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CHEMICAL KINETICS REACTION RATES BY JOANNE SWANSON.

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1 CHEMICAL KINETICS 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: 3H 2 + N 2  2NH 3 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 = -  [conc.] /  ime where  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: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) The rate expression could be shown as: rate = -  [N 2 O 5 (g) ] /  T  We would have an experimentally determined table of (concentration vs. time) data: Time (min) Conc. N 2 O 5 (M) Instantan- taneous rate mole L. min

6 Sample Question  Calculate the average rate of the decomposition of mole N 2 O 5 in a 1L container at 67 o C, measured at one minute intervals, (using the data table), between a) t=0 and t = 1min and b) t= 1 and t = 2 min. a) Rate = -      = M M = mol 1 min – 0 min L. min b) Rate = M – M = mol 2 min – 1 min L. min

7 Determining Change in Concentration: What is the change in conc. Of NO 2 in the first minute? (in the same reaction)  2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) Initial Final (0.047) ½(0.047) Therefore the  [NO 2 ] = 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 =  conc. /  T The slope of the tangent to the curve from the plotted concentration vs. time data is equal to the instantaneous rate.

9  y = slope = instantaneous rate  x  y  x

10 Reaction Order Looking at our earlier data table, you can see that as concentration 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. N 2 O 5 ) 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 N 2 O 5

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 N 2 O 5 :  2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) Is: rate = k [N 2 O 5 ] 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 / [N 2 O 5 ] Using our previous data, k = mol / L min mol /L k = 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 Rate 1 = 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: CH 3 CHO (g)  CH 4(g) + CO (g) DATA: CONC. CH 3 CHO RATE (MOLE / L. S) Now take two rate expressions and divide them to solve for ‘m’: Rate 2 = k(conc. 2 ) m, 0.34 = (0.20) m, 4 = 0.20 m Rate 1 = k(conc. 1 ) m (0.10) m 0.10

18 Therefore, 4 = 2 m and so m = 2. This reaction is second order with respect to the reactant. Rate = k [CH 3 CHO] 2 Now we can use the same data to determine ‘k’ for this reaction: k = rate k = mol / L. s [ CH 3 CHO] 2 (0.10 mol / L) 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 [CH 3 CHO] = 0.50M Rate = k [CH 3 CHO] 2 = 8.5 L 0.50 mol 2 mol. S L rate = 2.1 mol / L. s

20 At what concentration of CH 3 CHO is the rate of reaction = 0.20 mol / L. s ? EASY !!!Rate = k [conc.] mol / L. s = (8.5 L / mol. s) x [conc.] 2 Therefore: 0.20 mol / L s = [conc.] 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) + NO 2(g)  CO 2(g) + NO (g) Experimentally, this reaction was determined to have the following rate expression, above 600k: rate = k (conc. CO) x (conc. NO 2 ) Therefore this reaction is first order with respect to CO and first order with respect to NO 2 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 H NO (g)  N 2(g) + 2 H 2 O (g) The data tables follow. In table one, NO is held constant. In table two, H 2 is held constant. We can look at the data to determine the order. Then plug in data into the rate 2 / rate 1 equation to prove it.

24 Table one: TrialH 2 (M)NO(M)rate You can see that rate is directly proportional to [ H 2 ], therefore the reaction is first order with respect to H 2 (m=1).

25 Table two: H 2 NOrate 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 ] 2 and 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 (this is from rate 2 / rate 1 =……) Therefore, m = 1 From table 2: 0.40 = 0.20 n Therefore, 4 = 2 n and therefore, n = 2

27 We can also find the concentration of a reactant after a period of time (t) or the time required for 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 Orln (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 1.Calculate the [N 2 O 5 ] after 4.0 min, starting with a conc. Of 0.160M and k = 0.35 / min Log [N 2 O 5 ] 1 = kt [N 2 O 5 ] Log 0.160M = min x min 2.30 Log x = -1.40, therefore, x = 10 –1.40 = or 0.04

30 2.Calculate the time required for the concentration to drop from M to M. This time we will use ln rather than Log. Here we do not need the conversion factor of Ln [N 2 O 5 ] 1 = Kt [N 2 O 5 ] 2 ln – ln = (0.35min -1 ) t or ln / = (0.35min -1 ) t Either way you get, 0.47 = (0.35 min -1 ) t And therefore t = 1.34 min.

31 3.Calculate the time required for half of a sample of the N 2 O 5 to be decomposed. (note, this is called the half life and is represented by t 1/2 ). When ½ of the [N 2 O 5 ] 1 is decomposed, the [N 2 O 5 ] is then equal to [N 2 O 5 ] 1 / 2 Therefore, ln [ N 2 O 5 ] 1 = kt [ N 2 O 5 ] 1 2 ln 2 = Kt, = Kt, = (0.35 min -1 ) t 1.98 min = tYou should remember that ln 2 = and t 1/2 = / k for a 1 st order rxn.

32 Half life, t 1 / 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 t 1 / 2 ) Reactions of gases are most commonly second order overall. Look at your chart One can determine the order of a reaction by plotting t vs. [conc.], t vs. ln [conc.], or t vs. 1 / [conc.]. A straight line for each of these plots will indicate zero order, 1 st order, or 2 nd order, respectively.

33 Kinetic Models: Activation Energy

34 ACTIVATION ENERGY : E a 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 E a (Activation Energy) in order for the reaction to occur between them. If E a large, then reaction is slow.(requires a lot of activation energy, or, there aren’t enough collisions with enough energy). If E a small, then reaction is fast.

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

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, E a, for the reaction to occur is: f = e –E a / RT where e = inverse ln E a = activation energy R = J/mol. K T = temperature (Kelvin)

37 Since K and f are similar, that is, more particles with E a = 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 –E a / RT or ln K = ln A – (E a / RT), and since ln A is a constant, ln K = A – E a / 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 = -E a 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 – E a 2.30 RT Plotting this should give a straight line and we can solve for the activation energy, E a : Plot Log K vs. 1 / T (because Log K is y and 1 / T is x). Then m = Log K 2 – Log K 1 =  y = -E a 1/ T / T 1  x 2.30 R And since m = -E a / 2.30 R, Then E a = - ( m) (2.30) (R)

39 Using the Two Point form of the line equation gives us the Clausius – Clapeyron Equation: The two point form: Y 2 – y 1 = m (x 2 – x 1 ) This involves two rate constants and two temperatures. Log K 2 – Log K 1 = E a R T 2 T 1 This can be made simpler, the Clausius –Clapeyron equ. Log K 2 = E a T 2 - T 1 K R T 2 T 1

40 Sample problem: The activation energy of a reaction is 9.32 x 10 4 J at 27 o C, and K = 1.25 x L / mol. s. Calculate K at 127 o C. First convert Celsius to Kelvin: T 1 = = 300 k, T 2 = 127 o C = 400k. Log K 2 = 9.32 x x (8.314) 400 x 300 Log K 2 = 4.06 ; K 2 = ; K 2 =144 L/mol s 1.25 x x 10 -2

41 b.Calculate T 2 given that: K 2 = 2.50 x L / mol s K 1 = 1.25 x L / mol s T 1 = 300 K E a = 9.32 x 10 4 J Log 2.50 x = 9.32 x 10 4 T x (8.31) T 2 x 300 Solving the equation gives T 2 = 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: NO 2 = NO 2  NO 3 + NO RATE = K 1 [ NO 2 ] x [NO 2 ] = K 1 [NO 2 ] 2 Step 2: NO 3 + CO  NO 2 + CO RATE = K 2 [NO 3 ] x [ CO] continued…….

43 2.The slow step is the rate 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 = K 1 [NO 2 ] 2 3.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 O 3 to O 2 is a two step mechanism as follows: step 1: O 3  O 2 + Ofast step step 2: O 3 + O  2 O 2 slow step 2 O 2  3 O 2 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 = K 2 [O 3 ] 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]. K c = [O 2 ] [O]REMEMBER, THIS [O 3 ]COMES FROM PRODUCTS OVER REACTANTS. NOW SOLVE THAT EXPRESSION FOR [O]: K c [O 3 ] [O 2 ] = [O]NOW SUBSTITUTE: RATE = K 2 x K c x [O 3 ] 2 Since K 2 and K c [O 2 ] are constants…..

46 Rate = K [O 3 ] 2 [O 2 ] Note that since O 2 is in the denominator, the rate is inversely proportional to [O 2 ]. So if [O 2 ] 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 (E a ). 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 10 o C increase in temperature will double the reaction rate. For example: Food at room temperature (25 o C) will decay ~4 times faster than food in the refrigerator (5 o C). Interestingly, food cooked at 110 o C will cook twice as fast as food cooked at 100 o C. WHY????

50 1.Heating increases kinetic energy of particles. 2.In order for a reaction to occur, particles must collide with enough kinetic energy to reach the activation energy (E a ). 3.Heating, which increases K.E., also will increase the number of particles colliding with enough K.E. to react.

51 Summary: The rate of a reaction may be obtained from data plugged into the equation, rate = -  [X] /  T. From this equation, one can find rate, change in concentration of X, or change in time. The rate expression for a 1 st order reaction is: rate = K [X]. From this we can also calculate K. One can determine an instantaneous rate by plotting the conc. vs. time data and finding the slope to the tangent, which is the the change in concentration over the change in time. The differential rate expression is determined by the dependence of rate on concentration.

52 The integrated rate expression is determined by the dependence on concentration with time. For a first order reaction, a plot of ln [x] vs. time will give a straight line. If this graph does not give a straight line, the reaction is not first order in (X). This comes from the equation for the straight line, rearranged to give ln [X 0 ] =Kt This equation may be used to find [X] change in concentration over time. The half life equation for a first order reaction is: t 1/2 = / K To determine the order of a reaction, with regard to one reactant, use the equation: rate 2 = k [X 2 ] m solve for m rate 1 k [X 1 ] m

53 Once the order has been established, the rate expression may be used to determine K or rate by plugging in a concentration and the order: rate = K [X] m To determine the overall order of a reaction with two reactants, hold the concentration of one constant, while changing the other, then repeat for the other reactant: rate 2 = [X 2 ] m hold [y] rate 1 [X 1 ]constant Then repeat for [y 2 ] / [y 1 ] holding [X] constant.

54 For a second order reaction the rate expression is Rate = K [X] 2. The integrated form of the rate expression is: = Kt [X] [X 0 ] It can be used to solve for any of the variables within. The equation to find the half life (t 1/2 ) for a 2 nd order reaction is : t 1/2 = 1 / K [X 0 ] For a zero order reaction the rate expression is: Rate = K and the integrated form is: [X] – [X 0 ] = Kt and t 1/2 = [X 0 ] / 2K

55 Other equations studied were the Arrhenius equation: ln K = A – E a / RT which is useful when activation energy is involved in a problem. And the Clausius –Clapeyron equation (from the two point form of the line equation). It is useful for calculating K 1 or K 2 or T 1 or T 2 or E a The equation is:Log K 2 = E a T 2 – T 1 K R T 2 T 1


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