Presentation on theme: "18 - 1 Topic # 15 Chemical Kinetics Conditions of Reaction Rates Rate of a reaction General Rate Law Orders of Reactions Half-Life Reaction Mechanisms."— Presentation transcript:
18 - 1 Topic # 15 Chemical Kinetics Conditions of Reaction Rates Rate of a reaction General Rate Law Orders of Reactions Half-Life Reaction Mechanisms (Zumdahl Chapter 12)
18 - 2 Collision Theory A reaction will only take place if three conditions are met: 1. the reactants come into contact (collide) 2. the collision happens with enough energy, (activation energy) 3. the reactants hit at the best possible orientation to facilitate a reaction.
18 - 3 Factors that control the rate of a reaction A. Concentration putting more reactants in the same space will increase the collision frequency, producing a faster reaction rate. Same effect occurs by increasing the pressure in a gaseous environment. B. Temperature Increasing temperature will increase the rate of a reaction. GENERALLY, a 10 o C rise will result in approx. doubling the rate.
18 - 4 As the system is heated, the position of the average kinetic energy must shift to the right in order to reflect that change. As the system cools, the average kinetic energy must respond by shifting position to the left. At higher temperatures, the particles are more evenly distributed over a range of kinetic energy values.
18 - 5 Rate and temperature Reaction rates are temperature dependent. Here are rate constants for N 2 O 5 decomposition at various temperatures. T, o C k x 10 4, s -1 20 0.235 25 0.469 30 0.933 35 1.82 40 3.62 45 6.29 k x 10 4 (s -1 ) Temperature ( o C)
18 - 6 C. Solid particle size increase in surface area of the solid will increase the reaction rate D. Catalysis Catalysts are substances that increase the the rate of a reaction while remaining unchanged chemically. They work by providing an different reaction route that needs a lower activation energy.
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18 - 8 Rates of reactions Rate of a chemical reaction. The change in the quantity of a reactant or product that takes place in a period of time. rate = = concentration later - concentration earlier time later - time earlier [ ] t Molarity
18 - 9 Rates of reactions To study rates of reaction, you must: Identify the reactants and products. Carry out the reaction. Measure the concentrations of one of the reactants or products at known intervals. There MUST be DATA! There needs to be a way to measure at least one of the species involved.
18 - 10 The example reaction Decomposition of N 2 O 5 Dinitrogen pentoxide is known to decompose completely by the following reaction. 2N 2 O 5 (g) 2N 2 O 4 (g) + O 2 (g) This reaction can be conducted in an inert solvent like carbon tetrachloride. When N 2 O 5 decomposes, N 2 O 4 remains in solution and O 2 escapes and can be measured.
18 - 11 An example reaction We can easily measure the oxygen as dinitrogen pentoxide decomposes. Temperature must be maintained to within +0.01 o C. The reaction flask must be shaken to keep oxygen from forming a supersaturated solution. It is found that the reaction initially occurs very rapidly but gradually slows down.
18 - 12 An example reaction Gas buret Constant temperature bath
18 - 13 An example reaction Time (s) STP O 2 in mL 0 3001.15 6002.18 9003.11 12003.95 18005.36 24006.50 30007.42 42008.75 54009.62 6600 10.17 7800 10.53 Here are the results for an experiment.
18 - 14 An example reaction Volume, mL O 2 Time, s The rate of O 2 production slows down with time. The rate of O 2 production slows down with time.
18 - 15 Calculations with Average rates We can calculate the average rate of oxygen formation during any time interval as: Average rate of O 2 formation = V O 2 t Time, sRate O 2 * 0 3000.0038 6000.0034 9000.0031 12000.0028 18000.0024 24000.0019 30000.0015 * The rates shown here have units O 2 mL( at STP) /s. Notice how the rate decreases with time.
18 - 16 Looking at the N 2 O 5 reaction again….. Since we know the stoichiometry for our reaction, we can calculate the concentration of N 2 O 5 during the reaction. 2N 2 O 5 (g) 2N 2 O 4 (g) + O 2 (g) For each mole of O 2 produced, two moles of N 2 O 5 will have decomposed. The rate of reaction will be: rate of reaction = = - [O 2 ] t [N 2 O 5 ] t 1212
18 - 17 Using the equation: 2 A + B + 3 C + D The rate of reaction can be expressed in terms of D: d[D] dt = 1d[C] 3 dt = d[B] dt - = 1 d[A] 2 dt - rate = Note the use of coefficients and choosing a component to base the rates on! This method can be used for any equation…….
18 - 18 2 A + B + 3 C + D Use the Equation to solve for concentrations: d[C] dt = 0.35 mole/liter-min Find value of d[A] dt Relationship of [A] will be negative and ratio 2 3 - 2 [0.35] 3 1 = 0.23 mole/liter-min is being lost
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18 - 20 General Rate Equation A + B C + D For an equation: rate = k [A] x [B] y x & y are ORDER of reaction for that component they are NOT coefficients!! k is the rate constant Collected data is used to answer questions on rate orders. d[C] dt = k [A] x [B] y Actual Rate Called the rate law for the equation
18 - 21 Finding rate order Method of initial rates The order for each reactant is found by completing an experiment and using the data: Change the initial concentration of a reactant. Hold all other initial concentrations and conditions constant. Measure the initial (or starting) rates of reaction The change in rate is used to determine the order for that specific reactant. The process is repeated for each reactant.
18 - 22 N 2 O 5 example The following data was obtained for the decomposition of N 2 O 5. Experiment[N 2 O 5 ]Initial rate, M/s 1 0.100 3.62 x 10 -5 2 0.200 7.29 x 10 -5 We know that the rate expression is: rate = k [N 2 O 5 ] x Our goal is to determine what x (the order) is. 2N 2 O 5 (g) 2N 2 O 4 (g) + O 2 (g)
18 - 23 N 2 O 5 example For exp. 27.29 x 10 -5 M/s = k (0.200 M) x For exp. 13.62 x 10 -5 M/s = k (0.100 M) x We can now divide the equation for exp. two by the one for exp. one. 7.29 x 10 -5 M/s k (0.200 M) x 3.62 x 10 -5 M/s k (0.100 M) x which give2.01 = (2.00) x andx = 1 (first order reaction) = rate = k [N 2 O 5 ] 1
18 - 24 The data below for rate of reaction was obtained for the following reaction: A + B + C... Exp. [A] [B] [C] Initial rate, M/s 1 0.030 0.010 0.0501.7 x 10 -8 2 0.060 0.010 0.050 6.8 x 10 -8 3 0.030 0.020 0.0504.9 x 10 -8 4 0.030 0.010 0.1001.7 x 10 -8 Note that the concentrations double in some experiments. Notice the products are not needed!
18 - 25 rate = k[A] x [B] y [C] z A + B + C ……… Equation: Rate Law: Pick data so calculations for two of the reactants will cancel out……..
18 - 26 Order for A Use exps one and two since [B] and [C] are the same and would cancel out. 6.8 x 10 -8 M/s k(0.060 M) x 1.7 x 10 -8 M/s k(0.030 M) x 4.0 = (2.0) x x = 2 [A] is second order =
18 - 27 Order for B Use experiments one and three. 4.9 x 10 -8 M/s(0.020 M) y 1.7 x 10 -8 M/s (0.010 M) y 2.9 = (2.0) y The order is not obvious by inspection. You must take the natural logarithm of both sides and solve for y. ln 2.9 = y (ln 2.0) y = 1.54 or 3232 =
18 - 28 Order for C Use experiments one and four. Experiment [C] Initial Rate 10.0501.7 x 10 -8 40.1001.7 x 10 -8 Here the rate did not change when [C] was doubled. This is an example of a zero order reaction. z = 0
18 - 29 A more complex example We can now write the overall rate law. rate = [A] 2 [B] 3/2 [C] 0 since [C] has no effect on the rate: rate = [A] 2 [B] 3/2 The overall order for the reaction is: x + y + z = 2 + 3/2 + 0 = 3 1/2
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18 - 31 Graphing rate laws Graphical method (some form of [ ] versus time!) Using integrated rate law, one can produce straight line plots. The order for a reactant is assigned if the data produces a straight line. Rate integrated Graph Slope Order law rate law vs. time 0 rate = k [A] t = -kt + [A] 0 [A] t -k 1 rate = k[A] ln[A] t = -kt + ln[A] 0 ln[A] t -k 2 rate=k[A] 2 = kt + k 1 [A] 0 1 [A] t 1 [A] t
18 - 32 Graphing rate laws 0 order plot 1st order plot 2nd order plot As you can see from these plots of the N 2 O 5 data, only a first order plot results in a straight line. So N 2 O 5 is 1 st order! As you can see from these plots of the N 2 O 5 data, only a first order plot results in a straight line. So N 2 O 5 is 1 st order! Time (s) [N 2 O 5 ] 1/[N 2 O 5 ] ln[N 2 O 5 ]
18 - 33 Practice Graphing: Time (s) [NO 2 ] (mol/L) 00.500 1.2 x10 3 0.444 3.0 x 10 3 0.381 4.5 x 10 3 0.340 9.0 x10 3 0.250 1.8 x 10 4 0.174 Graph: [NO 2 ] vs. time Zero order ln[NO 2 ] vs. time First Order 1/[NO 2 ] vs. time Second Order Straight line tells the order !
18 - 34 Special First order reactions Reactions that are first order with respect to a reactant are of great importance. Describe how many drugs pass into the blood stream or used by the body. Often useful in geochemistry Radioactive decay Half-life (t 1/2 ) The time required for one-half of the quantity of reactant originally present to react.
18 - 35 Half-life From our N 2 O 5 data, we can see that it takes about 1900 seconds for the concentration to be reduced in half. It takes another 1900 seconds to reduce the concentration in half again. From our N 2 O 5 data, we can see that it takes about 1900 seconds for the concentration to be reduced in half. It takes another 1900 seconds to reduce the concentration in half again. Time (s) [N 2 O 5 ]
18 - 36 Half-life The half-life and the rate constant are related. t 1/2 = t 1/2 = Half-life (time) can be used to calculate the first order rate constant. For our N 2 O 5 example, the reaction took 1900 seconds to react half way so: k == = 3.65 x 10 -4 s -1 0.693k 0.693 t 1/2 0.693 1900 s
18 - 37 Fluorine -21 has a half-life of 5 seconds. What fraction of the original nuclei remain after 1 minute? The answer is solved by creating the fraction. Where n = number of ½ lives. In 1 minute there will be 12 half-lives. Answer is…….
18 - 38 Reaction mechanisms A detailed molecular-level picture of how a reaction might take place. activated complex = bonds in the process of breaking or being formed
18 - 39 Could also look like: R + S D but in reality… R B B + S D Or even: A X X + B C + D A + B C + D is really: Never see the activated complex in the equation!
18 - 40 Reaction mechanisms Molecularity The number of particles that come together to form the activated complex in an elementary process (each step). 1 - unimolecular 2 - bimolecular 3 - termolecular
18 - 41 Reaction mechanisms For some elementary processes, the exponents for each species in the rate law are the same as the coefficients in the equation for the step. For our earlier example, Bimolecular…. the rate law is: rate = k [NO] [O 3 ] (Bimolecular mechanism)
18 - 42 Reaction mechanisms In general, the rate law gives the composition of the activated complex, because that is what comes together to form the complex. The power of a species in the rate law is the same as the number of particles of the species in the activated complex. If the exponents in the rate law are not the same as the coefficients of the equation for the reaction, the overall reaction must consist of more than one step. Lets look at N 2 O 5 - again!
18 - 43 Reaction mechanisms Earlier we found that for: 2N 2 O 5 2N 2 O 4 + O 2 The rate law was: rate = k [N 2 O 5 ] According to the equation, it should be second order (coefficient of 2 in front) but the data shows it to be first order. The reaction must involve more than one step.
18 - 44 Reaction Mechanisms Consider the following reaction. 2NO 2 (g) + F 2 (g) 2NO 2 F (g) If the reaction took place in a single step the rate law might be: rate = k [NO 2 ] 2 [F 2 ] However, the experimentally observed rate law is: rate = k [NO 2 ] [F 2 ]
18 - 45 Reaction Mechanisms Since the observed rate law is not the same as if the reaction took place in a single step, we know two things. More than one step must be involved The activated complex must be produced from two species. A possible reaction mechanism might be: Step one Step oneNO 2 + F 2 NO 2 F + F Step two Step twoNO 2 + F NO 2 F Overall Overall2NO 2 + F 2 2NO 2 F
18 - 46 Reaction Mechanisms Rate-determining step. slow When a reaction occurs in a series of steps, with one slow step, it is the slow step that determines the overall rate. Step one Step oneNO 2 + F 2 NO 2 F + F Expected to be slow. It involves breaking an F-F bond. Step two Step twoNO 2 + F NO 2 F Expected to be fast. A fluorine atom is very reactive.
18 - 47 Reaction Mechanisms Since step one is slow, we can expect this step to be the determiner of the overall rate of the reaction. NO 2 + F 2 NO 2 F + F This would give a rate expression of: rate = k 1 [NO 2 ] [ F 2 ] This agrees with the experimentally observed results.
18 - 48 Let’s look again: R + S D but in reality… R B slow B + S D very fast The rate only depends on the concentration of R so the rate law only contains R! Rate = k[R] x The Slow process determines the rate law!
18 - 49 A + B C + D is really: A X Fast and equilibrium X + B C + D SLOW Here the slow step contains X and B, but the formation of X depends on A. Since X is an intermediate it isn’t in the rate law. X is substituted with A…. Rate = k[A][B] No intermediates are written in the rate law!
18 - 50 Take note: the sum of all steps needs to tally up to the reaction equation. R + S D R B slow B + S D very fast
18 - 51 Another mechanism example…… All gases: NO 2 + NO 2 NO 3 + NO SLOW NO 3 + CO NO 2 + CO 2 Fast Rate of Formation of NO 3 = ∆[NO 3 ] ∆t Overall rate = k[NO 2 ] 2 Bimolecular first step mechanism
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18 - 53 Arrhenius Equation ln k 1 = - Ea R 1 1 T 2 T 1 - k = rate constant Ea = Activation energy R = molar gas constant (8.314 J/mol K ) T = absolute temperatures ( Kelvin) ln k 2