Chemical Kinetics PDF

Chemical Kinetics 01006727 General Chemistry 01006708 Chemistry Brown Chemistry:The Central Science 13th Edition Chapter 14 (c) 2014  spontaneous ? Loo thermodynamics : Chemical Kinetics how fast ? In c...

Chemical Kinetics 01006727 General Chemistry 01006708 Chemistry Brown Chemistry:The Central Science 13th Edition Chapter 14 (c) 2014  spontaneous ? Loo thermodynamics : Chemical Kinetics how fast ? In chemical kinetics we study Factors affecting the rate the rate (or speed) at which 1) State of the reactants a chemical process occurs. 2) Reactant concentrations Kinetics also sheds light on 3) Reaction temperature the reaction mechanism, a molecular-level view of the 4) Presence of a catalyst inhibitor (path leading from reactants to products.S 2  1. Physical State of the Reactants Reactant must come together to react. The more readily the reactants collide, the more rapidly they react. Homogeneous reactions are often faster. Heterogeneous reactions that involve solids are faster if the surface area is * increased; i.e., a fine powder reacts fastersurface than a pellet or tablet. 3  2. Reactant Concentrations Increasing reactant concentration generally increases reaction rate. Since there are more molecules, more collisions occur. ferric oxide 4  3. Temperature Reaction rate generally increases with increased temperature. Kinetic energy of molecules is related to temperature. KEL T(k) At higher temperatures, molecules move more quickly, increasing numbers of collisions and the energy the molecules possess during the collisions. 5  4. Presence of a Catalyst that increase Catalysts agents are reaction rates without being used up. Catalysts affect rate without being in the overall balanced equation. Catalysts affect the change th kinds of collisions, mechanism changing the mechanism. Catalysts are critical in many biological reactions. 6  Reaction Rate reactants products Rate is a change in concentration over time 3) -Δ[R]/Δt or +Δ[P]/Δt – Δ means “change in.” + – [ ] means molar concentration. – t represents time. # 1  Three types of rates can be quoted  average rate (measured over a period of time  instantaneous rate  initial rate measured at any point during the ryn) (measured at the beginning of the rxn ). 7  Following Reaction Rates [reactants) reaction rate = rate = -change in change in time  The average rate is calculated by the -D[C4H9Cl] / Dt. L318 6 [reactant]/ Ot  The table shows the average rate for a variety of time intervals. 8  Chemical Kinetics Rate, or speed, refers to The rate can be followed something that happens in a using any of the reagents or unit of time. products in the reaction. – The rate of reaction is a – Rate of appearance positive quantity – it is necessary to divide all – Rate of disappearance rates by the appropriate stoichiometric coefficients reactants are disappearing while the products are appearing take the reciprocal - rate of reaction = k [A]x[B]y * X and Y not related to Not to be confused with the Rate law stoichiometric coefficients 9  Reaction Rates (a) How is the rate at which ozone reactant disappears related to the rate at which oxygen appears in the reaction below (b) product rate == Ms ∆[ ] ∆[ ] ∆[ ] Rate = = - = If the rate at which O2 appears, is 6.0x10-5 ∆ ∆ ∆ M/s at a particular instant, at what rate is O3 disappearing at this same time. ∆[ ] =- ∆[ ] = 6.0x10-5 M/s ∆ ∆ reaction Rate 2 O3(g) →3 O2(g)? balanced ? - rate ∆[ ] of - = (6.0x10-5 M/s) = 4.0x10-5 M/s ∆ (a) rate = -= Ms rate of appearin is (b) 40XM #3) = 2002 (6xMi = 10  Plotting Rate Data A plot of the data gives more information about rate of reaction. – The slope of the curve at one point in time gives the instantaneous rate. – The instantaneous rate at time zero is called the initial rate; this is often the rate of interest to chemists. Note: Reactions typically slow down over time! 11  Rates of Reaction Using the date provided, calculate the average rate at which C4H9Cl say disappears – Between t= 0 to 210s – Between t = 500 to 700s. Rate = EX[CpHaCI] # · ∆[ ] (. At. ) Rate= - =- = 1.9 𝑥 10 M/s - - = ∆ 10 06 (. ) - 0. 10) M = 1 9x18. Mil (218 - 0)5 ∆[ ] (.. ) Rate=- ∆ =- ( ) = 6.5 𝑥 10 M/s Rate = - D [CpHaCl] msb Ot T=0, [C4H9Cl]=0.10 = - 10 38-0 25)M.. = 6. 5x T=210, [C4H9Cl]=0.06 T=500, [C4H9Cl]=0.38 1500 - 700)S T=700, [C4H9Cl]=0.25 12  Determining Concentration Effect on Rate To determine the effect of For the table below: concentration on the rate we – Expt. 1-3 show how [NH4+] affects rate. keep all but one reagent – Expt. 4-6 show how [NO2−] affects rate. concentration constant. depends on the[reactants We then measure the impact The rate law, below shows the changes in concentration this relationship between rate and reagent has on the rate. concentration for all reactants: ammonium nitrite [NH4+] [NO2−] → N2 + 2 H2O · S = constantl Rate = k [NH4+] [NO2−] k [Reactant 1 * [Reactant = constant 37 Rate = constant 13 X, X : from experiment.  More about Rate Law not related to the coefficients in the reaction equ. The exponents tell the order Some rates depend only on of the reaction with respect to one reactant to the first power. each reactant. reaction – These are first order reactions. order of – The rate law becomes: Rate = k [NH4+]1[NO2−]1 – Rate = k [A]= 1st order total order sum of the individual = ordersfor reactant - The order with respect to each reactant is 1. The reaction is defined as second order Some rates depend only on a (1 + 1 = 2). We simply add up all of the reactants’ orders to get the reaction’s order. reactant to the second power. – These are second order The rate constant is a reactions. temperature-dependent quantity – The rate law becomes: specific for each reaction. – Rate = k [A]2 K - unique for every reaction oth order : Rate = kn - varies with Temperature Rate = k - determined experimentally 14  Reaction Rates What are the overall reaction orders for the reactions described in equations (1) below? · coeficent a) The rate of the reaction is first order for N2O5 and first order overall. rate of the ryn ↓ (a) 2 N2O5(g) → 4 NO2(g) + O2(g) The is 1st order for N2O5 & b) The rate of the reaction is first order for both H and I , and E from experimental data overall. 2 2 Rate = k[N2O5] second order overall. ist order c) The reaction is first order in CHCl3 and one-half order in Cl2. The overall reaction order is three halves. & Iz (b)11 H2(g) + I2(g) → 2 HI(g) the storder for both H2 is Rate = k[H2The rate of rxn ][I2] overall order (1 + 1) & end order overall. (c) CHCl3(g) + Cl2(g) → CCl I 4(g) + HCl(g) (2) Rate = k [CHCl3][Cl2]1/2 1st CHCs & order (a) order in The rate of the rxn is k= / = 𝑠 -1 in C2. The overall / rxn order is 3 K (b) k = = M-1s-1 What are the units of the rate constant for the rate law in Equation A? Ist order / 2nd ordera Ms (c) k = = M-1/2Mis s-1 kCA- (b) k / = rate = = (a) Ms = rate--OAot M Ms = ↓ (d) k = M 15 = =  Determining the Rate Law The initial rate of a reaction A + B. C was measured for several different starting concentrations of A and B, and the results *From the table we can deduce visually or from below (i.e) (a) & 2 -5 (A) n constant compare expts 14.0x10 are as follows: Using these data, determine = , = 0.1 (a) the rate law for the reaction, (b) the rate 16.0x10-5 0.2n rate (n = Di constant, (c) the rate of the reaction when = [A] = 0.050M and [B] = 0.100 M. 0.25 = 0.5n A (POX) = n=2 L Rate = k[A] m[B] n~ 1 (0 5)" k[A]2[B]00 5M Rate 0= 25 =.. =. H = 0 [B] coust rate 1 [A1]n keep m =2. = n 2 rate 2 [A2] rate ↓[A]2[BJO k[A] = = 4.0x10−5 M/s= KCA72-2nd order k = [A]2 (0.100 M)2 4.0x10-3 M -1s-1 = (b) k =t Ms P xs = rate law rate = (0 1 M) 2. rateRate = k[A]2 = (4.0x10-3 M -1s-1)(0.050 M)2 KCAs s = ↑ * =1.0x10-5 M/s 3 = (4 0x13M's /(0.. 050M) - = 1.0x105 Mst 16  Determining the Rate Law k =he SH2] Determine the rate law for the reaction below. > b) Calculate the rate constant. and the rate when [NO] = 0.050 M and [H2] = 0.150 M 2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g). deduceMs 23x153 *From the table= we1 can that: [NO2 CH27 Rate = k[A]m[B]n = 1 23x103Ms[H2] Rate = k[NO] 2 overall reaction order 3. = (0 1 M) 2 (0 1 M). answers : (a) rate = KNO]2 CH27 " 1.23x10−3 M/s k = [NO]2[H ] =(0.10 M)2(0.10 M)= 1.23 M -2s-1 2 1 23 M-2 gl =. Rate = krate (e) [NO]2[H = kCWOR (H2] 2] = (1.23 M s )(0.050 M) (0.150 M) -2 -1 2 =4.6x10-4 M/s - = 1 23. M 2" (0 0SM)310 150M).. - recei = 4. 6x104M5 17  Integrated Rate Equations Zero Order 1st order 2nd order -[] Solution Solution Solution O Plot Ipy YS. t - y = mx + C + k = Slope plot t - InCA] t [A) vs. linear equations XS. slope = -k - -slope--k 18  Integrated Rate Equations Obtaining data at multiple reagent conditions can be used to determine reaction order. slope = -k Straight line plot: [A] vs t slope = -k Straight line plot: ln[A] vs t slope =+ k Straight line plot: 1/[A] vs t 19  Half-life: 1st order te of 50 % its original amount The half-life is the amount reduction. a of time it takes for one-half of a reactant to be used up in a chemical reaction. As we have seen before, for 50% 1st half-life a 1st order reaction:...... · 1st order  ln [A] = − kt + ln [A]o & + +1  = - ln ([A]o/2) = − k t½ + ln [A]o  − ln ([A]o/2) + ln [A]o = k t½  ln ([A]o / [A]o/2) = k t½  ln 2 = k t½ t½ = 0.693/k ist order rxn. te endtlz does not 27 depend 20 1st Order Reaction: Conversion of Methyl Isonitrile to Acetonitrile The equation for the reaction: * CH3NΞC → CH3CΞN [AJt vs t Ek &. vst + Xath order Is it 0th, 1st order or 2nd order?. Rate = k [CH3NC] The integrated rate law for a 1st order reaction is: slope -k = ln [A] = − k t + ln [A]o – The plot of ln[A] vs ln[A]o will give a O straight line. Its slope will equal −k. S 21  2nd Order Reaction: Decomposition of NO2  NO2 decomposition must be 1st 2nd order because it is linear for 1/[NO2], not linear ln[NO2]. NO2 → NO + ½ O2 Ist order For a 2nd order reaction the t1/2 is derived as follows: – 1/[A] = 1/[A]o + k t 2nd – 1/([A]o/2) = 1/[A]o + k t½ +k – 2/[A]o −1/[A]o = k t½ [A]t time any – t½ = 1 / (k [A]o) ~ initial * half-life is a concentration dependent quantity for second order reactions 22  Zero Order Reactions Occasionally, rate is independent of the concentration of the reactant: – These are zero order reactions. ⑧ – These reactions are linear in concentration. - ↓ – Rate = k half-life depends on S - initial concentration of reactant - rate constant 23  Half Life Using the date provided, calculate the &........... ∆[ ] (.. ) Rate= - ∆ =- ( ) = 1.9 𝑥 10 M/s Calculate the rate constant from the t to 350s half-life: T=0.. st order/ k= = k = 1.98x10-3 s-1 0 693 Agee : =. = ti 98x183g tz 1 k = [I. - T=0, [C4H9Cl]=0.10 en T=210, [C4H9Cl]=0.06 T=500, [C4H9Cl]=0.38 T=700, [C4H9Cl]=0.25 t 24  Integrated Rate Equation rate law = k[AIM Y The following data were obtained for NO2(g) → NO(g) + ½ O2(g). the gas-phase decomposition of nitrogen dioxide at 300 °C. Is the reaction 1st or 2nd order in NO2? ath order[At]=[Ao]SAIz : - kt 0th order CASo-At = ln[At]=ln[Ao] - kt 1st order 1/[At]=1/[Ao] + kt 2nd order 1st order : enCA)t = InCAJo It- 2and nd order process shows a straight line graph order : SASt + It = co ln[conc] vs t Slope = =k 2nd order : slope = +h slope = OX k=(= (2154)M" ) = W ( ) E (300 100)s - 1/[conc] vs t + k k= 545Mig Im = 0 2. = k=0.545 M-1s-1 25  Integrated Rate Equation The decomposition of a A rate k[A]" = & InCAlt (a) InCAjo -kt = chemical follows first-order kinetics with a rate constant (k) of 1.45 yr-1. The initial ln[At]= ln(5.0*10-7 g/cm3) - (1.45 yr-1)(1 yr) = -1.45 +(-14.51) = -15.96 concentration = 5.0*10-7 g/cm3. [At] = e-15.96 = 1.2x10-7 g/cm3 (a) What is the concentration of the insecticide after 1 year? 0x1 ga3) (a) InCA]f = In (5. - (1. 45 yr's (1yr) (b) t ? = = - 15 96. (b) How long will it take for the [Alt 15 96) 1 2x10"gams concentration to reach 3.0x10-7 = exp( -. =. ln(3.0x10 g/cm )= ln(5.0*10-7 g/cm3) - (1.45 yr-1)(t) -7 3 g/cm3? -15.02 = -14.51 - (1.45 yr-1)(t) In 3. 0x109gcm3) = In (5. 0x10Tgcm3) - (1. 45 yrtt = 1.450.51yr −1 =0.35 yr Rate = k[A] - 15 02. = - ln[At]=ln[Ao]-kt 14 51. - (1. 45 jr(t) t = y = 0 35. yr 26  Factors That Affect Reaction Rate 1. Temperature 2. Frequency of collisions 3. Orientation of molecules 4. Energy needed to start the reaction (activation energy) Generally, as temperature increases, rate increases. The rate constant is temperature dependent Rate constant approx. doubles with each 10 ºC rise. 27  Frequency of Collisions The collision model is based on the kinetic molecular theory. 1. Molecules must collide to react. 2. If there are more collisions, more reactions can occur. 3. If there are more molecules, the reaction rate is faster. 4. If the temperature is higher, molecules move faster, causing more collisions and a higher rate of reaction. 28  The Collision Model In a chemical reaction, bonds are broken & new bonds are formed. – Molecules can only react if they collide with each other. – Molecules can often collide without forming products. – Molecules can collide in different ways, leading to different products – Aligning molecules properly can lead to chemical reactions. For a reaction to occur, the molecules: – Must react in the correct orientation. – Must have sufficient energy to overcome the barrier to reaction. This is called activation energy. /Ea] barrier energy properangle effective collision proper amount of energy 29 Transition State (Activated Complex) Reacting particles must have sufficient energy to reach the TS maximum energy state. – This is called the transition state (or activated complex). – The energy needed to form this state is called the activation energy. (Ea) methyl isocyanide isomerization crearrangement) 30  Transition State (Activated Complex) R transition state P * = Eproduct Freactant - Plots are used to show the energy required by reacting particles. – Reactions can be either r.... DE =endothermic or exothermic after this point. GE ⑦ = 31  Distribution of the Energy of Molecules The energy possessed by particles in a reaction are dictated by the temperature – Gases have an average temperature, but each individual molecule has its own energy. – At higher energies, more molecules possess the energy needed for the reaction to occur. Boltzmann distribution of modules as a function of KE. activation energy sconstant for each reaction ( number of modules that have sufficient energy. 32  Relationship Between EA & T Arrhenius noted relationship between activation energy and temperature: ↓k k = Ae−Ea/RT A (frequency factor) is a constant for a particular reaction determined is at eveare – Activation energy can be determined graphically by reorganizing the equation: ln k = −Ea/RT + ln A 1-2 X rearrange equ. Ink = Ea + MA - Q enke = -Ette B O endothermic exothermic GE = Eprod-Ereact 33  Activation Energy Calculate the activation energy for the 1st order reaction described to the right mX + b Y - ------------ O 8 Anhenius equ, -euk ~ ---------- slope = Slope = -Ea/R = E I I 10 589 ( 7 370) ↓ - -. = (−10.589) − (−7.370). - 𝑆𝑙𝑜𝑝𝑒 = (2 160 x 103 ) 1 986x153)k-1 (2.160𝑥10 ) − (1.986𝑥10. -. Fa 𝑆𝑙𝑜𝑝𝑒 = −3.219 = 0.000174 /1 85x10. = −1.85𝑥104 𝐾 = 85x104k - K + Ea 8 Ea= -slope*R ==−(−1.85𝑥104. 3145mol 1 K)*(8.315 kJ/mol.K). = 153.8x103 J/mol = 154 kJ/mol = 153. 8x103Jmol" = 154kTmol 34  Activation Energy K What is the rate constant at 430.0 K given the following information for a 1st order reaction? O ⑧ - EA = 154 kJ/mol (from the last step) - lnK = -10.589 (from the Table, T=189.7 oC) - Ink = 10 589 - K= e-10.589 = 2.5𝑥10. 𝑀 𝑠 k = exp( - 10 58a) = / 2. 5x155M's ln = −... ⁄... en(2551) = M 30 = - ln = 3.06. si 19x10dj7. k1= 2.5𝑥10 𝑀 𝑠= expl 𝑒 3 06)ky = 1.87𝑥10 𝑠 -. = 1.. 2 35  theory is explanation of natural phenomena. (it is not prediction) Law vs. Theory scientific laws refer to rules for how written as nature will behave under certain conditions , usually an equation. Kinetics gives what happens. We call the description the rate law. Why do we observe that rate law? We explain with a theory called a mechanism. explanation why we observe something. A mechanism is a series of stepwise reactions that show how reactants become products. 36  Reaction Mechanisms Reactions may occur all at once or through several discrete steps. a sequence of reactions. Each of these processes is known as an elementary reaction or elementary process. moleularity the number of molewles coming together to react in elementary reaction : an. 1 I 1 O 11 11 Yunlikely reaction order : an empirical quantity (obtained from the experimental rate law) 37  Rate Determining Steps Ter-molecular steps require The overall reaction three molecules to cannot occur faster than simultaneously collide with the slowest reaction in the proper orientation and the mechanism. the proper energy. less chance – This is called the rate- – These are probabilistically RDS determining step. less likely than uni-or bi- molecular steps. Nearly all mechanisms use only unimolecular or bimolecular reactions. 38  Determining the Rate Law write the rate law from the RDSP (the slow step · The rate law must be able to be devised from the rate-determining step. taken from the slow multistep process-rate law step overall is the summation rxn The stoichiometry must be obtained when all - all the multistep process. steps are added up. of – Each step must balance, like any equation. does not appear in the balanced equation. formed & consumed in next step – All intermediates are made and used up. – Any catalyst is used and regenerated.. Ea & increasing the reaction rate reducing 39  Mechanism With a Slow Initial Step the step that determines the rate law If the first step is the rate-determining step, the coefficients on the reactants side are the same as the order in the rate law! probable mechanism : overall run : NO2 + CO → NO + CO2 ~ NOz + N- + NO Rate law: Rate = k [NO2]2 , 2 ⑱ + [0 - N + CO2 experiment I The actual reaction is believe to occur in two steps: Nez is an Intermediate – The easiest way to complete the first step is to make a product: if stept is slow - NO + NO2 → NO + NO3 2 Rate = k [NO2]g2 with expt to compare. – NO3 is an intermediate. It must be used in a more rapid, 2nd step. > - is slow NO3 + CO → NO2 + CO2 if step 2 determine which is correct Rate = k [NO3][CO] 40  Mechanism With a Fast Initial Step If the first step is not the rate-determining step, the coefficients in the rate law are dependent on some other step! ~ nitrosyl bromide 2 NO + Br2 - - 2 NOBr order 3 = Rate law: rate = k [NO]O 2 [Br ] 2 "experimental rate law Because termolecular processes are rare, this rate law suggests a multistep mechanism. – Since the first step is the slowest step, it gives the rate law. – Add up all of the individual steps to get the stoichiometry. This is a plausible 2 step mechanism. equilibrium constant ~ K (1) NO + Br2 NOBr2 fast equilibrium , k2 (2) - NOBr 2 + NO → 2 NOBr Slow (RDS) S - * rate law rate = kz [NOBre] [NO] intermediate - [NO) [Br2] or 41  What is the Rate Law? The rate of the overall reaction Substituting for the forward and depends upon the rate of the slow reverse rates: step which would be (2) k1 [NO][Br2] = k−1 [NOBr2] Rate = k2[NOBr2] [NO] instep(1) [NOBreS [NOSCOR] = – But how can we find [NOBr2]? – NOBr2 can react two ways: With NO to form NOBr. Solve for [NOBr2], then substitute Decomposition to reform NO & Br2. into the rate law: Step 2 : Rate = Kz [NOBr2][NO] The reactants and products of the Rate = k2 (k1/k−1) [NO][Br2][NO] first step are in equilibrium with each other. Thus, the rate of This gives the observed rate law! forward & reverse steps are equal: Rate = k [NO]2 [Br2] Ratef = Rater Rates = k , [NO] (Br] the same as experimental rate law Rate = k _ [NOBR] (the mechanism is possible 1 42 -  Rate Laws If the following reaction occurs Consider the reaction below. The 1st step in the reaction follows the in a single elementary following rate law: Rate=k[A][B]. Which J reaction, predict its rate law: of the 5 possible elementary steps are compatible with this situation. H2(g) + Br2(g) → 2 HBr(g) 2 A + B →X + 2 Y it to assume bimolerlar leading is assume it is unimolecular leading We would reasonably to a 2nd order reaction a 2nd order run. rate = k[A] 2 (a) A + A → Y + Z Rate=k[H2][Br2] rate = k[Hz](Ori] (b) A → X + Z rate = k[A] (c) A + A + B → X + Y + Y However in reality, the reaction must proceed via a single rate = [A]2(8) elementary the (d) B → X + Y Note :I n reality , reaction must rxnstep proceed (e) A + B → X + Z rate = k[B] via single elementary step a Rate=k[H2][Br2]1/2 Rate = KLAJCE] e, depends on both A and B, forming product X and an Rate = KCHz)[Bre] 2 intermediate Z which can form Y in a later step. 43  Catalysis Catalysis is the increase in rate of a reaction that results from the addition of a catalyst Ea – Homogeneous catalysts are , in the same phase as the Ea reacting species. - change the mechanism & lower Ea – Consider the conversion of O3 to O2 in the presence and intermediate absence of Cl atoms (ozone depletion in the atmosphere) Three types of catalysts 1) Homogeneous catalysts 2) Heterogeneous catalysts 3) Enzymes (they speed things up without beingused up in the process 44  Catalysts Homogeneous Heterogeneous The reactants and catalyst are The catalyst is in a different in the same phase. phase than the reactants. – Many times, reactants and – Often, gases are passed over catalyst are dissolved in the a solid catalyst. same solvent, as seen below. – The adsorption of the reactants is often the rate- determining step. 45  Enzymes decomposition of H202 by blood enzyme is catalyst a protein. Enzymes are biological catalysts. – They have a region where the reactants attach called the active site. – The reactants are referred catalyzed by to as substrates. enzyme catalase In the enzyme–substrate rapid decomposition model, the substrate fits into the active site of an enzyme, much like a key fits into a lock. – They are highly specific. 46