Chapter 1 Reaction Kinetics PDF

CHAPTER 1 REACTION KINETICS 1 2  CHAPTER 1 REACTION KINETICS REACTION KINETICS Chemical kinetics, also known as reaction kinetics, is the study of rates of chemical processes. Chemical kinetics includes investigations of how different experimental...

CHAPTER 1 REACTION KINETICS 1 2  CHAPTER 1 REACTION KINETICS REACTION KINETICS Chemical kinetics, also known as reaction kinetics, is the study of rates of chemical processes. Chemical kinetics includes investigations of how different experimental conditions can influence the speed of a chemical reaction and yield information about the reaction's mechanism and transition states. Rates of chemical reaction depend on the inherent characteristics or nature of the reactants. That is the reason, why some reactions are fast while others are slow, e.g., oxidation of ferrous ion by KMNO4 (potassium permanganate) in acidic medium is fast, while theoxidation of oxalate ion by the same reagent is slow.Apart from the nature of the reactants, the rate of a reaction may be affected by other factors such as temperature, light, catalyst, concentration of reactants and nature of solvent. A drug, in a pharmaceutical product, is a chemical compound that undergoes chemical changes with time. It may react with the environment and other excepients in the formulation. Application of study of chemical kinetics to the pharmaceutical field involve: 1. Stabilization of drug products and prediction of shelf life and optimum storage conditions. 2. Drug release from dosage forms in dissolution studies 3. Absorption, distribution, and elimination studies on the drugs in the body 4. Explaining the drug action at the molecular level when the response is a rate process. 3  CHAPTER 1 REACTION KINETICS In all the cases the concepts are the same and the equations are also same. HOW DOES A CHEMICAL REACTION OCCUR? Reactions usually require collisions between reactant molecules or atoms. The formation of bonds requires atoms to come close to one another. New bonds can form only if the atoms are close enough together to share electrons. Some collisions are not successful. These are called ineffective collisions. The particles simply hit and then rebound. Fig. 1-1 Effective and ineffective collision Collisions that lead to products are called effective collisions. In order for the collision to be effective, the molecules have to have: 4  CHAPTER 1 REACTION KINETICS 1. Correct Particle Orientation The particles have to collide the right way around. In the reaction between ethene and hydrogen chloride: The reaction can only happen if the hydrogen end of theH- Cl bond approaches the carbon-carbon double bond. Any other collision between the two molecules doesn't work. Fig. 1-2 Correct and incorrect orientation for the reactionbetween ethene and hydrogen chloride 2. Activation Energy (Ea) Even if the species are orientated properly, a reaction won't occur unless the particles collide with a certain minimum energy called the activation energy of the reaction. 5  CHAPTER 1 REACTION KINETICS Fig. 1-3 Activation energy REACTION RATES Reaction rate is the velocity with which a reactant orreactants undergo a chemical change. Rate of reaction is defined as the change in concentration with time. (It is measured in mol ml-1sec-1) This change may be a decrease in concentration of reactants or an increase in concentration of products change in time change in time 6  CHAPTER 1 REACTION KINETICS Fig. 1-4 Change of concentration of reactant A and product Bwith time (A B) Reaction rate could be expressed in 3 different ways1. Average Rate The ratio of change of concentration of a reactant to the time needed for this change to occur. concentration of A at time t2 − concentration of A at time t1 Rate = t2 − t1 d[ A] Rate = dt Note: The rate will always be defined as a positivequantity Example 1 In this reaction, the concentration of NO2 was measured atvarious times, t. 2NO2 (g) → 2NO(g) + O2 (g) 7  CHAPTER 1 REACTION KINETICS Time NO2 (sec) (mol/L) 0 0.0100 50 0.0079 100 0.0065 The average rate over the first 50 seconds: [NO2 ]t =50 −[NO2 ]t =0 = 50 s − 0 s.0079 mol / L − 0.0100 mol / L = = −4.210−5 mol / L. s 50 s Since concentration of reactants decrease with time, rate expression involving a reactant will include a negativesign. The rate from 0 to 50 sec is: d[NO2 ] Rate = − dt = −(−4.210−5 )mol / L.s = 4.210−5 mol / L.s Example 2 C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) In this reaction, the concentration of butyl chloride,C4H9Cl, was measured at various times, t. 8  CHAPTER 1 REACTION KINETICS The average rate was calculated for each time interval.The average rate was found to decrease as the reaction proceeds because as the reaction goes forward, there are fewer collisions between reactant molecules. 2. Instantaneous Rate Rate of change of concentration at any particular instant during the reaction. To determine the instantaneous rate we have to plotconcentration versus time. If we take the data of the previous example, we can plotthe curve as follows: 9  CHAPTER 1 REACTION KINETICS Fig. 1-5 Plot of concentration versus time showing the determination of instantaneous and initial rate The slope of a line tangent to the curve at any point is the instantaneous rate at that time. 3. Initial Rate: Initial rate is the instantaneous rate at t = 0, that is whenthe reactants are first mixed 10  CHAPTER 1 REACTION KINETICS REACTION RATES AND STOICHIOMETRY For the example: C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) The ratio of C4H9Cl to C4H9OH is 1:1. Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH. Rate of reaction = - Δ C4H9Cl Δ C4H9OH = Δt Δt What if the ratio is not 1:1?For the example: A+2B → 3C The stoichiometric coefficient for species B is twice thatof species A; thus the concentration of B will declinetwice as fast as that of species A. Similarly, theconcentration of species C increases three times as fast as the concentration of A decreases. Conceptually there should be a single, definite rate for a reaction. How might such a rate be defined given thehighly varied rates of change for the various species in the reaction? The rate of reaction, r, is defined to be the change in concentration over change in time for a species divided by the stoichiometric coefficient of that species. Additionally, if the species is a reactant, the negative sign is used, because a positive rate is desired. 11  CHAPTER 1 REACTION KINETICS For the previous example (A + 2B → 3C) the rate of reaction is calculated as follows: d[A] Rate of reaction = - =- 1 d[B]= 1 d[C] dt 2 dt 3 dt Similarly: H2(g) + I2(g) 2 HI(g) To generalize: For the reaction aA + bB cC + dD where a, b, c and d are the stoichiometric coefficients forthe reactants A, B, C and D respectively. This has the advantage that the same rate of reaction is obtained, no matter which reactant or product is studied. Problem: H2 + Br2 2 HBr If the change in Br2 in 0.01 second is - 0.001 mol/L. Calculate v , d[H2]/dt, d[Br2]/dt and d[HBr]/dt 1 Solution: v= 12  CHAPTER 1 REACTION KINETICS d[H 2 ] 1 = d[Br2 ] 1 = d[HBr ] −1 dt −1 dt 2 dt 13  CHAPTER 1 REACTION KINETICS v = 0.1 mol L-1 sec-1 d[H2]/dt = - 0.001/0.01 = - 0.1 mol L-1 sec-1 d[Br2]/dt = - 0.001/0.01 = - 0.1 mol L-1 sec-1 d[HBr]/dt = 0.002/0.01 = 0.2 mol L-1 sec-1 REACTION RATE AND REVERSIBILITY In case of reversible reactions the change in the concentration of reactant or product depends on the difference in the rates of the forward and reversereactions. This complication can be avoided if we study the rate of a reaction under conditions where the reverse reaction makes only a negligible contribution. RATE LAWS In many reactions, the rate of reaction changes as thereaction progresses. Initially the rate of reaction isrelatively large, while at very long times the rate of reaction decreases to zero (at which point the reaction is complete). In order to characterize the kinetic behavior ofa reaction, it is desirable to determine how the rate of reaction varies as the reaction progresses. A rate law is a mathematical equation that describes the progress of the reaction. In general, rate laws must be determined experimentally i.e. it is not possible to predict the rate law from the overall chemical equation. There are two forms of a rate law for chemical kinetics: the differential rate law and the integrated rate law. 14  CHAPTER 1 REACTION KINETICS DIFFERENTIAL RATE LAW (RATE LAW) It expresses the relationship between concentration of reactants and rate of the reaction. The differential rate law is usually just called “the rate law.” For the following reaction: The effect of changing the concentration of NH4+ (the concentration of NO2- was kept constant) on the rate of reaction was studied and it was found that the rate is directly proportional to the concentration of NH4+. The same study was done for NO2- and was also found that therate of reaction was directly proportional to the concentration of NO2-, thus Rate α [NH4+], andRate α [NO2-], thus Rate α [NH4+][NO2-], thusRate = k [NH4+][NO2-] This equation is called the rate law and k is the rate constant. (note: square brackets stand for concentration in moles/liter) Generally: Consider the following reaction nA + mB → products the rate of the reaction is related to the concentrations of A and B in this way: 15  CHAPTER 1 REACTION KINETICS where: k = reaction rate constant a, b =orders of reaction with respect to A & B respectively [A], [B] = molar concentration of A & B respectively The rate constant (k) − The units of k depend on the overall order of reaction − The value of k is constant for a specific reaction, but, it’s value varies with change in temperature and catalyst (not a true constant) − The value of k is determined experimentally. Reaction orders − It gives you information about how concentrations affect the rate of the reaction. − They cannot be deduced from the stoichiometric coefficients in the balanced overall equation, they have to be found experimentally. − The overall reaction order is the sum of the exponents. − The exponent can have integer values of 0, 1, 2 or 3 and fractional values like 1/2 or 3/2. However the orders of reaction are likely to be 0, 1 or 2. 16  CHAPTER 1 REACTION KINETICS Let us discuss the different orders: For the equation: nA → product, the rate law will be: rate = k [A]a In the case of: 1. Zero order: The exponent = 0, any number raised to the power of zero(x0) is equal to 1, thus, rate = k [A]o = k Rate of reaction is a constant, i.e. the rate does not dependon the concentration of that reactant 2. First order: For an exponent of 1, the rate law will be rate = k [A]1 That means that rate is directly proportional to the concentration of that reactant, i.e. if concentration is doubled, rate doubles. 3. Second Order For an exponent of 2, the rate law will be rate = k [A]2 That means that if the concentration of that reactant is doubled rate is quadrupled. Examples Each of the following examples involves a reaction between A & B, i.e. nA + mB → product 17  CHAPTER 1 REACTION KINETICS For each example the rate law is given and the order of reaction of A and B individually and the overall reaction order should be found. Example 1: rate = k [A] [B] Solution The order of reaction with respect to both A and B is 1. The overall order of reaction is 2, found by adding up the individual orders. Example 2: rate = k [B]2 Solution This reaction is zero order with respect to A because the concentration of A doesn't affect the rate of the reaction. The order with respect to B is 2. The reaction is second order overall (because 0 + 2 = 2). Example 3: rate = k [A] Solution This reaction is first order with respect to A Zero order with respect to B, because the concentration of B doesn't affect the rate of the reaction. The reaction is first order overall (because 1 + 0 = 1). Determining the Form of the Rate Law The determination of the form of the rate law, i.e. finding the power to which each reactant concentration is raised in the rate law, is done experimentally. 18  CHAPTER 1 REACTION KINETICS There are three methods: 1. Experimental determination 2. Method of initial rates 3. Isolation method 1. Experimental Determination The concentration of the reactant at different timeintervals is determined and the graph of concentration versus time is plotted. e.g. decomposition of dinitrogen pentoxide 2N2O5 (so ln) → 4NO2 (so ln) + O2 (g) For this reaction the rate law will be: rate = k [N2O5]n Fig. 1-6 Plot of N2O5 concentration versus time showing thedetermination of instantaneous at two different concentrations The reaction rate at two different concentrations is evaluated by taking the slopes of the tangents to the curve at these points 19  CHAPTER 1 REACTION KINETICS (Instantaneous rate). Rate (mol/L.s) Note that when [N2O5] is halved, the rate is also halved. This means that the rate of this reaction depends on the concentration of N2O5 to the first power. rate = k [N2O5]1= k [N2O5] 2. Method of Initial Rates The initial rate of a reaction is the instantaneous rate determined just after the reaction begins. Several experiments are carried out using different initial concentrations, and the initial rate is determined for each run. The results are then compared to see how the initial rate depends on the initial concentrations. e.g. NH 4 + (aq) + NO2 − (aq) → N2 (g) + 2H 2O(l) Initial Initial Initial Rate Experiment Concentration Concentration (mol/L.sec) of NH4+ of NO2- 1 0.100 M 0.005 M 1.35 × 10-7 2 0.100 M 0.010 M 2.70 × 10-7 3 0.200 M 0.010 M 5.40 × 10-7 20  CHAPTER 1 REACTION KINETICS The general form for the rate law for this reaction: Rate = k [NH + 4]n [NO − 2]m 1. Determination of the order of [NO2-] We choose experiments 1 and 2 as the concentrations ofNH + are constant 4 while that of NO - are changed. 2 From experiment 1 1.35 × 10-7 mol/L.sec = k (0.100)n(0.005)m From experiment 2 2.70 × 10-7 mol/L.sec = k (0.100)n(0.010)m By division Rate 2 2.70 10−7 mol / L. sec k (0.100 mol / L)n (0.010 mol / L)m = = Rate1 1.3510−7 mol / L. sec k (0.100 mol / L)n (0.005 mol / L)m Rate 2 = 2.0 = (2.0)m Rate1 Therefore, m = 1 2. Determination of the order of [NH4+] We choose experiments 2 and 3 as the concentrations ofNO - are constant 2 while that of NH + are changed. 4 From experiment 2 2.70 × 10-7 mol/L.sec = k (0.100)n(0.010)m From experiment 3 5.40 × 10-7 mol/L.sec = k (0.200)n(0.010)m By division 21  CHAPTER 1 REACTION KINETICS Rate 3 5.40 10−7 mol / L. sec (0.200 mol / L)n = = Rate 2 2.70 10−7 mol / L. sec (0.100 mol / L)n n  0.200  = 2.0 =   = (2.0)n  0.100  Therefore, n = 1 The rate law will thus be: Rate = k [NH + ]1 [NO − ]1 4 2 The rate constant k can be calculated as follows 1.35 × 10-7 mol/L.sec = 2.7 x 10-4 L/mol.sec k= (0.1 mol/L) (0.005 mol/L) Problem: Determine the rate law and calculate the rate constant forthe following reaction from the following data: 2- - S2O (aq) + 3I 8 (aq) 2SO42- (aq) + I3- (aq) [S2O 82-] Initial Rate Experiment [I-] (M/s) 1 0.034 0.08 2.2 x 10-4 2 0.017 0.08 1.1 x 10-4 3 0.017 0.16 2.2 x 10-4 Solution: rate = k [S2O82-]x[I-]y 22  CHAPTER 1 REACTION KINETICS Double [I-], rate doubles (experiment 1 & 2), thus y = 1 Double [S2O 2-], 8rate doubles (experiment 2 & 3), thusx=1 Rate law: rate = k [S2O82-]1[I-]1 2.2 x 10-4 M/s k= = 0.08 M-1 s-1 (0.08 M)(0.034 M) 3. Isolation Method: For complicated reactions where many reactants are involved, the order of the reaction is determined by the method of isolation. In this method, the dependence of the rate on each reactant can be found by isolating each reactant in turnand keeping all other reactants in large excess (As a general rule, a minimum of a 20-fold stoichiometric excess is necessary. A 50-fold or 100-fold stoichiometric excess is preferable.) To see how this method works, consider the following reaction: A + B → product, thus r = k [A]m [B]n Suppose the reaction were performed [B]>>[A] e.g. starting with 0.100 mole/L B and 0.00100 mole/L A. When the reaction is finished, the solution will contain 0.099 mole/L B and essentially no A. Notice that the concentration of B is often comparable to or smaller than the experimental error. Hence, by the end of the reaction, [B] would not have changed that much, and can be regarded as a constant. 23  CHAPTER 1 REACTION KINETICS Since A is the reactant that changes, then the rate becomes dependent on A, and the rate law becomes: r = kobs [A]m , where kobs = k[B] n 0 By taking the logarithm of this equation a straight line could be obtained. log r = log kobs + m log [A] A plot of log r versus log [A]t gives a straight line with slope = m and intercept = log kobs. log r slope = m log A Fig. 1-7 Plot of log rate versus logA with m as a slope With the roles of A and B reversed, n can be found in a similar manner, k can then be evaluated using any data set along with the known values of m and n. MECHANISM OF REACTION A balanced chemical equation represents the overall result of a chemical reaction. However, at the molecular level, more than one reaction step might be involved. Most 24  CHAPTER 1 REACTION KINETICS reactions can be broken down into a sequence ofelementary reactions. This breaking down of a reaction into a sequence of elementary reactions means elucidating the reactionmechanism e.g., the reaction of nitrogen dioxide (NO2) with carbon monoxide (CO) is given by the equation. This gas phase reaction is believed to occur in two steps with the formation of nitrogen trioxide (NO3) as the intermediate. The presence of NO3 is inferred from the visible spectrum of NO3. The rate law was found to be, r = k [NO2]2. The proposed reaction mechanism agrees with the rate law. So,the reaction of nitrogen dioxide with carbon monoxide, most probably occurs via the two elementary reactions mentioned above. What is a reaction mechanism? In any chemical change, some bonds are broken and new ones are made. Quite often, these changes are too complicated to happen in one simple stage. Instead, the reaction may involve a series of small changes one after the other. A reaction mechanism describes the one or more steps involved in the reaction in a way which makes it clear exactly how the various bonds are broken and made. The following example comes from organic chemistry. 25  CHAPTER 1 REACTION KINETICS This is a reaction between 2-bromo-2-methylpropane and the hydroxide ions from sodium hydroxide solution: The overall reaction replaces the bromine atom in theorganic compound by an OH group. The first thing that happens is that the carbon-bromine bond in a small proportion of the organic compound breaks to give ions: Carbon-bromine bonds are reasonably strong, so this is a slow change. If the ions hit each other again, the covalent bond will reform. The curly arrow in the equation shows the movement of a pair of electrons. If there is a high concentration of hydroxide ions present, the positive ion stands a high chance of hitting one of those. This step of the overall reaction will be very fast. A new covalent bond is made between the carbon and the oxygen, using one of the lone pairs on the oxygen atom. 26  CHAPTER 1 REACTION KINETICS Because carbon-oxygen bonds are strong, once the OH group has attached to the carbon atom, it tends to stay attached. The mechanism shows that the reaction takes place in two steps and describes exactly how those steps happen interms of bonds being broken or made. It also shows that the steps have different rates of reaction - one slow and one fast. The Rate Determining Step The overall rate of a reaction (the one which you would measure if you did some experiments) is controlled by the rate of the slowest step. In the example above, the hydroxide ion can't combine with the positive ion untilthat positive ion has been formed. The second step is in a sense waiting around for the first slow step to happen. The slow step of a reaction is known as the rate determining step. As long as there is a lot of difference between the rates of the various steps, when you measure the rate of a reaction,you are actually measuring the rate of the rate determiningstep. Reaction mechanisms and orders of reaction The examples we use at this level are the very simple ones where the orders of reaction with respect to the various substances taking part are 0, 1 or 2. These tend to have the slow step of the reaction happening before any fast step(s). 27  CHAPTER 1 REACTION KINETICS Example 1 Here is the mechanism we have already looked at. Howdo we know that it works like this? By doing rate of reaction experiments, you find this rate equation: The reaction is first order with respect to the organiccompound, and zero order with respect to the hydroxide ions. The concentration of the hydroxide ions isn't affecting the overall rate of the reaction. If the hydroxide ions were taking part in the slow step of the reaction, increasing their concentration would speed the reaction up. Since their concentration doesn't seem to matter, they must be taking part in a later fast step. Increasing the concentration of the hydroxide ions will speed up the fast step, but that won't have a noticeable effect on the overall rate of the reaction. That is governed by the speed of the slow step. 28  CHAPTER 1 REACTION KINETICS In a simple case like this, where the slow step of the reaction is the first step, the rate equation tells you what is taking part in that slow step. In this case, the reaction is first order with respect to the organic molecule - and that'sall. Example 2 At first sight this reaction seems identical with the lastone. A bromine atom is being replaced by an OH group in an organic compound. However, the rate equation for this apparently similar reaction turns out to be quite different. That means thatthe mechanism must be different. The reaction this time is first order with respect to boththe organic compound and the hydroxide ions. Both of these must be taking part in the slow step of the reaction. The reaction must happen by a straightforward collision between them. The carbon atom which is hit by the hydroxide ion has a slight positive charge on it and the bromine a slight negative one because of the difference in their electronegativities. 29  CHAPTER 1 REACTION KINETICS As the hydroxide ion approaches, the bromine is pushed off in one smooth action. It is important to understand the difference between molecularity and the order of a reaction. The terms were sometimes used carelessly as if they mean the same thing - they don't! Order of Reaction The important thing to realize is that this is something which can only be found by doing experiments. It gives you information about which concentrations affect the rate of the reaction. You cannot look at an equation for a reaction and deduce what the order of the reaction is going to be - you have to do some practical work! Having found the order of the reaction experimentally,you may be able to make suggestions about the mechanism for the reaction - at least in simple cases. Molecularity of a reaction This starts at the other end! If you know the mechanismfor a reaction, you can write down equations for a series of steps which make it up. Each of those steps has amolecularity. The molecularity of a step simply counts the number of species (molecules, ions, atoms or free radicals) takingpart in that step. For example, going back to the mechanisms we've been looking at: 30  CHAPTER 1 REACTION KINETICS This step involves a single molecule breaking into ions. Because only one species is involved in the reaction, it has a molecularity of 1. It could be described as unimolecular. The second step of this mechanism, involves two ions reacting together. This step has a molecularity of 2 - a bimolecular reaction.The other reaction we looked at happened in a single step: Because of the two species involved (one molecule and one ion), this reaction is also bimolecular. Unless an overall reaction happens in one step (like this last one), you can't assign it a molecularity. You have to know the mechanism, and then each individual step has its own molecularity. INTEGRATED RATE LAW Rate laws can also be expressed to relate the concentration of reactants to the time of the reaction. 31  CHAPTER 1 REACTION KINETICS Such an expression is called an integrated rate lawbecause it is derived from the rate law by its integration. For reactions involving one reactant A → Product All of which have a rate law of the form: d[ A] Rate = = k[ A]ndt We will develop the integrated rate laws for:n=0 (zero-order) n=1 (first order) n=2 (second- order) Zero Order Reaction When the rate is independent of the concentration of the reactant, the reaction is said to be a zero order reaction. For the reaction: A → Product The differential rate law will be: Rate = -d[A]/dt = k [A]o = k The integral rate law will be:[A]t = [A]o - kt where: [A]t = concentration of A at time t [A]o = concentration of A at zero timek = reaction rate constant t = time 32  CHAPTER 1 REACTION KINETICS The integral rate law is derived from the differential ratelaw by integration as follows: rate = -d[A]/dt = k rate = d[A]/dt = - k d[A] = - k dt [A]t t [A]o∫d[A] = - k o ∫dt [A]t – [A]o = - k (t – 0) [A]t = [A]o - kt Or simply Ct = Co – kt When Ct is plotted versus time a straight line is obtained with slope equal to – k and y-intercept equal to Co intercept = Co slope = - k Time Fig. 1-8 Plot of concentration versus time for a zero orderreaction 33  CHAPTER 1 REACTION KINETICS The Unit of the Zero-order Rate Constant Co - Ct mole/liter k= = second = mole.liter-1.s-1 = M/s t Half-life of a Zero-order Reaction Half life (t½) of a chemical reaction is the time requiredfor the initial concentration of the reactant to get reduced to half, i.e. Ct = ½ Co Substituting this in the above equation ½ Co = Co – k t½Co –½ Co = k t½ ½ Co = k t½ Co t½ = ½Co/k = 2k The half life of a zero order reaction is directly proportional to the initial concentration, i.e. the half life period decreases as the zero order reaction proceeds with time. The time needed for 100% to be reduced to 50% is double that needed for 50% to go to 25%, i.e. Each half-life ishalf the preceding one. Shelf-life of a Zero-order Reaction An expression of importance in the pharmaceutical field is t90% or t0.9 which is the time required for the initial concentration of the reactant to get reduced to 90%, i.e. Ct = 0.9 Co 34  CHAPTER 1 REACTION KINETICS Substituting this in the above equation 0.9 Co = Co – k t0.9Co –0.9 Co = k t0.9 0.1 Co = k t0.9 0.1 Co t0.9 = 0.1Co/k = k Examples for Zero Order Reaction 1. Photochemical reactions In many photochemical reactions the rate of reactiondepends on the light intensity rather than the concentration e.g. loss of color of sulfa drug 2. Catalyzed reactions This can occur if the rate is limited by the concentration of a catalyst. In this case k is proportional to the catalyst. When the Pt surface (catalyst) is completely covered with N2O molecules, an increase in the concentration of N2O has no effect on the rate. Pt wire 2N2O(g) N2(g) + 3H2(g) Fig. 1-9 When surface of platinum wire is saturated the rate of reaction is independent of concentration of N2O 35  CHAPTER 1 REACTION KINETICS Problem The color intensity of a drug preparation exposed at 40°C decreased from 1.42 to 1.4 in 3 months. Estimate thereaction rate constant assuming that the color fading follows zero order reaction. Solution Co = 1.42 Ct = 1.4 t = 3 months = 90 days Since the color fading follows zero order reaction, then Ct = Co – k t or Co - Ct k= t 1.42 – 1.4 k= = 0.00022 day-1 90 First Order Reaction When the rate is directly proportional to the first power of the concentration of a single reactant, the reaction is saidto be a first order reaction. For the reaction: A → Product The differential rate law will be: Rate = -d[A]/dt = k [A]1 = k [A] The integral rate law will be:ln [A]t = ln [A]o - kt The integral rate law is derived from the differential ratelaw by integration as follows: 36  CHAPTER 1 REACTION KINETICS rate = -d[A]/dt = k [A] 37  CHAPTER 1 REACTION KINETICS rate = d[A]/dt = - k [A] d[A]/[A] = - k dt [A]t t [A]∫od[A]/[A] = - k o∫dt ln [A]t – ln [A]o = - k (t – 0) ln [A]t = ln [A]o - k t Or simply ln Ct = ln Co – kt Converting to common logarithmic form: log Ct = log Co – kt/2.303 In exponential form, the equation becomes: Ct = Co e– kt When Ct is plotted vrs t on Cartesian paper (rectangularpaper) a curve is obtained Fig. 1-8 Plot of concentration versus time for a first order reaction 38  CHAPTER 1 REACTION KINETICS When lnCt is plotted versus time a straight line is obtained with slope equal to – k and y-intercept equal to lnC0 lnC intercept = lnC0 slope = - k Time Fig. 1-9 Plot of ln concentration versus time for a first order reaction When log Ct is plotted versus time a straight line is obtained with slope equal to – k/2.303 and y-intercept equal to log C0 The Unit of the First-order Rate Constant lnC0 - lnCt ln (C0/Ct) k= = second = s-1 t Half-life of a First-order Reaction Ct = ½ C0 Substituting this in the above equation ln ½ C0 = ln C0 – k t½ln C0 –ln ½ C0 = k t½ 39  CHAPTER 1 REACTION KINETICS ln [C0/½ C0] = k t½ ln 2 = k t½t½ = ln2/k = 0.693 k The half life of a first order reaction is a constant valueand independent of the initial concentration, i.e. the time needed for 100% to be reduced to 50% is equal to that needed for 50% to go to 25%. Shelf-life of a First-order Reaction Ct = 0.9 Co Substituting this in the above equation ln0.9 Co = lnCo – k t0.9lnCo – ln0.9 Co = k t0.9 ln [Co/0.9Co] = k t0.9 t = ln1.11/k = 0.105 0.9 k Examples for First Order Reaction Decomposition of hydrogen peroxide, decomposition of cyclopropane to propene, decomposition of nitrous pentaoxide are all examples for first order reactions. First-order rate processes are not restricted to chemical reactions. The passive diffusion of drugs through biological membranes and processes of drug absorption, distribution, metabolism and excretion often can be described by first-order rate processes. 40  CHAPTER 1 REACTION KINETICS Rate of growth of microorganisms and rate of killing of microorganisms by heat or chemical agents are usually first- order processes. Problem A solution of drug contained 680 mg/ml when prepared. It was analyzed after a period of 60 days and was found to contain 400 mg/ml. Assuming the decomposition is first order, at what time will the drug have decomposed to one half of its original concentration. Solution Co = 680 mg/mlCt = 400 mg/ml t = 60 days Since the reaction follows first order reaction, then ln Ct = ln Co – kt or k = lnCo - lnCtt k = ln [680/400] 60 k = 8.8 x 10-3 day-1 0.693 t½ = 0.693 = = 78 days k 8.8 x 10-3 41  CHAPTER 1 REACTION KINETICS Second Order Reaction A reaction is said to be a second order reaction if the rate of reaction is proportional to the concentration of each of two reactants or second power of the concentration of one reactant. For the reaction: A → Product The differential rate law will be: Rate = -d[A]/dt = k [A]2 The integral rate law will be: 1/[A]t = 1/[A]o + kt The integral rate law is derived from the differential rate law by integration as follows: rate = -d[A]/dt = k [A]2 rate = d[A]/dt = - k [A]2 d[A]/[A]2 = - k dt [A]t t [A] ∫d[A]/[A] = - k ∫dt 2 o o 1/[A]t = 1/[A]o + k t Or simply 1/Ct = 1/Co + kt When 1/Ct is plotted versus time a straight line is obtained with slope equal to k and y-intercept equal to 1/Co 42  CHAPTER 1 REACTION KINETICS 1/C slope = k intercept = 1/CoTime Fig. 1-10 Plot of reciprocal of concentration versus time for a second order reaction The Unit of the Second-order Rate Constant 1/mole/liter 1/Ct – 1/Co = liter/mole.second = M-1 s- k= = 1 t second Half-life of a Second-order Reaction Ct = ½ Co Substituting this in the above equation 1/½Co = 1/Co + k t½2/Co –1/Co = k t½ 1/Co = k t½ t½= 1/k Co 43  CHAPTER 1 REACTION KINETICS The half-life is inversely proportional to the initial concentration. Thus when C0 is reduced by a factor of 2, the half-life is doubled (in fact each successive half-life is double the preceding one), i.e. the time needed for 100%to be reduced to 50% is half that needed for 50% to go to 25%. Shelf-life of a Second-order Reaction Ct = 0.9 Co Substituting this in the above equation 1/0.9 Co = 1/Co + k t0.9 1/0.9 Co – 1/Co = k t0.9 1/9Co = k t0.9 t0.9 = 1/9 k Co Examples for Second Order Reaction Example of reaction involving one reactant:2HI → H2 + I2 r = k [HI]2 Example of reaction involving 2 reactants: CH3COOC2H5+NaOH → CH3COONa+C2H5OHr = k [CH3COOC2H5][NaOH] Problem In a saponification reaction of ethyl acetate at 25°C, the concentration of sodium hydroxide remaining after 1 hr 44  CHAPTER 1 REACTION KINETICS was 0.00684 M. If the initial concentration of the ester as well as the base was 0.01 M, calculate the second order reaction rate constant and half life of the reaction. Solution Co = 0.01 M Ct = 0.00684 M t = 1 hr Since the reaction follows second order reaction, then 1/Ct = 1/Co + k t or 1/0.00684 - 1/0.01 k= 1 k = 46.2 M-1hr-1 t½= 1/k C0 t½= 2.16 hr Pseudo- Order Reactions 1. Pseudo-Zero Order The rate of reaction in case of suspension is pseudo zero order as there is an excess of the drug in the suspension (which act as a reservoir). As a result, the rate of reaction will not depend on the reactant concentration because the reservoir will compensate the consumed concentration and concentration will seem to be constant all the time. When the suspension particles are totally consumed the reaction will convert to 1st order kinetic. 45  CHAPTER 1 REACTION KINETICS 2. Pseudo–First Order A chemical reaction obeys pseudo first order kinetics when the rate of process is proportional to the concentration of only one of the reactants even though the reaction involves several reactants. For example consider the hydrolysis of ethyl acetate: CH3COOC2H5 + H2O CH3COOH + C2H5OH The reaction is a second order reaction and the rate of reaction is expressed as: Rate = k [CH3COOC2H5] [H2O] However in a dilute aqueous solution of ethyl acetate,[H2O] is very large compared to [CH3COOC2H5] and hardly alters during the course of the reaction. [H2O] can be taken as a constant and incorporated into the second order rate constant, k'. Rate = k' [CH3COOC2H5] Where, k' = k [H2O] Thus the reaction is in fact first order with a rate constant k'. This applies also when the concentration of hydroxide ion in the saponification of an ester is in great excess compared to the concentration of the ester, or if a buffer system is employed to control hydroxide-ion concentration, then the concentration of hydroxide ion is essentially invariant throughout the experiment, and the reaction is considered pseudo-first order. 46  CHAPTER 1 REACTION KINETICS DETERMINATION OF REACTION ORDER 1- Substitution Method In this method, the data obtained from a kinetic experiment is substituted in the relevant integrated rate equation. The equation is found in which k values remains constant. Example Determine the order of the reaction A → 2B + C from the following data obtained for [A] as a function of time. Time [A] (min) 0 0.80 M 8 0.60 M 24 0.35 M 40 0.20 M For zero order reaction: Co - Ct k= t k = 0.8 – 0.6 = 0.025 M/min 8 k = 0.8 – 0.35 = 0.019 M/min 24 k is not constant, thus the reaction is not zero order 47  CHAPTER 1 REACTION KINETICS For first order reaction: lnCo - lnCt k= t ln [0.8/0.6] k= = 0.035 min-1 8 ln [0.8/0.35] 24 k= = 0.034 min-1 k is constant, thus the reaction is first order 2- Graphical Method In this method, the data obtained from a kinetic experiment (concentration versus time) is plotted graphically to find the order of the reaction. If a straight line is obtained by plotting concentration versus time, the reaction is zero order. If plot of ln concentration versus time yield a straight line, the reaction is first order. Similarly for second order reaction, a plot of 1/C versus time results in a straight line. 3- Half-life Method Experimental data are collected and t1/2 is determined. Experiments are done by using different initial concentrations to determine the order of the reaction 1 t1/ 2 n−1 C Where n is the order of the reaction If t1/2 is proportional to the initial concentration thereaction is said to be zero order. For first order reactions 48  CHAPTER 1 REACTION KINETICS the t1/2 is constant and is not dependant on the initialconcentration. In the case of second order reaction t1/2 is inversely proportional to Co. The studied parameters for the studied reaction orderscould be summarized as follows: Order zero 1st 2nd Rate law rate = k rate = k[A] rate = k[A]2 Integrated Ct= C0−kt lnCt= lnC0−kt 1/Ct=1/C0+ kt rate law Straight- C vs. t lnC vs. t 1/C vs. t line plot Slope −k −k k Units of k M/s s-1 M-1s-1 Half-life Co/2k 0.693/k 1/kC0 49