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CHEMICAL KINETICS

CHEMICAL KINETICS. REACTION RATES BY JOANNE SWANSON. TOPICS TO COVER. Define the reason for studying kinetics Average and Instantaneous rates of reaction Rate laws Order of reaction Kinetics models Mechanism of reaction Catalysis. Why Study Reaction Rates?.

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CHEMICAL KINETICS

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  1. CHEMICALKINETICS REACTION RATES BY JOANNE SWANSON

  2. TOPICS TO COVER • Define the reason for studying kinetics • Average and Instantaneous rates of reaction • Rate laws • Order of reaction • Kinetics models • Mechanism of reaction • Catalysis

  3. Why Study Reaction Rates? • Thermodynamics tells us whether a reaction will occur spontaneously. • Spontaneity means the reaction will proceed on its own. It does not mean instantaneous. • A reaction may be spontaneous, but very slow. • Kinetics tells us about the speed at which a reaction will occur. • For example: 3H2 + N2 2NH3 occurs spontaneously, but extremely slowly. This reaction, called the Haber process, is needed to produce fertilizer by the millions of tons. Kinetics helps us to find how to make the reaction proceed at a faster rate in order to be useful.

  4. Reaction Rates • Rate is defined as the change in concentration of reactants or products over time. • Things that effect rate are Temperature, Concentration, and Catalysts. • The rate equation is: rate = -D[conc.] /D Time where D stands for final – initial conc. or time. • The units for concentration are Molarity; moles / L • The units for time may be seconds, minutes, hours, days or years, (easily interconverted). • So the units for rate will be something like: Moles / L. s, or Moles / L. h

  5. How to measure Rate • For the reaction: 2N2O5(g) 4NO2(g)+ O2(g) The rate expression could be shown as: rate = - D [N2O5(g)] / DT • We would have an experimentally determined table of (concentration vs. time) data: Time (min) 0 1 2 3 4 Conc. N2O5 (M) 0.160 0.113 0.080 0.056 0.040 Instantan- taneous rate mole 0.056 0.039 0.028 0.020 0.014 L.min

  6. Sample Question • Calculate the average rate of the decomposition of 0.160 mole N2O5 in a 1L container at 67 oC, measured at one minute intervals, (using the data table), between a) t=0 and t = 1min and b) t= 1 and t = 2 min. • Rate = -D [N2O5] / DT = - 0.113 M - 0.160 M = 0.047 mol 1 min – 0 min L. min • Rate = - 0.080M – 0.113 M = 0.033 mol 2 min – 1 min L. min

  7. Determining Change in Concentration: What is the change in conc. Of NO2 in the first minute? (in the same reaction) • 2N2O5(g)  4NO2(g) + O2(g) Initial 0.160 0 0 Final 0.1132(0.047)½(0.047) 0.047 Therefore the D[NO2] = 0.094M and the rate = 0.094mol / L.min

  8. This work gave us the average rate.Normally we are interested in instantaneous rate: • If the data were plotted we would get a curve with the slope of the tangent to that curve = Dconc. /DT • The slope of the tangent to the curve from the plotted concentration vs. time data is equal to the instantaneous rate.

  9. 0.0160 0.120 Dy = slope = instantaneous rate Dx 0.080 Dy 0.040 Dx 0 1 2 3 4

  10. Reaction Order Looking at our earlier data table, you can see that asconcentration decreases by a factor of 2, (0.160 to 0.080), the reaction rate decreases by ½, (0.056 to 0.028), also a factor of 2. This shows that in this reaction, concentration is directly proportional to rate. Therefore we can say that rate = K (conc. N2O5) where ‘K’ is a proportionality constant called the rate constant. When the rate is directly proportional to the concentration, the reaction is a FIRST ORDER REACTION. In this case the reaction is first order with respect to the N2O5

  11. If we plotted concentration vs. rate (instead of concentration vs. time, like before), we would get a straight line for this reaction, showing the direct correlation between concentration and rate. rate Conc.

  12. Therefore the rate expression for the decomposition reaction of N2O5 : • 2N2O5(g)  4NO2(g) + O2(g) Is: rate = k [N2O5] Note that for a fast reaction ‘k’ is large and for a slow reaction ‘k’ is small. Note that as temperature increases, ‘k’ increases.

  13. Note that we can also calculate ‘k’ given the rate and the concentration simply by rearranging the equation. k = rate / [N2O5] Using our previous data, k = 0.056 mol / L min 0.160 mol /Lk = 0.35 / min Take note of the unit cancellation and what units you are left with for ‘k’. The same result would be obtained from any of the data points from our table, with some experimental error. (because all data points were taken at the same temperature).

  14. Rate expressions have been established for many different types of reactions. We have looked at a first order reaction. For the reaction of a single reactant, A (g) products The rate was expressed as rate = k (conc. A)m where ‘m’ describes the order of the reaction. In the reaction we studied, m=1, for a first order reaction. The power, m, could also be zero, for a zero order reaction, and m = 2 for a second order reaction. .

  15. m = 0, zero order reaction, rate is independent of reactant concentration m = 1, first order reaction, rate is directly proportional to concentration. m = 2, second order reaction, rate is proportional to the concentration squared, (ex. 2x conc. = 4x rate)

  16. To mathematically determine the order of a reaction: Rate 2 = k (conc.2)m Rate1 = k (conc.1)m plug in data from a data table and solve for the power ‘m’.

  17. Sample problem: Determine the order with respect to the reactant, for the following reaction: CH3 CHO(g) CH4(g) + CO(g) DATA: CONC. CH3 CHO 0.10 0.20 0.30 0.40 RATE (MOLE / L.S) 0.085 0.34 0.76 1.4 Now take two rate expressions and divide them to solve for ‘m’: Rate2 = k(conc.2)m , 0.34 = (0.20)m , 4 = 0.20 m Rate1 = k(conc.1)m 0.085 (0.10)m 0.10

  18. Therefore, 4 = 2m and so m = 2. This reaction is second order with respect to the reactant. Rate = k [CH3 CHO] 2 Now we can use the same data to determine ‘k’ for this reaction: k = rate k = 0.085 mol / L. s [CH3 CHO] 2 (0.10 mol / L) 2 8.5 L / mol . s (notice the units)

  19. The units of k = L / mol . s are the units when the reaction is second order. So far we have seen k have units of k = s-1 for a first order reaction and k = L / mol .s for a second order reaction. We can now determine the rate of the reaction at [CH3 CHO] = 0.50M Rate = k [CH3 CHO] 2 = 8.5 L 0.50 mol 2 mol . S L rate = 2.1 mol / L. s

  20. At what concentration ofCH3 CHO is the rate of reaction =0.20 mol / L. s ? EASY !!! Rate = k [conc.] 2 0.20 mol / L. s = (8.5 L / mol. s) x [conc.] 2 Therefore: 0.20 mol / L s = [conc.] 2 8.5 L / mol s Therefore concentration = 0.15 mol / L

  21. Now we are ready to work with reactions that have more than one reactant. aA + bB  products The general rate expression is : rate = k (conc. A)m x (conc. B)n Where ‘m’ is the order of the reaction with respect to reactant ‘A’ and ‘n’ is the order with respect to reactant ‘B’. The overall rate of the reaction is the sum of m + n

  22. Sample problem: For the reaction: CO(g) + NO2(g) CO2(g) + NO(g) Experimentally, this reaction was determined to have the following rate expression, above 600k: rate = k (conc. CO) x (conc. NO2) Therefore this reaction is first order with respect to CO and first order with respect to NO2 and is thus second order, overall.

  23. To determine the order of a reaction with more than one reactant, one reactant concentration is held constant while the other is changed. Determine the order for the reaction below, given the data that follows: 2 H2 + 2 NO(g) N2(g) + 2 H2 O(g) The data tables follow. In table one, NO is held constant. In table two, H2 is held constant. We can look at the data to determine the order. Then plug in data into the rate2 / rate1 equation to prove it.

  24. Table one: Trial H2(M) NO(M) rate 1 0.10 0.10 0.10 2 0.20 0.10 0.20 3 0.30 0.10 0.30 4 0.40 0.10 0.40 You can see that rate is directly proportional to [ H2 ] , therefore the reaction is first order with respect to H2 (m=1).

  25. Table two: • H2 NO rate • 0.10 0.10 0.10 • 0.10 0.20 0.40 • 0.10 0.30 0.90 • 0.10 0.40 1.6 • Here you can see that the rate of reaction is proportional to the concentration squared, therefore the reaction is second order with respect to NO, (m = 2). • So the rate expression for the reaction is: • Rate = K [H 2 ]1 x [ NO ]2and the reaction is third order overall.

  26. We can prove this the way we found reaction order before: From table 1: 0.20 = 0.20 m 0.10 0.10 (this is from rate2 / rate1 =……) Therefore, m = 1 From table 2: 0.40 = 0.20 n 0.10 0.10 Therefore, 4 = 2 nand therefore, n = 2

  27. We can also find the concentration of a reactant after a period of time (t)or the time requiredfor a reactant to drop to a certain concentration. The equation: ln (conc. Initial) = Kt (conc. Final) Is derived from the straight line equation, y = mx + b Ln (conc. Initial) = ln (conc. Final) – Kt Or ln (conc. Initial) = - Kt + ln (conc. Final) you can see that K is the slope. Subtracting the ln (conc. Final) from both sides gives the equation. Using Log, base ten, we must divide Kt / 2.30, a conversion factor.

  28. So, the equation, using Log is as follows: Log (conc. Initial) = kt Log (conc. Final) 2.30 This is the concentration vs. time relationship for a first order equation. Lets try an example………….

  29. Calculate the [N2O5] after 4.0 min, starting with a conc. Of 0.160M and k = 0.35 / min • Log [N2O5] 1 = kt • [N2O5] 2 2.30 • Log 0.160M = 0.35 4.0min • x min 2.30 • Log x = -1.40, therefore, x = 10 –1.40= 0.039 or 0.04

  30. Calculate the time required for the concentration to drop from 0.160 M to 0.100 M. • This time we will use ln rather than Log. Here we do not need the conversion factor of 2.30. • Ln [N2O5] 1 = Kt • [N2O5] 2 • ln 0.160 – ln 0.100 = (0.35min-1) t • or ln 0.160 / 0.100 = (0.35min-1) t • Either way you get, 0.47 = (0.35 min-1) t • And therefore t = 1.34 min.

  31. Calculate the time required for half of a sample of the N2O5 to be decomposed. (note, this is called the half life and is represented by t1/2 ). • When ½ of the [N2O5] 1 is decomposed, the [N2O5] is then equal to [N2O5] 1 / 2 • Therefore, ln[ N2O5 ] 1 = kt • [ N2O5 ] 1 • 2 • ln 2 = Kt, 0.693 = Kt, 0.693 = (0.35 min-1) t • 1.98 min = t You should remember that ln 2 = 0.693 and t1/2 = 0.693 / k for a 1st order rxn.

  32. Half life, t1 / 2 , is inversely proportional to K. • If a reaction is fast, then K is large and the reaction will have a short half life (or small t1 / 2 ) • Reactions of gases are most commonly second order overall. • Look at your chart • One can determine the order of a reaction by plottingt vs. [conc.],t vs. ln [conc.] , or t vs. 1 / [conc.]. A straight line for each of these plots will indicate zero order, 1st order, or 2nd order, respectively.

  33. Kinetic Models:Activation Energy

  34. ACTIVATION ENERGY : Ea HAS UNITS OF kj Not every collision between particles leads to a reaction. The particles must collide with enough force to break bonds. The total kinetic energy between colliding particles must at least equal the Ea (Activation Energy) in order for the reaction to occur between them. If Ea large, then reaction is slow.(requires a lot of activation energy, or, there aren’t enough collisions with enough energy). If Ea small, then reaction is fast.

  35. Ea is independent of temperature and concentration. Activated complex Ea = 134 Kj reactants Ea’ = 360Kj DH = -226Kj products The difference between Ea and Ea’ = DH rxn DH rxn = Ea - Ea’ The activated complex is an intermediate, unstable species formed before the products.(pg 588 text)

  36. Since we know that the rate of a reaction increases with temperature increase, we can see that K will also increase with an increase in temperature. (K becomes larger as rate increases). An expression from the kinetic theory for showing the fraction of molecules that collide with enough energy, Ea, for the reaction to occur is: f = e–Ea / RTwhere e = inverse ln Ea = activation energy R = 8.314 J/mol. K T = temperature (Kelvin)

  37. Since K and f are similar, that is, more particles with Ea = greater K, then K = constant x f, or K = cf. C is just a proportionality constant and is usually represented by ‘A’, therefore, K = Af, or K = A e –Ea/ RT or ln K = ln A – (Ea / RT), and since ln A is a constant, ln K = A – Ea / RT This is called the Arrhenius equation

  38. Once again, we can use the equation for a line, y = mx + b and see that ln k = -Ea 1 + A where we can see that R T A is b,the y -intercept. Remember, whenever using Log instead of ln, you must divide by 2.30 to convert. Log K = A – Ea 2.30 RT Plotting this should give a straight line and we can solve for the activation energy, Ea: Plot Log K vs. 1 / T (because Log K is y and 1 / T is x). Then m = Log K2 – Log K1 = Dy = -Ea 1/ T2 - 1 / T1 Dx 2.30 R And since m = -Ea / 2.30 R, Then Ea = - ( m) (2.30) (R)

  39. Using the Two Point form of the line equation gives us the Clausius – Clapeyron Equation: The two point form:Y2 – y1 = m (x2 – x1) This involves two rate constants and two temperatures. Log K2 – Log K1 = Ea 1 - 1 2.30 R T2 T1 This can be made simpler, the Clausius –Clapeyron equ. Log K2 = Ea T2 - T1 K1 2.30 R T2 T1

  40. Sample problem: The activation energy of a reaction is 9.32 x 104 J at 27 oC, and K = 1.25 x 10-2L / mol . s. Calculate K at 127 oC. First convert Celsius to Kelvin: T1 =27 + 273 = 300 k, T2 = 127 oC = 400k. Log K2 = 9.32 x 104 400 - 300 1.25 x 10-2 2.30 (8.314) 400 x 300 Log K2 = 4.06 ; K2 = 104.06 ; K2 =144L/mol s 1.25 x 10-2 1.25 x 10-2

  41. Calculate T2 given that: K2 = 2.50 x 10-2 L / mol s • K1 = 1.25 x 10-2 L / mol s • T1 = 300 K • Ea = 9.32 x 104 J Log 2.50 x 10-2 =9.32 x 104 T2 - 300 1.20 x 10-2 2.30 (8.31) T2 x 300 Solving the equation gives T2 = 306 K

  42. Reaction Mechanisms – the path or sequence of stepls involved when a chemical reaction occurs. A rate expression can sometimes be determined from a reaction mechanism. THERE ARE THREE RULES: 1. The coefficient of each reactant in each step of the mechanism is its order: example, Step 1: NO2 = NO2 NO3 + NO RATE = K1[ NO2] x [NO2] = K1[NO2]2 Step 2: NO3 + CO  NO2 + CO RATE = K2 [NO3] x [ CO] continued…….

  43. The slow step is therate determining step. • In the above reaction, step one is slow, therefore it is the rate determining step. (Think of it as who will determine which team wins a relay race, the fastest runner or the slowest runner?) • therefore; Rate = K1 [NO2]2 • The final rate expression can include only the species for the overall balanced chemical reaction, not the intermediates. • Lets do a sample problem:

  44. The decomposition of O3 to O2 is a two step mechanism as follows: step 1: O3 O2 + O fast step step 2: O3 + O  2 O2 slow step 2 O2  3 O2 Since the rate of a reaction is determined by the slowest step, step 2 is the rate determining step. Therefore the rate expression would be: Rate = K2 [O3] x [O] But, [O] is an intermediate and not allowed in the overall rate expression. Here is what we do……….

  45. Write an equilibrium expression for the first step, and then solve it for [O]. Then we can plug that solution into the rate expression in place of [O]. Kc = [O2] [O] REMEMBER, THIS [O3] COMES FROM PRODUCTS OVER REACTANTS. NOW SOLVE THAT EXPRESSION FOR [O]: Kc [O3] [O2] = [O] NOW SUBSTITUTE: RATE = K2 x Kc x [O3]2Since K2 and Kc [O2] are constants…..

  46. Rate = K [O3]2 [O2] Note that since O2 is in the denominator, the rate is inversely proportional to [O2]. So if [O2] is increased, reaction rate is slowed (LeChatelier’s Principle).

  47. Some notes on Catalysis: A catalyst speeds up a reaction by lowering the activation energy (Ea). It does this by providing another pathway for the reaction. There are two types of catalysts: HOMOGENEOUS AND HETEROGENEOUS. Homogeneous – The reaction takes place in one phase. The catalyst is often a solute dissolved in a reaction solution (thus, they are all in the aqueous phase).

  48. Heterogeneous – The catalyst is in a different phase from the reactants. Often in industry, a solid metal may be a catalyst for a gaseous reaction to take place on its surface. The bond between the metal surface and an atom of the gaseous molecule, weakens the molecular bonds in the gas, thus making it easier for the reaction to proceed. IMPORTANT NOTE ABOUT CATALYSTS – THEY DO NOT AFFECT EQUILIBRIUM. THAT IS, A REATION WILL NOT SHIFT LEFT OR RIGHT UPON THE ADDITION OF A CATALYST. THEY ONLY AFFECT RATE.

  49. In biological systems, enzymes are catalysts. There are specific enzymes for each step of individual biochemical reactions. Temperature affects on reaction rate: An approximate and general rule is that a 10oC increase in temperature will double the reaction rate. For example: Food at room temperature (25oC) will decay ~4 times faster than food in the refrigerator (5oC). Interestingly, food cooked at 110 oC will cook twice as fast as food cooked at 100oC. WHY????

  50. Heating increases kinetic energy of particles. • In order for a reaction to occur, particles must collide with enough kinetic energy to reach the activation energy (Ea). • Heating, which increases K.E., also will increase the number of particles colliding with enough K.E. to react.

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