CHE 156 Physical Chemistry 1 PDF

CHE 156 Physical Chemistry 1 Topic Rate of reaction and factors that affect rate. This topic is about Reaction Kinetics: Rates of Reactions. The study of reaction kinetics is concerned with the measurements of the velocities or rates of reaction together with the factors which affect these rates and reaction mechanism; The rate of reaction describes how fast the reaction goes; An explanation of how a reaction occurs is mechanism of the reaction; Factors which affect the velocity of a chemical reaction are: concentration of the reactants, temperature of the reacting system, nature of the reacting species and presence of catalysts or inhibitors; There will be two quizzes and one test. Nelson OBI-EGBEDI Key Concepts Velocities/speed; reactants and products; initial rate; determining/limiting steps Unit 1: Objectives In this Unit, you will learn about factors that affect the rate(speed) of a reaction and how to explain the rate of reactions. The Unit will enable you to understand that some reactions occur very rapidly, whereas others take hours, days, or even years. Many chemical reactions occur when we prepare food and our recipes require heating to make those reactions occur. You heat an egg so that necessary chemical reactions will happen. The length of time required for the completion of a reaction depends on factors. Expected Outcome(Unit 1) At the end of Unit 1, you should be able to : Define/explain rate or velocity and mechanism of reaction; List the factors that affect the rate of reaction. REACTION KINETICS UNIT 1 Explanation on rate of chemical reactions and factors that affect rate It is not enough to understand the stoichiometry and thermodynamics of reaction, We must also understand the FACTORS that govern/affect the rate of reaction. Chemical kinetics/Reaction kinetics is the study of how reaction rates change under varying conditions and what molecular events occur during the OVERALL REACTION. The study of the rates of chemical reaction is called KINETICS. Question? How fast does a reactions go? And what is the mechanism of the reaction. RATE is (# of events per unit time) WHY MEASURE REACTION RATES? Below are some of the reasons: How fast a chemical product will ‘work’ How quickly, and economically, a product can be made. A knowledge of the reaction mechanism helps the industrial chemist to modify a product, thereby increasing its effectiveness. Learn a great deal about how reactants change into products. FACTORS THAT AFFECT REACTION RATES Chemical nature of the reactants; Ability of the reactants to come in contact with each other; Concentrations of the reactants: increasing the concs.; Temperature of the reaction system and Availability of agents called CATALYSTS. The rate or velocity with which a chemical reactant is converted under varying conditions into a product and the mechanism by which such a reaction occurs constitute the field of study known as Reaction Kinetics. The rate of such a chemical reaction is determined by measuring the change in the concentration of either the reactant or the product with time. What is Mechanism of a reaction? The mode by which a reactant is transformed into a product is the mechanism of the reaction. Collection of individual kinetic processes/ elementary steps involved in the transformation of reactants into products Explanation of how a reaction takes place/occurs. The molecular EVENTS which occur during a reaction. The SERIES of individual steps leading to the overall observed reaction. For example: consider the thermal decomposition of dinitrogen pentoxide. The STOICIOMETRIC equation for the reaction is: The reaction can take place in stages, probably as follows: 2 N 2O5 → 4 NO2 + O2 N2O5 ⎯⎯⎯ → NO2 + NO3 slow NO2 + NO3 ⎯⎯→ NO2 + NO + O2 fast One Way NO + NO3 ⎯⎯→ 2NO2fast 2 N2O5 → 4 NO2 + O2 Rates of Chemical Reaction [Conc ] [time]  Equations (1a &1b) indicate that the rate of a reaction may be defined in two major ways, either in terms of the disappearance of the reactants or the appearance of the products, depending upon whether the reactant or the product is being observed or monitored. What one observes experimentally in determining the rate of a reaction is the change in conc. , of one or more of the reactants or product as a function of time, t. Ci In practice, the observation of a change in a physical property (spectroscopic, hydrogen ion concentration, etc) of the reactant or product is used to measure the extent of change as a function of time. Consider a hypothetical homogeneous reaction, which is unidirectional in the direction of the arrow in Equation (2), with reactant concentration as the only monitored variable A(g) + B(g) C(g) + D(g) (undefined) (2) The conc. of A or B decreases during the time interval, therefore the change in A or B is a negative quantity. The rate of a reaction is a positive quantity, so a minus sign is needed in the rate expression to make the rate positive. In Equation (2), all the stoichiometric coefficients of the reactants and the products are the same. When they are not, the stoichiometric factors have to be taken into account in the definition of the rate of reaction. Consider a general reaction aA + bB → cC + dD (4) Take an arbitrary reaction A B d [ A] [ A] slope = − = cons tan t dt Conc. [ B] d [ B] slope = = cons tan t dt Time / s linear plot C D [C ] Conc. [ D] Time/s Non-linear plot. The rate depends on reaction conc. and time. Quiz # 1 1. In 100 seconds, the conc of A(g) falls from 2.400 x 10-2 mol dm-3 to 2.000 x 10-2 mol dm-3 by the reaction A B , what is the rate of reaction? 2. Which term describes the measure of the increase in the concentration of a product per unit time? A. activation energy B. kinetics C. reaction rate D. reaction time Express the rate of 2A → 3C in terms of change of the concentrations. A. Rate = = - 1 d[C] 3 dt 2 dt B. Rate = = - 1 d[A] d[C] dt 1 d[A] 2 dt C. Rate = - = - 1 d[C] 3 dt 1 d[A] 2 dt D. Rate = = 1 d[C] 3 dt 1 d[A] 2 dt Reference: Fundamental Physical Chemistry(Revised Edition). Edited by Iweibo I., Okonjo K. & Obi-Egbedi N. CHE 156 Physical Chemistry 1 Nelson Okpako OBI-EGBEDI Unit2 Topic: Concentration, the rate law and experimental determination of rate expression Objectives In this Unit, you will learn how to write the rate law and use it to predict the speed of reactions. Experimental determination of rate law. Expected Outcome At the end of Unit 2, you should be able to : Know how concentration affects rate of reaction; Determine the order of a reaction; Write the rate law/expression; and Do some calculations. The rate of reaction is a function of factors such as the concentrations of the reactants, the temperature of the reaction system, the identity of the reactants and the catalyst used. Let us consider one of the factors: Concentration Concentration and the Rate Law The rates of most reactions change when the concentrations of the reactants change. Chemical reactions are usually faster when concs. of the reactants are increased and slower when decreased. In general, the rate is proportional to the concentration of the reactants raised to appropriate powers/indices/exponents. The rate is proportional to the molar concentrations of the reactant raised to an appropriate power. Consider the following chemical equation: xA + yB Product(s) (6) The above equation is like :H2 + 1/2O2 → H2O In Equation (7), k is the specific reaction rate constant. The exponents n and m are called reaction orders, and the sum (m + n) is the overall reaction order. Equation (7) is known as the rate law. The rate law is an expression for the rate as a function of concentration at a fixed temperature. Equation (7) is determined experimentally. Experimental Determination of Rate Laws Most chemical changes consist of several steps called elementary steps. Different reactants can affect the rate of a particular reaction differently. Consequently, Equation (7) cannot be written from the balanced chemical equation but from experimental data. To determine the order with respect to each reactant in expression (7), it is necessary to study the mechanism by which changes in the concentration of each reactant affect the initial rate of the reaction. A + B → Product(s) Let us conduct a hypothetical experiment for the above-named reaction in order to determine its rate law, using the initial rates method. How the initial rate depends on the starting concs is preferable. Why? It is because …. as the reaction proceeds, the concs of the reactants decrease and it may become difficult to measure the change accurately. PROCEDURES Mix the solutions of the reactants A and B with initial concentrations [A]0 and [B]0, respectively. The temperature of the mixture is fixed. At convenient time intervals, withdraw some aliquots of this mixture, stop the reaction or slow it down drastically, analyze the withdrawn sample chemically to determine the change in concentration of the reactant A which occurs after reactants have been mixed. Instead of monitoring the change in concentration of reactant A, the change in its physical property as a function of time could be measured. For example, its optical absorbance, fluorescence, optical activity or refractive index. Perform other experiments with different values of the initial concentration of A at the same fixed temperature, keeping the initial concentration of reactant B constant. Determine the initial rate, at (time = 0), for each experiment from a plot of concentration against time. Compare the initial rates obtained from these experiments. If there is no change in rate, the order with respect to A is zero. If the rate of reaction doubles when the initial concentration of A is doubled, the order with respect to A is one. If the rate quadruples when the initial concentration of A is doubled, then the order with respect to A is two. Similarly, varying the initial concentration of reactant B and keeping the initial concentration of reactant A and the temperature of the reaction constant, the effect of the concentration of reactant B on the initial rate of the reaction can also be investigated and the order with respect to B determined. These experimentally determined orders with respect to each reactant will enable us to write Equation (7), the rate law. For example, consider the reaction: A(g) + B(g) →C(g) The experimental rate law of the above reaction is: Rate = k [A] [B]0 What is the overall order of the reaction? Since the overall order is the sum of the exponents, we obtain (1 + 0) = 1 as the reaction order. For a hypothetical reaction in which B and C react to form the product, D, the following data were obtained from three experiments. B+C D Experiments [B] [C] (Rate of formation of D) 1. 0.15M 0.075M 3.5 x 10-3mol dm-3min-1 2. 0.30M 0.15M 1.4 x 10-2mol dm-3 min-1 3. 0.15M 0.15M 7.0 x 10-3mol dm-3 min-1 Suppose we have the reaction: oxidation of Bromide ions by Bromate ions in acid solution. BrO3− (aq) + 5Br − (aq) + 6 H + (aq) 3Br2 (aq) + 3H 2O(l ) Mixture [Bro3] [Br-] [H+] Relative initial rate of formation of Br2 I 0.05 0.25 0.30 1 II 0.05 0.25 0.60 4 III 0.10 0.25 0.60 8 IV ? ? ? ? R = k[ BrO3− ]l [ Br − ]m [ H + ]n 1 = k[0.05] [0.25] [0.30] l m n 4 = k[0.05] [0.25] [0.60] l m n 1 [0.30]n = 4 [0.60]n 1 1 2 = n 2 2 n= 2 The exponents in the rate law appear to be unrelated to the coefficients in the overall chemical equation. See Equation 7 There is, in fact, no way to know for sure what the exponents in the rate law of an overall reaction will be without doing expts to determine them. Sometimes, however, the coefficients and the expts are the same by COINCIDENCE, as in the case of elementary reactions. In general, the order of a reaction must be determined by expt; it cannot be deduced from the overall balanced equation. Quiz #2 1. A + 2B D The rate of formation of D in the above reaction is found experimentally to be independent of the concentration of B and to quadruple when the concentration of A is doubled. Write the rate law for the reaction. 2. For the general rate law, Rate = K[A][B]2 , what will happen to the rate of reaction if the concentration of A is tripled? [A] The rate will be halved [B] The rate will be doubled [C] The rate will be tripled [D] The rate will remain the same 3. Methanol can be produced by the following reaction: CO(g) + 2H2(g) CH3OH(g). How is the rate of disappearance of hydrogen gas related to the rate of appearance of methanol? 4. What is the overall reaction order for the reaction represented by the rate law: Rate = k[H2][NO]2? [A] zero [B] first [C] second [D] third Unit 3 Topic: Molecularity, Mechanism ,Predicted Rate Law and Units of Rate Constant Expected Outcome: Obtain the units of the specific rate constant for a given order An overall balanced chemical equation does not tell us much about how a reaction takes place. 2H + O → 2H O 2 2 2 The chemical equation represents the sum of a series of simple reactions called the elementary steps/elementary reactions. They represent the progress of the overall reaction at the molecular level. The sequence of elementary steps that leads to product formation is called the reaction mechanism. What is Mechanism of a reaction? The mode by which a reactant is transformed into a product is the mechanism of the reaction. Collection of individual kinetic processes/ elementary steps involved in the transformation of reactants into products Explanation of how a reaction takes place/occurs. The molecular EVENTS which occur during a reaction. The SERIES of individual steps leading to the overall observed reaction. For example: consider the thermal decomposition of dinitrogen pentoxide. The STOICIOMETRIC equation for the reaction is: 2 N 2O5 → 4 NO2 + O2 The reaction can take place in stages, probably as follows: Molecularity, Mechanism and Predicted Rate Law Most chemical reactions consist of elementary reactions or steps. One of these steps, which is much slower than the others, determines the rate of the reaction and it is referred to as the rate- determining or rate-limiting step. An elementary reaction can be characterized by the number of reactant molecules involved in it. This number of molecules or species on the reactant(s) side / in an elementary reaction /step is known as the molecularity of the reaction. For the elementary process, the order(s) in its rate law are the same as the stoichiometric factors, or equal to the coefficients of the reactants in the chemical equation for the process. An elementary step in which only one reacting molecule participates is UNIMOLECULAR Two molecules is BIMOLECULAR. TERMOLECULAR reaction involves the participation of three molecules in one elementary step. The rate of an elementary reaction is proportional to the product of the concentrations of the reactants that are in that step. Therefore, the rate law for an elementary process can be predicted from the chemical equation for the process. Consider the reaction: PCl5 →PCl3 + Cl2 The above reaction can be explained by a two-step mechanism: Step (1) is the slow (rate limiting) step. PCl5 →PCl4 + Cl. slow Step 2 PCl4 →PCl3 + Cl. fast Net: PCl5 →PCl3 + Cl2 The experimentally determined rate law is: Rate = k [PCl5] and, from the rate-limiting step, the predicted rate law is Rate k [PCl5] The elementary steps must satisfy two requirements: The sum of the elementary steps must give the overall balanced equation for the reaction. The rate-determining step, which is the slowest step in the sequence of steps leading to product formation, should predict the same rate law as is determined experimentally. The order of multistep reaction must be obtain experimentally. For example, the conversion of A + C D A B slow, RDS B+ C D very fast A+C D The rate at which D is formed depends on how quickly A gives B, not how quickly B reacts with C. The rate equation would be: Rate = k[A]x Consider a hypothetical reaction: 2C + D → E + F What is the rate law of the rate determining step in terms of the concentrations of the reactants in the overall balanced equation? Fast step: rate1 (forward reaction) Rate2 (backward reaction) (i) (ii) The rate of the slow step: Rate (iii) Sub. (ii) into (iii) yields the overall rate as Rate (iv) UNITS OF RATE CONSTANTS The units of the specific rate constant depend on the overall order of the reaction. The units are worked out from the units of the individual terms in the rate equation. Units of initial rate Units of rate constants = (Units of initial concentration) x (9) The units of rate are mol dm-3 time-1 and the units of [B] are mol dm-3 Substituting these units into Equations (9)/ (10), and assuming that time is in seconds and x = 1, gives the units of k as s-1 or time-1. [ ] = mol dm-3 M = mol dm-3 Quiz # 3 1. A + 2B D The rate of formation of D in the above reaction is found experimentally to be independent of the concentration of B and to quadruple when the concentration of A is doubled. Write the rate law for the reaction. 2. For the general rate law, Rate = K[A][B]2 , what will happen to the rate of reaction if the concentration of A is tripled? [A] The rate will be halved [B] The rate will be doubled [C] The rate will be tripled [D] The rate will remain the same 3. Methanol can be produced by the following reaction: CO(g) + 2H2(g) CH3OH(g). How is the rate of disappearance of hydrogen gas related to the rate of appearance of methanol? 4. What is the overall reaction order for the reaction represented by the rate law: Rate = k[H2][NO]2? [A] zero [B] first [C] second [D] third Reference: Fundamental Physical Chemistry(Revised Edition). Edited by Iweibo I., Okonjo K. & Obi- Egbedi N. CHE 156 Physical Chemistry 1 Nelson Okpako OBI-EGBEDI Unit 4 Topic: ORDER, KINETICS EQUATIONS/CONCS- TIME EQUATIONS Expected Outcome: At the end of this Unit 4, you should be able to: 1. Derive kinetic equations for ﬁrst and second orders reactions 2. Explain the meaning of half-life of a reaction 3. Do calculations using the kinetic equations ORDER OF REACTION Reaction can be classiﬁed by its order/ we can classify reactions into: Zeroth/Zero order First order Second order Third order(rarely) Fractional order Negative order KINETICS EQUATIONS/CONCS-TIME EQUATIONS The rate law tells us how the speed/rate of a reaction varies with concentrations of the reactants at a particular moments. The relationship between the concentration of a reactant and TIME can be obtained from the rate law of a reaction. This can be done by integration of the rate expression/ transform a rate law into a mathematical relationship between conc. and time. ) x x t t dx  x 0 b x   t 0 kd t Separating the variables and integrating Equation (13) between the limits x = 0 at t = 0 and x = x at some time t, we have (14) x x t t [  ln(b  x )] x 0  k [ t ] t 0 (15) b ln  kt b x 1  x dx  l n x  l og e x We can have the equation as: b ln  kt b x b  ex p  kt  bx (b  x )  b ex p  kt  [ B ]t  [ B ] 0 ex p( kt ) l og(b - x ) S l op e = - k /2.3 0 3 -3 M ol d m T i m e/s L i n ear p l ot for a fi r st or d er r eacti on Second-order Reactions let us consider a situation in which the reactants are present with initial equimolar concentrations. Consider a reaction : C+D → Products [C] = [D] i.e., the initial conc. of [C]=[B] d[C ]  k [ C ] [ D ] OR dt d [ C ] d [ D ] dx    k2[ c  x] 2 OR dt dt dt x t dx  (c  x) 0 2  k 0 2 dt (17) x  1  k 2t   (18) c  x 0 1 1 k 2t   (19) cx c Equation (19) is the second-order kinetic equation. k2 can be evaluated graphically. Or by substitution of the initial concentration and the concentration of a reactant at time t into Equation (19). 1 /c- x S l op e = k 2 T i m e/s L i n ear p l ot for a secon d or d er r eacti on Half- life of zero order: Second-order Half life: METHODS FOR ORDER The order of a reaction can be determined by: Graphical method Substitution method- constant ks Half-life method. Half-Life Method Example The rate constant for the ﬁrst-order decomposition of N2O5 at temperature t0C is 2.2 x 1 1 10- s-. If the initial concentration of N2O5 is 3.6 x 10-3 mol dm-3, what will the concentration be after four half-lives? For a ﬁrst-order reaction, the half-life is independent of the initial concentration of the reactant. After the ﬁrst half-life, the initial concentration would have decreased by half. Therefore, after the fourth half-life, the concentration will be 2.25 x 10-4 mol dm-3. Unit 5 Topic: The Temperature Dependence of Reaction Rates: The Arrhenius Equation THEORIES OF REACTION RATES The collision Theory The Transition-State Theory Expected Outcome: At the end of this Unit 5, you should be able to: 1. How temperature affects the rate of reaction 2. Use the Arrhenius equation to explain the effect of temperature 3. Explain the collision and transition-state theories 4. Do calculations using the Arrhenius equation. The Temperature Dependence of Reaction Rates: The Arrhenius Equation The rate of reaction, or the rate constant, increases if the temperature of the reacting system rises. This arises from the increase in the collision frequency and the fraction of molecules with energy equal to or greater than Ea (the activation energy). For ﬁxed conc. of reactants, a reaction proceeds at different rates at different temperatures. The increase in the speciﬁc rate constant, k, with rise in temperature varies according to the following equation. k = A exp – (Ea/RT) where k = the speciﬁc rate constant; A = the pre-exponential factor/frequency factor; Ea = the activation energy in J; R = the gas constant (8.314J K-1mol-1) T= the absolute temperature in K (0C + 273); The effect of temperature on the reaction rate can be seen in Equation (21). A reaction will always be faster at a higher temperature because many more molecules will possess the activation energy on collision. The higher the activation energy is, the slower the reaction at any given temperature. k  A ex p (  E a ) RT Consequences of Arrhenius equation. TheE of the reaction can be calculated. a The k can be calculated. Knowing the activation energyE , ,the a exponential factor can be calculated. k2 Ea  T 2  T1  log    k1 2.3 0 3 R  T 2 T1  THEORIES OF REACTION RATES The collision Theory The rate constant, k, varies with temperature and one of the theoretical models used to explain the variation of k with temperature is the collision theory. For a bimolecular gaseous reaction with colliding molecules possessing the required energy of impact and the molecules properly oriented, the collision theory predicts that the speciﬁc rate constant, k, at a temperature T0 Kelvin, is a product of the following the collision frequency (Z) the orientation factor P, and the activation state factor exp (-Ea/RT) From these three requirements, the rate constant can be deﬁned as k = ZP exp (- Ea/RT) (27) Equation (27) is in the same form as Equation (21) if the pre-exponential factor is given by A = ZP. Equation (27) indicates that the rate of a reaction is proportional to the number of collisions between reactant molecules per unit time. The Transition-State Theory The transition-state theory explains that at any given temperature the rate of a reaction depends on the concentration of the activated complex, which is in thermodynamic equilibrium with the molecules of the reactants, and on the rate of decomposition of the activated complex to form product(s). Slow A+B= [A-B]‡ product Consider a bimolecular reaction A and B to form product(s) According to the theory, a reaction proceeds as follows: The equilibrium constant, K  for the reaction producing the activated complex can be A + B = [A-B]‡ Slow product(s) 𝐾‡ = [ 𝐴 −written ‡ 𝐵] as ሾ𝐴 ሿ[ 𝐵 ]  K  K K  Ea k  A ex p (  ) RT From equation (35), the expression for k can be written as: k = A exp- (Ea/RT) The introduction of an entropy of activation, ΔS±, in Equation (35), leads to a better interpretation of reaction rates. Unit 6 Topic: CATALYSIS Expected Outcome: At the end of this Unit 5, you should be able to: 1. To explain the effects a positive and negative catalysts 2. Explain the meaning of Heterogeneous and Homogeneous catalyses 3. Do calculations using the effects of a catalyst Catalysis A catalyst is a substance that alters the speed of a chemical reaction without itself undergoing any net chemical change. A positive catalyst causes a reaction to proceed more rapidly and a negative catalyst, such as an inhibitor, retards the rate of a reaction. A positive catalyst performs some or all of the following functions: It increases the collision frequency (Z) It orientates the reactant molecules to obtain the correct collision geometry, It provides an alternative pathway or a mechanism with a lower activation energy. Consider, for example, a bimolecular reaction catalyzed by a positive catalyst at 298K. A + B → Product(s), with a rate expression: Rate of reaction = k [A]x [B]y (37) In Equation (37), k is the speciﬁc rate constant, x, and y are the orders with respect to A and B, respectively. If the concentration of the reactants and the temperature of the reaction medium are kept constant, for uncatalyzed and catalyzed reactions, Equations (21), (27) and (37) indicate that a positive catalyst can speed up a reaction by a combination of all or some of the enumerated functions. k  A ex p (  E a ) (21) RT k = ZP exp (- Ea/RT) (27) Rate of reaction = k [A]x [B]y (37) Equations (21), (27) and (37) indicate that a positive catalyst can speed up a reaction by a combination of all or some of the enumerated functions. The effect produced by a positive or negative catalyst is referred to as catalysis. The changes in the thermodynamic functions of state, ΔG, ΔH and ΔS, that are characteristic of the reaction are not affected by catalysis. There are two types of catalysis: Homogeneous catalysis, and Heterogeneous catalysis Homogeneous catalysis A reaction is said to be kinetically homogeneous if it takes place in one phase only. That is, the catalyst is present in the same physical state as the reactant(s). For example, the oxidation of sulphur dioxide to sulphur trioxide in the presence of nitric oxide in the lead-chamber process of sulphuric acid. Heterogeneous catalysis In heterogeneous catalysis or surface contact catalysis, the reacting species and the catalyst are in different physical states. In this type of catalysis, the catalyst is solid and the reactants are gases or sometimes liquids. For example, the synthesis of ammonia from hydrogen and nitrogen by the Haber process. In this case, the catalyst, iron (in ﬁnely divided form) in the presence of molybdenum is a solid and the reactants are gaseous. Reference: Fundamental Physical Chemistry(Revised Edition). Edited by Iweibo I., Okonjo K. & Obi-Egbedi N.

CHE 156 Physical Chemistry 1 PDF

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