Integrated Rate Laws Chemica Kinetics Part 2 The rate aw we have discussed thus far is the differentia rate aw. Let us consider the very simpe reaction: a A à products The differentia rate reates the rate of the reaction to concentrations of reactants: Chapter 16 When the rate aw is integrated, the resut is an equation reating concentration and time. Note: We wi consider ony n= 0,1, or 2 Start with the instantaneous rate equation: Zero-order integration: First Order Integration Integrate over zero to time t: Zero-order integrated rate aw equation:
Integrated Rate Laws Zero-order reaction: [A] = -kt + [A] 0 First-order reaction: n[a] = -kt + n[a] 0 or or Graphica determination (continued) An experiment that measures time versus the concentration of the reactant must be designed. Based on which of the foowing graphs yieds a straight ine, the order of the reaction may be determined. [reactant] vs. time à Zero order n [reactant] vs. time à 1st order 1 / [reactant] vs. time à 2nd order Second-order reaction: Note that a of the above reactions are in the form: y = m x + b Consider the first-order rate aw equation and reate it to the equation of a ine: y = mx + b where: y = n[a] x = t m = -k b = n[a] 0 From the sope of the straight-ine graph, we can then determine k for the reaction. Exampe: Graphica determination of the order of a reaction Consider the reaction: time (s) [N2O5] (M) n [N2O5] 1/[N2O5] (M -1 ) 0 0.1-2.303 10 50 0.0707-2.649 14.14 100 0.05-2.996 20 200 0.025-3.689 40 300 0.0125-4.382 80 400 0.00625-5.075 160 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) 1. Write the rate aw equation in the integrated form. 2. Cacuate k. 3. Write the rate aw equation in the differentia form. Graph of [N 2 O 5 ] vs. time The pot of concentration vs. time resuts in a curve. The reaction is not zero order for N 2 O 5.
Graph of n[n 2 O 5 ] vs. time 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) n [N 2 O 5 ] vs. time The pot is a straight ine; therefore, we concude the reaction is ikey first order. The pot the inverse of concentration vs. time aso yieds a curve, confirming that the reaction is NOT 2 nd order. We concude that the reaction is 1 st order based on the graph on the previous side. Note: In an experiment, a three graphs wi be potted and reported. time (s) n [N 2O 5] 0-2.303 50-2.649 100-2.996 200-3.689 300-4.382 400-5.075 Exampe (cont.) Since it is a first order reaction, we know that in the first order integrated rate aw that k is equa to the negative of the sope of the ine. Decomposition of N 2 O 5 rate aw cacuations Consider a reaction system in which [N 2 O 5 ] o = 0.080 M 1. How ong wi it take for [N 2 O 5 ] to fa to 0.050 M? 2. When the [N 2 O 5 ] = 0.050 M, what wi [NO 2 ] and [O 2 ] be? 3. How ong wi it take for 25 % of the N 2 O 5 to decompose? 4. What wi be the initia rate of the reaction? 5. What wi be the [N 2 O 5 ] after 12 minutes? The sope can aso be taken from the equation for the inear regression ine on the pot of n[n 2 O 5 ] vs. time: y = -0.006931x - 2.302638 6. What wi be the rate of the reaction after 12 minutes? Note that this method yieds the same vaue for k. In genera, the k vaue determined from the sope of the ine is the best approximation of the rate constant.
Exampe: Determining rate aws Exampe: Determining rate aws Butadiene reacts to form a dimer by the equation: 2 C 4 H 6(g) à C 8 H 12(g) Write both the differentia rate aw expression and the integrated rate aw expression for this reaction based on the foowing data: (assume 25 C for a of the foowing questions) [C 4 H 6 ] (M) time ( 1s) 0.01000 0 0.00625 1000 0.00476 1800 0.00370 2800 0.00313 3600 0.00270 4400 0.00241 5200 0.00208 6200 We must determine the rate aw using the graphica method with integrated rates. We wi require the foowing data: [C 4 H 6 ] (M) time ( 1s) n [C4H6] 1/[C4H6] 0.01000 0-4.605 100 0.00625 1000-5.075 160 0.00476 1800-5.348 210 0.00370 2800-5.599 270 0.00313 3600-5.767 319 0.00270 4400-5.915 370 0.00241 5200-6.028 415 0.00208 6200-6.175 481 Graph suggests that the reaction is not zero order in butadiene. Graph suggests that the reaction is not first order in butadiene.
[Butadiene] Rate Law Cacuations 1. What is the k vaue for the reaction? For the foowing questions, the [C 4 H 6 ] o = 0.125 M 2. What wi be the rate ( [C 4 H 6 ]/ t) at time = 0? 3. What wi be the rate at time = 45.0 min? 4. What time wi be required for [C 4 H 6 ] to decrease to 0.025 M? 5. What wi be the [C 8 H 12 ] when the [C 4 H 6 ] = 0.025 M? Graph suggests that the reaction is second order in butadiene. There are two types of rate aws: Rate Laws: Summary 1. The differentia rate aw (often caed simpy the rate aw) shows how the rate of a reaction depends on concentrations. 2. The integrated rate aw shows how the concentrations of species in the reaction depend on time. Because we typicay consider reactions ony under conditions where the reverse reaction is unimportant, our rate aws wi usuay invove ony the concentrations of reactants. Because the differentia and integrated rate aws for a given reaction are reated in a we-defined way, the experimenta determination of either of the rates aws is sufficient. Experimenta convenience usuay dictates which type of rate aw is determined experimentay. Knowing the rate aw for a reaction is important because it aows us to make predictions about concentrations and rates, and it aows us to infer the individua steps invoved in the reaction from the specific form of the rate aw. Haf-Life of a First-Order Reaction Haf-ife is defined as the time required for a reactant to reach haf of its origina concentration. In the foowing sides we wi examine a graphica view of the meaning of haf-ife for the reaction: 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g)
A pot of [N 2 O 5 ] vs. time for three haf-ives Haf-Life of a First-Order Reaction 0.0600 t 0 After t 1/2 Use the first order integrated rate aw to determine a genera equation for the ½-ife of a 1st-order reaction. t 1/2 = haf-ife of the reaction k = rate constant 0.0500 0.0400 After 2t 1/2 [N 2 O 5 ] 0.0300 After 3t 1/2 0.0200 0.0100 0.0000 t 1/2 t 1/2 t 1/2 0 24 48 72 Time (min) Note: For a first-order reaction, the haf-ife does not depend on concentration. Exampe (1 st order Haf-Life) A certain first-order reaction has a haf-ife of 20.0 minutes. 1. Cacuate k for the reaction. 2. How much time is required for this reaction to be 75 % compete? #2: If the reaction is 75% compete that means there is 25% of the reactant eft. Cacuating haf-ife (t ½ ) from the integrated rate aw: In a previous exampe, the foowing reaction was determined to be 1 st order: 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) If [N 2 O 5 ] 0 = 0.010 M, and at t = 240 s, [N 2 O 5 ] = 0.0035 M. A. Cacuate t ½ for the reaction at this temperature. B. What time woud be required for the [N 2 O 5 ] = 1/32 its origina concentration?
Summary of Haf-Lives of Zero, 1 st, and 2 nd - Order Reactions Zero Order: 1 st Order: t 1/2 = haf-ife of the reaction k = rate constant A o = initia concentration of A 2 nd Order: First order haf-ife is not dependent on the initia concentration. Zero & 2 nd order haf-ife is dependent on the initia concentration. Svante Arrhenius In 1889, Svante Arrhenius discovered the reationship between the rate constant (k) and the Kevin temperature (T) of reacting partices. He proposed that there is a threshod energy (activation energy, E a ) which must be overcome in order for the chemica reaction to take pace. k = A e -E a/rt In his equation, R is the universa gas constant and e is the root of the natura ogarithm. A is the frequency factor, and it is reated to the coision frequency (Z) and the orientation probabiity factor, (p). The factor p is reated to the structure of reacting species. A = p Z Arrhenius and Factors that Affect the Rate of a Chemica Reaction If we consider the Arrhenius Equation, we can easiy see how various factors affect the rate of a chemica reaction. k = A e -E a/rt Concentration Temperature Activation energy Moecuar orientation A= p Z The Arrhenius equation natura og form: The Arrhenius equation 2 point form:
Reaction Mechanisms The sequence of events at the moecuar eve that contro the speed and outcome of a reaction. A chemica equation does not te us how reactants become products - it is a summary of the overa process. C in the atmosphere (from industria chemicas) with Br from biomass burning destroys stratospheric ozone. Reaction: 2 O 3 à 3 O 2 Step 1: Br + O 3 à BrO + O 2 Step 2: C + O 3 à CO + O 2 hu Step 3: BrO + CO Br + C + O 2 Reaction Mechanism Vocabuary Overa reaction: the overa transformation for reactants to products. Eementary Step: a reaction whose rate aw can be written from its moecuarity. An eementary step describes which species must coide in order for bonds to break and new bonds form. Moecuarity: the number of species that must coide to produce the reaction indicated by that step. Intermediate: formed in one step and used up in a subsequent step and so is never seen as a product. Catayst: a species that participates in the reaction mechanism, but is not changed by the overa reaction. Exampe: Reaction of NO 2 & CO NO 2(g) + CO (g) à NO (g) + CO 2(g) From experimentation: Rate = k[no 2 ] 2 Therefore, this overa reaction does not describe an eementary step. Consider this possibe mechanism: Termoecuar reactions are unusua because of the ow probabiity of 3 partices coming together with a of the necessary aspects of an effective coision.
Exampe: Reaction of NO 2 & CO (continued) For any reaction mechanism proposed, the sum of the eementary steps must give the overa baanced equation for the reaction. NO 2(g) + NO 2(g) NO 3(g) + NO (g) NO 3(g) + CO (g) NO 2(g) + CO 2(g) --------------------------------------------------- NO 2(g) + CO (g) NO (g) + CO 2(g) Exampe: Reaction of NO 2 & CO (continued) In a mutistep reaction, the sow step wi determine the rate of the reaction, and is caed the rate-imiting, or rate-determining step. In our exampe, the moecuarity of the first step matches the observed rate aw. The first step is bimoecuar for NO 2. The experimenta rate aw is Rate = k[no 2 ] 2 NO 3(g) is a reaction intermediate it is produced in step 1 and consumed in step 2. This indicates that the first step is the sow step in the mechanism. The mechanism must agree with the experimentay determined rate aw. NO 2(g) + NO 2(g) NO 3(g) + NO (g) NO 3(g) + CO (g) NO 2(g) + CO 2(g) sow fast This rate expression agrees with the experimentay determined rate aw. C 2 H 5 Br + OH - C 2 H 5 OH + Br - Consider the reaction above and its energy diagram beow. =E a A) Based on the diagram, is this is a one-step or twostep reaction mechanism? Is the reaction written an eementary step? B) What is the rate aw for the reaction if that mechanism is correct? C) If the reaction is reversibe, what is the activation energy for the reverse reaction?
2-step reaction energy diagram Which step is the sow step? How many transition states are in the overa process? Which E a represented on the graph is the E a for the overa reaction? Reaction Mechanism and Rate Laws The overa reaction: has two different possibe mechanisms suggested by different research groups. Determine what the rate aw woud be for each of the proposed mechanisms. Mechanism 1: Is this an endergonic or exergonic reaction? Mechanism 1: Reaction Energy diagram Write the rate aw: Reaction Mechanism and Rate Laws (cont.) Mechanism 2: If the overa reaction is endothermic, draw a reaction-energy diagram consistent with the first reaction mechanism. If you write the rate aw based on the sow step, there wi be an intermediate in the rate aw expression. Intermediates may not appear in a rate expression. Must use the equiibrium step to account for intermediate (Steady State Approximation).
Equiibrium (forward and reverse reactions) Equiibrium is a state in which there is baance. Chemica Equiibrium is a state in which the rate of the forward and reverse reactions are equa. This is a dynamic equiibrium Reaction Mechanism, Equiibrium, and the Steady State Approximation (SSA) Appy the SSA to write a rate aw expression for Mechanism 2: Consider the one-step reaction mechanism: A (g) k 1 k -1 2 B (g) At equiibrium: rate forward = rate reverse (definition) k 1 [A] = k -1 [B] 2 Mechanism 2: Reaction Energy diagram Draw a reaction-energy diagram consistent with the second reaction mechanism. (As indicated previousy, the reaction is endergonic.) Proposed Mechanisms and the Experimenta Rate Laws From an earier exampe, the experimenta rate aw for this reaction was determined to be: Rate = k[no] 2 [H 2 ] The rate aw for mechanism 1 does NOT match the experimenta rate aw, therefore, mechanism 1 is incorrect. The rate aw derived from mechanism 2 is consistent with this rate aw (they are the same); therefore mechanism 2 may be the actua mechanism of reaction.
Mechanism & Reaction Energy Diagram Write a mechanism and the overa reaction for the reaction-energy diagram on the eft. Cataysts A catayst changes the mechanism by which a reaction proceeds. The activation energy for the catayzed reaction is ess than the activation energy for the uncatayzed pathway as a resut, the catayzed pathway is faster. Cataysts may be in the same phase as the reactants and products (homogeneous) or in a different phase (heterogeneous). A homogeneous catayst wi usuay appear in the rate aw expression for a catayzed reaction. Heterogeneous Catayst (hydrogenation of ethyene) H 2 Heterogeneous Cataysis Meta surface Separated H atoms Ethene (C 2 H 4 ) A heterogeneous catayst is one that exists in a different phase reative to the reactants, usuay a soid. Heterogeneous cataytic reactions typicay invoves : Adsorption of the reactants Migration of the adsorbed reactants on the surface Reaction of the adsorbed substances Escape, or desorption, of the products 1 H 2 adsorbs to meta surface 3 After C 2 H 4 adsorbs, one C H forms. 2 4 Rate-imiting step is H H bond breakage. Ethane (C 2 H 6 ) Another C H bond forms; C 2 H 6 eaves surface.
Homogeneous Catayst A homogeneous catayst is one that exists in the same phase as the reacting moecues gas phase or in soution. An exampe of this is the breakdown of ozone (O 3 ) by its reaction with free O atoms: O (g) + O 3(g) 2 O 2(g) This reaction occurs as a singe eementary step and is very SLOW. The rate aw for this mechanism is: Catayzed Mechanism However, when free chorine atoms are present, it occurs more rapidy. Chorine atoms are reeased into the atmosphere from CC 2 F 2 (Freon-12) when it reacts with ight. These chorine atoms then go on to break down ozone by the foowing 2-step mechanism: C (g) + O 3(g) CO (g) + O 2(g) (sow) O (g) + CO (g) C (g) + O 2(g) (fast) O (g) + O 3(g) 2 O 2(g) The rate aw for this catayzed pathway is: Reaction-Energy Diagram: Catayzed vs. Uncatayzed Pathways Enzyme Cataysis Enzymes are proteins that catayze chemica reactions in iving systems. They do not change the equiibrium for chemica reactions, but provide an aternate means by which one or more chemicas may react.
Two modes of cataysis Enzyme Cataysis Kinetic imit: Space-fiing modes: