mo(CO)6 + P(CH3)3 → Mo(CO)5P(CH3)3 + CO the proposed mechanism is: Mo(CO)6 → Mo(CO)5 + CO Mo(CO)5 + P(CH3)3 → Mo(CO)5P(CH3)3 is the proposed mechanism consistent with the equation for the overall reaction? What is the molecularity of each step of the mechanism? Identify the intermediate(s) yes, the 2 equations add to yield the equation for the reaction unimolecular; bimolecular Mo(CO)5

Rate Determining Steps aka rate-limiting step slowest intermediate step the rate of a fast step that follows the RDS does NOT speed up the overall rate the products of the slow step are immediately consumed in the fast step if the slow step is not first, the faster preceding steps produce intermediate products that build up before being consumed the intermediate is a reactant in the slow step Why can’t the rate law for a reaction generally be deduced from the balanced equation? because the reaction can have more than one step, each with its own rate and the slowest step determines the overall rate, not the sum of the rates

Rate Determining Steps the overall rate of a reaction is determined by the molecularity of the slow step The decomposition of nitrous oxide is believed to occur in a 2-step mechanism: N2O(g) → N2(g) + O(g) (slow) N2O(g) + O(g) → N2(g) + O2(g) (fast) write the equation of the overall reaction write the rate law for the overall reaction 2N2O→ 2N2 + O2 rate = k[N2O] sample 14.14 p. 586

Rate Determining Steps Given: O3(g) + 2NO2(g) → N2O5(g) + O2(g) The reaction is believed to occur in two steps: O3(g) + NO2(g) → NO3(g) + O2(g) NO3(g) + NO2(g) → N2O5(g) The experimental rate law is rate = k[O3][NO2]. What can you say about the relative rates of the 2 steps of the mechanism? the first reaction is the slow step because it equals the rate law, the second reaction is the fast step “the rate law conforms to the molecularity of the first step, that must be the rate determining step. The second step must be much faster than the first one.”

Rate Determining Steps Choose the reaction mechanism for 2X + Y → Z; Rate = k[Y] X+Y → M (slow) X+M → Z (fast) 2. X+X ←> M (fast) equilibrium Y+M → Z (slow) 3. Y → M (slow) M+X → N (fast) N+X → Z (fast) #3 is correct check to see if the intermediates cancel to give the reaction, then check for the molecularity of the slow step

Catalysis Catalyst: changes speed of the reaction -catalyst remains unchanged -enzymes in body homogeneous catalyst: same phase as the reactants example: H2O2(aq) → 2H2O(l) + O2(g) With bromide ion catalyst: 2Br-(aq) + H2O2(aq) + 2H+(aq) → Br2(aq) + 2H2O(l) is the bromide ion a catalyst according to the definition above? No, but the Br2 also reacts with the H2O2: Br2(aq) + H2O2(aq) → 2Br-(aq) + 2H+(aq) + O2(g) combine both reaction and you get the original reaction of the decomposition of hydrogen peroxide