AP CHEMISTRY CHAPTER 12 KINETICS

Transcription

1 AP CHEMISTRY CHAPTER 12 KINETICS Thermodynamics tells us if a reaction can occur. Kinetics tells us how quickly the reaction occurs. Some reactions that are thermodynamically feasible are kinetically so slow as to be imperceptible Reaction Rate- change in concentration of a reactant or product per unit time. [A] = concentration in mol/l Rate = [A] t If the rate expression involves a reactant: Rate = [A] t (negative because [ ] decreases) This gives the average rate. To get an instantaneous rate, we can compute the slope of a line tangent to the curve at that point. Rate = (slope of the tangent line) The rate of a reaction is not constant but changes with time because concentrations change with time. We will only work with reaction rates that are initial rates (reverse reaction is negligible) Differential Rate Law or Rate Equation For the reaction aa+bb + gg + hh + Rate = k[a] m [B] n [A] & [B] represent molarities The exponents are positive or negative, integers or fractions. -usually positive integers (small whole numbers) k = specific rate constant -value depends on reaction, temperature and presence of a catalyst -faster the reaction, larger the k value Kristen Jones 11/3/2013 page 1

2 The exponents determine the order of the reactants. The sum of the exponents is the overall order of the reaction. R = k[a][b] 2 is first order in A, second order in B and third order overall. The units of k can be calculated by reaction orders and units of concentration and rate. For example, if rate is in mol/l. s in the above rate law, we can find the units for k as follows: Rate = k so k = mol This simplifies to: k = L 2 or M 2 s 1 [A][B] 2 L. s mol 2. s mol mol 2 L L 2 The shortcut to determining units of k is as follows: k = L x /mol x. time. The value of x will be one less than the order of the reaction. If the reaction is 3rd order, k = L 2 /mol 2. time. If the reaction is 2nd order, k = L/mol. time If the reaction is 1st order, k = 1/time Determining Differential Rate Laws from Experimental Data If doubling the initial [ ] of a reactant causes the initial rate to double, the reaction is first order in that reactant. If doubling the initial [ ] of a reactant causes the initial rate to quadruple, the reaction is second order in that reactant. If doubling the initial [ ] of a reactant does not change the initial rate, the reaction is zero order in that reactant and that reactant is removed from the rate law. Ex. 2A + B 2C A + 2B 2C [A] [B] Rate [A] [B] Rate Rate = Rate = overall reaction order = overall reaction order = A B A + B C [A] Rate Exp [A] [B] Rate Determine the rate law: Determine the rate law: Kristen Jones 11/3/2013 page 2

3 NH NO 2 N 2 + 2H 2 Exp [NH + 4 ] [NO 2 ] Rate(mol/Lmin) Determine the rate law: Determine the value of k and its units: [A] [B] Initial Rate Experiment (mol/l) (mol/l) (mol L 1 sec 1 ) Determine the rate law and justify your answers: Determine the value of k and its units: Integrated Rate Law -expresses how the concentration of the reactant depends on time -instead of changing initial concentrations and using multiple experiments, one experiment is done and concentration changes over time are measured. 1 st order integrated rate law [A] 0 = initial concentration or mass of reactant [A] t = final concentration or mass of reactant (at time t) Or ln[a] t ln[a] 0 = kt A plot of ln[a] vs t always gives a straight line for a 1 st order reaction. The slope = k. Kristen Jones 11/3/2013 page 3

4 Ex. At 400 o C, the 1 st order conversion of cyclopropane into propylene as a rate constant of s 1. If the initial concentration of cyclopropane is mol/l at 400 o C, what will its concentration be 24.0 hrs after the reaction begins? Radioactive decay is first order. While nuclear chemistry is not "tested" on the AP test, the use of nuclear decay examples for kinetics may be seen. Half-life (t 1/2 ) is the length of time required for the concentration of a reactant to decrease to half of its initial value. t 1/2 = 0.693/k A fast reaction with a short t 1/2 has a large k. A slow reaction with a long t 1/2 has a small k. Example: The decomposition of SO 2 Cl and Cl 2 is a first order reaction with k = s 1 at 320 o C. Determine the half-life of this reaction. Example: The decomposition of N 2 O 5 dissolved in CCl 4 is a first order reaction. The chemical change is: 2N 2 O 5 4NO 2 + O 2 At 45 o C the reaction was begun with an initial N 2 O 5 concentration of 1.00 mol/l. After 3.00 hours the N 2 O 5 concentration had decreased to mol/l. What is the half-life of N 2 O 5 expressed in minutes at 45 o C? Second order integrated rate law Rate = k[a] 2 A plot of 1/[A] t versus t produces a straight line with slope k. Zero Order Rate = k This equation is not found on the AP formula sheet and will not have to be calculated. Make sure that you understand the graphing. [A] t [A] o = kt A plot of [A] versus t produces a straight line with slope k. Kristen Jones 11/3/2013 page 4

5 Reaction Mechanisms intermediate - a species that is neither a product nor a reactant in the overall equation -is used up in a subsequent step elementary step- a reaction whose rate law can be written from its molecularity (balanced equation) molecularity- the number of species that must collide to produce the reaction represented by the elementary step. Unimolecular step- a reaction step involving only one molecule Bimolecular step- a reaction step involving the collision of two molecules (Rate law always 2 nd order) Rate-determining step -slowest step Elementary Reaction -agrees with the balanced equation Reaction mechanisms must: 1. Add up to the overall balanced equation. 2. Agree with the rate law We can t prove a mechanism absolutely. We can only come up with a possible mechanism. Ex. Elementary Rxn : NO + N 2 O NO 2 + N 2 Rate Law: Rate = k[no][n 2 O] Ex. The reaction 2NO 2 (g) + F 2 (g) 2NO 2 F(g) is thought to proceed via the following two-step mechanism: NO 2 + F 2 NO 2 F slow F + NO 2 NO 2 F fast Determine the rate law for the reaction: When an intermediate is a reactant in the rate-determining step, the derivation of the rate law is more difficult. Ex. NO 2 + CO NO + CO 2 Mechanism: NO 2 + NO 2 NO 3 + NO k 1 k 1 NO 3 + CO CO 2 + NO 2 Both fast with equal rates slow Slow step determines rate law (rate-determining step) Rate law: Rate = k[no 3 ][CO] But, NO 3 was an intermediate. We must come up with something equal to NO 3 to substitute. k[no 2 ] 2 = k[no 3 ][NO] [NO 3 ] = [NO 2 ] 2 [NO] Rate = k [NO 2 ] 2 [CO] NO Kristen Jones 11/3/2013 page 5

6 Ex. Cl 2 + CHCl 3 HCl + CCl 4 Mechanism k 1 Cl 2 2Cl k 1 fast Cl + CHCl 3 HCl + CCl 3 slow Cl + CCl 3 CCl 4 fast Rate Law: Rate = k[cl][chcl 3 ] Cl is an intermediate k[cl 2 ] = k[cl] 2 [Cl 2 ] 1/2 = [Cl] Rate = k[cl 2 ] 1/2 [CHCl 3 ] Increasing temperature increases reaction speed. Rate and rate constants often double for every 10 o increase in temperature. Molecules must collide to react. Only a small portion of collisions produce a reaction. Activation energy (E A )- energy that must be overcome to produce a chemical reaction. Rate of reaction depends on E A, not E. E has no effect on rate of reaction. The higher the E A, the slower the reaction at a given temperature. Molecules and collisions have varying energies. As temperature increases, more collisions will have sufficient energy to overcome the activation energy. As temperature doubles, the fraction of effective collisions increases exponentially. Reaction rate is smaller than would be predicted from the number of collisions having sufficient energy to react. This is because of molecular orientations. 2 factors: 1. sufficient energy 2. proper orientation Kristen Jones 11/3/2013 page 6

7 Arrhenius Equation- -can be used to calculate E A ln k = ln A - E A k = rate constant RT R = J/K mol T = Kelvin temperature A = frequency factor (constant as temperature changes) E A = activation energy As E A increases, k decreases. The Arrhenius equation describes a line. We can plot 1/T vs ln k and get a straight line whose slope is equal to E A /R. A variation of the Arrhenius equation can be used to calculate E A or to find k at another temperature if E A is known: k 1 EA 1 1 ln k2 R T2 T1 Calculations involving the Arrhenius equation are not tested on the AP test. Catalysis Catalysts, such as enzymes in a biological system or the surfaces in an automobile's catalytic converter, increase the rate of a chemical reaction. catalyst- substance that speeds up a reaction without being consumed. -produces a new reaction pathway with a lower activation energy -A catalyst lowers the E A for both the forward and reverse reaction. Some catalysts change the mechanism of the reaction greatly and others simply change the rate-determining step. enzymes- biological catalysts homogeneous catalyst- present in the same phase as the reacting molecules (usually liquid phase) Kristen Jones 11/3/2013 page 7

8 heterogeneous catalyst- exists in a different phase -usually involves gaseous reactants being adsorbed on the surface of a solid catalyst (such as a car s catalytic converter) Absorption involves penetration. Either a new reaction intermediate is formed or the probability of successful collisions is increased. This is sometimes called a surface catalyst. How does the mechanism for the catalyzed reaction differ from the mechanism of the uncatalyzed reaction? Kristen Jones 11/3/2013 page 8

9 Kristen Jones 11/3/2013 page 9

10 Kristen Jones 11/3/2013 page 10