Calculating Rates of Substances. Rates of Substances. Ch. 12: Kinetics 12/14/2017. Creative Commons License

Transcription

1 Ch. 2: Kinetics An agama lizard basks in the sun. As its body warms, the chemical reactions of its metabolism speed up. Chemistry: OpenStax Creative Commons License Images and tables in this file have been used from the following sources: OpenStax: Creative Commons Attribution License 4.0. ChemWiki (CC BY-NC-SA 3.0): Unless otherwise noted, the StatWiki is licensed under a Creative Commons Attribution- Noncommercial-Share Alike 3.0 United States License. Permissions beyond the scope of this license may be available at copyright@ucdavis.edu. Principles of General Chemistry (CC BY-NC-SA 3.0): 2 Reaction Rates Reaction rate (or speed) = change in amount of substance per unit of time. 2. Reaction Rates Chemical Kinetics is the study of how fast reactions take place. Reactions can be fast (reaction of sodium metal with water) or very slow (rusting of iron). What factors can we change in lab to alter the rate of a reaction? Concentration Temperature Catalysts Surface area of solids 3 Why do you need to study kinetics? Predict rate/speed of reactions Drug delivery: time-release capsules Understanding reaction mechanisms Understanding metabolism regulation 4 Calculating Rates of Substances For the reaction: A B, we can measure the rate of disappearance of reactants: (assigned negative so rate is a positive value) and appearance of products: Reaction: A B Rates of Substances Image from Principles of General Chemistry -v.0/s8-chemical-kinetics.html, CC-BY-NC-SA 3.0 license 5 Image from Principles of General Chemistry -v.0/s8-chemical-kinetics.html, CC-BY-NC-SA3.0 license 6

2 Rates of Substances 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) What is the rate of disappearance of H 2 O 2 for each time interval? Rates of Substances 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) What is the rate of disappearance of H 2 O 2 for each time interval? Units: M = mol/l = mol L - Time (hr) [H 2 O 2 ], M [H 2 O 2 ], M t (hr) Rate of H 2 O 2, M/hr Time (hr) [H 2 O 2 ], M [H 2 O 2 ], M t (hr) Rate of H 2 O 2, M/hr Rates of Substances 2 H 2 O 2 (aq) 2 H 2 O(l) + O 2 (g) Rates are related by coefficients. Which curve is the reactant? How can you tell the difference between the curves of the products? Rates of Substances, Reaction 2 H 2 O 2 (aq) 2 H 2 O(l) + O 2 (g) Concentration (M) (M) 2 H2 H 2O 2 H 2O + 2 O 2 --> 2 H 2 O 2 + O Time (hr) (hr) 9 This is called the rate expression. We can use this expression (or equation) to determine the rate of each substance and the rate of reaction over a given time period. Notes about rate expressions: ) Reactants (-); Products (+); 2) Coefficients are in the denominator; 3) Every component in equation is equal to all others (each one calculates the rate of reaction). 0 Rates of Substances, Reaction 2 H 2 O 2 (aq) 2 H 2 O(l) + O 2 (g) Notes about rate expressions: ) Reactants (-); Products (+); 2) Coefficients are in the denominator; 3) Every component in equation is equal to all others (each one calculates the rate of reaction). Write the balanced equation for the rate expression below: Rates of Substances, Reaction 2 H 2 O 2 (aq) 2 H 2 O(l) + O 2 (g) If H 2 O 2 disappears at 0.20 M/s, ) What is the rate of appearance of H 2 O? 2) What is the rate of appearance of O 2? 3) What is the rate of reaction? Two methods: a) rate expression and b) stoichiometry a) b) 0.20 mol H 2 O 2 /L/s x (2 mol H 2 O / 2 mol H 2 O 2 ) Rate of reaction = rate of substance / its coefficient 2 2

3 Rates of Substances 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) What are the rates of appearance of H 2 O and O 2 from 0 6 hours? M/hr, M/hr (mol L - hr - ) From 6-2 hours? M/hr, M/hr Rates of Substances, Reaction Write the rate expression and calculate rates for the reaction: SiO 3 + 4C SiC + 3CO ) If the rate of disappearance of SiO 3 is 0.25 M/s, what are the rates of disappearance of C, appearance of SiC and CO? Time (hr) [H 2 O 2 ], M [H 2 O 2 ], M t (hr) Rate of H 2 O 2, M/hr ) What is the rate of the reaction? Examples: 2., Calculating Rates Practice 5Br - (aq) + BrO 3- (aq) + 6H + (aq) 3Br 2 (aq) + 3H 2 O(l) If the rate of appearance of Br 2 is.2 x 0-3 M/s, what is the rate of disappearance of Br -?.2 x 0-3 M/s * (5 mol Br - /3 mol Br 2 ) = 2.0 x 0-3 M/s C 6 H 2 O 6 (aq) 2C 2 H 5 OH(aq) + 2CO 2 (g) If rate of appearance of ethanol is 4.2 x 0-2 M/s, what is rate of disappearance of glucose? 2. x 0-2 M/s SiO 3 + 4C SiC + 3CO If the rate of appearance of CO is 0.78 M/s, what is the rate with respect to C?.0 M/s 2.3 Rate Laws; Reaction Orders Average Rate: rate of substance over a given time interval Rate = (M f M i ) / (t f t i ) Instantaneous rate: rate at a specific time Determined by measuring slope of tangent at a point Initial rate: rate at time = 0 Determined by measuring slope of tangent at t = 0 These rates will all have different values. Rates are not constant; they change over time. If they were constant, graphs would be straight lines. 5 6 Reactant Concentrations & Rate Laws Rate law - relationship between concentration of each reactant and the reaction rate. aa + bb cc + dd rate = k[a] x [B] y Rate is the rate of reaction k is the rate constant, with units [A] and [B] are reactant concentrations x and y are orders of reactants (NOT necessarily the same as coefficients). 7 Rate Laws Rate Laws have to be determined from experimental data (concentrations and reaction rates) using initial rates (at time = 0) Br 2 (aq) + HCOOH(aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) rate = k[br 2 ] [HCOOH] 0 = rate= k[br 2 ] Examples of balanced equations and rate laws: Reaction Equation Rate Law 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) rate = k[n 2 O 5 ] 2 H 2 O 2 (aq) + I - (aq) 2 H 2 O (l) + O 2 (g) rate = k[h 2 O 2 ][I - ] H 2 (g) + I 2 (g) 2 HI (g) rate = k[h 2 ][I 2 ] 8 3

4 Determining a Rate Law 5Br - (aq) + BrO 3- (aq) + 6H + (aq) 3Br 2 (aq) + 3H 2 O(l) Experiment [Br - ](M) [BrO 3- ](M) [H + ](M) Initial Rate (M/s) Rate Laws 5Br - (aq) + BrO 3- (aq) + 6H + (aq) 3Br 2 (aq) + 3H 2 O(l) Experiment [Br - ](M) [BrO 3- ](M) [H + ](M) Initial Rate (M/s) x x x x 0 3 Need two trials where only one reactant concentration changes (all others are constant). Compare change in rates to change in concentration to determine the order of the reactant (the exponent in the rate law) x 0 3 [Br 2 - ] Rate x 0 triples triples x 0 [BrO 3 3- ] [H + ] constant constant x 0 3 [Br - ] triples, rate triples: Br - is st order (x = ) Which two trials should be used for BrO 3-? Which two for H +? 20 Rate Laws Experiment [Br - ](M) [BrO 3- ](M) [H + ](M) Initial Rate (M/s) x x x x ? rate = k[br - ][BrO 3- ] 2 Now we can solve for the rate constant. How? What information do we need to solve for k? Rate = 20 M -2 s - [Br - ][BrO 3- ] 2 What is the rate of reaction for experiment 5? 2 Rate Laws Three important things to remember about rate laws: ) A rate law is an equation that is always defined in terms of reactant concentrations, never products. 2) The exponents (orders of reactants) in a rate law must be determined from a table of experimental data. 3) Comparing changes in individual reactant concentrations to changes in rate allows you to find the order of the reactant. 22 Orders of reactants Shortcut for determining orders: If the concentration doubles and the rate doesn t change, then the concentration has no effect on rate and the reactant is 0 order If the concentration doubles and the rate doubles, then the reactant is st order. What would happen to rate if the concentration of a st order reactant tripled? If the concentration doubles and rate quadruples, then the reactant is 2 nd order. What would happen to rate if the concentration of a 2 nd order reactant tripled? 23 Orders of reactants What do orders of reactants mean? They are exponents that indicate by what factor the rate of reaction will change when concentration changes. For a zero order reactant, raising the ratio of concentrations to the 0 th power results in no change for rate. Raising the ratio of concentrations to the 0 th power equals. For a first order reactant, the change in concentration (double, triple, etc.) is the same as the change in rate. Raising the ratio of concentrations to the st power results in the same ratio for rate. For a second order reactant, raising the ratio of concentrations to the 2 nd power is the same as squaring. If a reactant is 2 nd order and its concentration is tripled, how will the rate change? 24 4

5 Rate Laws Practice S 2 O 8 2 (aq) + 3I (aq) 2SO 4 2 (aq) + I 3 (aq) Determine the rate law and calculate the rate constant, including its units. Experiment [S 2 O 8 2 ](M) [I ](M) Initial Rate (M/s) x x x 0 4 rate = 0.08 M - s - [S 2 O 8 2 ] [I ] 25 Rate Laws Practice Determine the rate law and calculate the rate constant (with units): 2NO + Cl 2 2NOCl. Calculate the rate of reaction when [NO] and [Cl 2 ] = 0.5 M. Experiment [NO] [Cl 2 ] Rate (M/s) 0.0 M 0.0 M.7x M 0.20 M 6.7x M 0.30 M.5x M 0.30 M 3.0x M 0.30 M 4.5x0-3 rate = 0.7 M -2 s - [NO] [Cl 2 ] 2 ; rate = 5.7x0-4 M/s Rate Laws Practice 2.4 Integrated Rate Laws Determine the rate law and calculate the rate constant for: 2O 3 (g) 3O 2 (g) [O 3 ] (M) Rate (M/s) x x x 0-8 What is the rate if [O 3 ] = M? Hint: What other information is needed to solve this? Rate = M - s - [O 3 ] 2 ; rate = 3.49 x 0-7 M/s Examples Concentration/Time relationship: st order reaction ( A B ), Rate also can be written, rate = k[a] Setting them equal: Integrate to get st order integrated rate law: 28 Integrated Rate Laws IRL - First Order First order integrated rate law: ln is natural logarithm Linear format: y = mx + b Math note: ln(5/7) = ln 5 ln 7 ln x = What is x? x =.3 29 Data from slide #8: 2 H 2 O 2 (aq) 2 H 2 O (l) + O 2 (g) [H 2 O 2 ] and ln [H 2 O 2 ] [H2O2] [H 2 O] vs Time y = x R² = 0.87 y = -0.55x R² = Time (hr) 30 5

6 Half-life: the time it takes for the amount of reactant to decrease by one half (50%). For st order reactions, each half-life is the same length. ln (50/00) = -kt /2 First Order Half-Life IRL - Second Order Two options for rates of second order reactions: Rate = k[a] 2 OR Rate = k[a][b] 2 nd option is too complex, we ll only work with the first example. Rate = k[a] 2 = - [A]/ t Integrate: /[A] t = kt + /[A] o (linear eqn) 3 32 IRL - Second Order Second Order Half-Life Concentration [A] Concentration Functions vs Time [A] /[A] Linear ([A]) Linear (/[A]) Half-life: the time it takes for a reactant to decrease by half (50%). For 2 nd order reactions the length of each half life doubles in length. Concentration [A] Concentration Functions vs Time 4 y = 0.078x R² = Time (s) [A] /[A] Log. ([A]) Log. ([A]) Linear (/[A]) Time (s) Second Order Half-Life Half-life: the time it takes for a reactant to decrease by half (50%). For 2 nd order reactions, the length of each half life doubles in length. Concentration [A] Concentration Functions vs Time y = 0.078x R² = Time (s) [A] /[A] Half-life Linear (/[A]) 35 IRL s Graphing Integrated Rate Laws can also be used to graph functions of concentrations over time. The function that gives the most linear graph determines the order of the reactant (the other functions will be curved). [A] vs. time (zero order) ln [A] vs time (first order) /[A] vs. time (second order) Consider the reaction 2NO 2 (g) 2 NO(g) + O 2 (g). Time and concentration are plotted to find the most linear plot. 36 6

7 IRL s Graphing IRL s Graphing Consider the reaction 2NO 2 (g) 2 NO(g) + O 2 (g) Plot of concentration vs time (zero order): Time (s) [NO 2 ], M ln [NO 2 ] /[NO 2 ], M -.00E-02 [NO 2 ] versus time 0.00x x x E E E E x x E E E x E x E E IRL s Graphing IRL s Graphing Plot of ln concentration vs time (first order): Plot of /concentration vs time (second order): -4 ln [NO 2 ] versus time 400 /[NO 2 ] versus time IRL Practice The half life for the reaction 2A B is 8.06 minutes at 0 o C. The reaction is first order in A. How long will it take for A to decrease from.25 M to 0.7 M? k = 8.598x0-2 min - ; t = 6.6 minutes Image from Principles of General Chemistry 4 -v.0/s8-chemical-kinetics.html, CC-BY-NC-SA 3.0 license 42 7

8 IRL Practice If the half life is 8.2 min for a first order reactant with initial concentration of 2.4 M, what is the concentration of A after 7.5 minutes? IRL Practice A second order reaction has a half life of 699 seconds. If the initial reaction concentration is M, what will the reactant concentration be after 855 seconds? k = min -, [A] t =.8 M If a first order reaction has a rate constant of 4.73 x 0 2 s -, how long would it take for 75% to react? t = 2.9 x 0-3 s Examples k = M - s -, [A] t = M Collision Theory Collision Theory Most reactions occur as a result of collisions between reacting molecules. Collision theory: rate of reaction is directly proportional to the number of molecule collisions per time. Effective Collisions: molecules must collide with sufficient kinetic energy and in the correct orientation in order to make products. 2 CO (g) + O 2 (g) 2 CO 2 (g) As temperature and concentration increase, more collisions occur and therefore rates of reaction increase Collision Theory Activation Energy NO + O 3 NO 2 + O 2 Activation Energy (E a ): energy barrier molecules have to overcome in order to react. The high energy state is the transition state. A + B C + D Image from Principles of General Chemistry v.0/s8-chemical-kinetics.html, CC-BY-NC-SA 3.0 license 48 8

9 Activation Energy Only a small fraction of molecules have enough kinetic energy to overcome the activation energy. E a is different for each reaction. Reaction Energy Profile Enthalpy ( H) is heat of reaction: Endothermic: Energy absorbed, H > 0 Exothermic: Energy released, H < 0 E a = T.S. R (for forward reaction) E a = T.S. P (for reverse reaction) Reactions with low E a are faster because more molecules can overcome the E a. At higher temperatures, a larger fraction of molecules have enough kinetic energy to overcome E a. 49 Image from Principles of General Chemistry v.0/s8-chemical-kinetics.html, CC-BY-NC-SA 3.0 license Reaction Energy Profile Is the reaction shown on the energy profile below endo- or exothermic? What is the value of E a? What is H? Reaction Energy Profile Draw an energy profile diagram that has an enthalpy value of +5 kj and an Activation Energy = +40 kj. 60 Potential Energy (kj) reaction progress Example Example Arrhenius Equation The Arrhenius Equation can be used to relate the temperature of a reaction to the rate constant. Original form: k: rate constant A: frequency factor (we will not use or calculate this value) E a : activation energy (kj/mol) R: gas constant (8.34 J/mol K) T: temperature (Kelvin) Arrhenius Equation The initial equation can be modified to give a linear variation. This variation works well for sets of 4 or more data points. y = m x + b Rate constant changes with temperature; Temperature increases, rate increases, k increases

10 Arrhenius Equation Worked Example 2.: Rate constants for the reaction 2HI(g) 2H 2 (g) + I 2 (g) were measured at five different temperatures. The data are shown in the table below. Determine the activation energy for this reaction. (This will be used in Kinetics lab!) T (K) k (M - s - ) /T (/K) ln k x x x x x Arrhenius Equation: Example 2. ln k Arrhenius Plot y = x /T (/K) m = -E a / R E a = m x -R = K x J / K mol E a = 85,294 J/mol = 85 kj/mol (3 s.f. because of T and k) 56 Arrhenius Equation The Arrhenius Equation can also be used to determine: ) the rate constant after a change in temperature (i.e., 2 data points) or 2) E a given two k s and T s: Arrhenius Equation Practice Using the activation energy just calculated (E a = kj/mol) for the system 2HI(g) 2H 2 (g) + I 2 (g), determine the rate constant at 325 o C, using k = 3.52x0-7 M - s - and T = 555 K. Convert E a to J/mol (or R to kj/mol) Convert T to Kelvin k 2 = 6.32x0-6 M - s Arrhenius Equation Practice k = 3.52x0-7 M - s -, T = 555 K, E a = J/mol 2.6 Reaction Mechanisms Reaction Mechanism: the sequence of reaction steps that describes the pathway from reactants to products. A balanced chemical equation does not indicate how a reaction actually takes place. A chemical reaction takes place through a series of simpler steps called elementary steps (typically 2 or more). These elementary steps determine the stepwise reaction mechanism for the overall reaction. k 2 = 6.32x0-6 M - s

11 Reaction Mechanisms Given elementary steps in a mechanism, we can: Determine the molecularity and rate law for each step Write the overall (or net) equation Use relative rates (slow or fast) to identify the ratedetermining step Identify the intermediate(s) Is produced and then consumed in a later step Identify the catalyst (if it exists) Is consumed and then produced in a later step (recycled) 6 Reaction Mechanisms Each step in a mechanism is defined by its molecularity: the number of reactant molecules. Rate laws for elementary steps are written based on the molecularity. unimolecular (one reactant molecule) O 3 O 2 + O rate = k[o 3 ] st order bimolecular (two reactant molecules) NO 2 + CO NO + CO 2 rate = k[no 2 ][CO] 2 nd order 2 HI H 2 + I 2 rate = k[hi] 2 2 nd order termolecular (three reactant molecules) rare! 2 NO + Cl 2 2 NOCl rate = k[no] 2 [Cl] 3 rd order 62 Multistep Mechanisms One step in a multistep mechanism will be slower than the others. The slow step is the rate-determining step (RDS). Step (slow): 2 NO 2 NO 3 + NO Step 2 (fast): NO 3 + CO NO 2 + CO 2 ) Write the overall equation. 2) What is the molecularity of each step? 3) Which step is the rate-determining step? 4) What is the intermediate? The catalyst? 5) Write the rate law for each step. 63 Rate Laws: Elementary Steps If the first step is slow (RDS), the rate law of the first step is the rate law of the overall equation. If the rate law for the overall equation does NOT match the rate law for the first step, the first step is NOT the slow step (RDS). If the second (or third or fourth or ) step is slow, the rate law of the overall equation is more difficult to determine. 64 Rate Laws: Mechanisms If the st step is slow, the rate law of the overall reaction is the same as the rate law of the first step. If the 2 nd step is slow, rate laws are more complicated to determine. Step : Tl 3+ + Fe 2+ Tl 2+ + Fe 3+ Step 2: Tl 2+ + Fe 2+ Tl + + Fe 3+ Overall rate law: Which step is rate-determining? Second 65 Mechanisms Practice Group Work: Step (slow): S 2 O I - 2SO I + Step 2 (fast): I + + I - I 2 Which step is rate determining? Step What is the intermediate? I + What is the rate law for each step? What is the overall rate law? rate = k[s 2 O 8 2- ][I - ]; rate = k[i + ][I - ] overall: rate = k[s 2 O 8 2- ][I - ] 66

12 Mechanisms Practice Step : 2 NO N 2 O 2 Step 2: N 2 O 2 + H 2 N 2 O + H 2 O Step 3: N 2 O + H 2 O N 2 + H 2 O Determine: overall equation, molecularity of each step, intermediate, and catalyst. Overall: 2 NO + H 2 N 2 + H 2 O; all 3 bimolecular; Intermediates: N 2 O 2, H 2 O, N 2 O; Catalyst: N/A Write the rate law for each step. Rate = k[no] 2 Rate = k[n 2 O 2 ][H 2 ] Rate = k[n 2 O] If the overall rate law is rate = k[no] 2 [H 2 ], which step is the rate-determining step? Not the first step 67 Rate Laws: Mechanisms Step : Step 2: (slow) H 2 O 2 + I H 2 O 2 + IO H 2 O + IO H 2 O + O 2 + I Overall reaction: 2H 2 O 2 2H 2 O + O 2 Which step is slow? Rate = k[h 2 O 2 ][I ] Catalysis A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. Step : (slow) Step 2: H 2 O 2 + I H 2 O 2 + IO H 2 O + IO H 2 O + O 2 + I Catalysis Energy profiles with catalysts. Figures 2.2 and Overall reaction: 2H 2 O 2 2H 2 O + O 2 A catalyst speeds up a reaction by providing a set of elementary steps with more favorable kinetics than those that exist in its absence; it usually does this by lowering the activation energy Catalysts Catalysts can be classified as heterogeneous: the catalyst and reactants are in different phases (Figure 2.26) Catalysts Practical Application: Catalytic Converter There are four steps in the catalysis of the reaction C 2 H 4 + H 2 C 2 H 6 by nickel. ) Hydrogen is adsorbed on the surface, breaking H H bonds and forming Ni H bonds. 2) Ethylene is adsorbed on the surface, breaking the π-bond and forming Ni C bonds. 3) Atoms diffuse across the surface and form new C H bonds when they collide. 7 4) C 2 H 6 molecules escape from the nickel surface, since they are not strongly attracted to nickel. A catalytic converter allows for the combustion of all carboncontaining compounds to carbon dioxide, while at the same time reducing the output of nitrogen oxide and other pollutants in emissions from gasoline-burning engines. 72 2

13 Catalysts Catalysts can also be classified as homogeneous: the catalyst and reactants are in the same phase. The advantages of homogeneous catalysts: Reactions can be carried out under atmospheric conditions Can be designed to function selectively Are generally cheaper Catalysts Enzymes: large protein molecules that act as catalysts for biological reactions. Digestive enzymes allow us to process starch (e.g., potatoes) but not cellulose (e.g., grass)