Name: UNIT 5 KINETICS NOTES PACEKT #: KINETICS NOTES PART C

Transcription

1 KINETICS NOTES PART C IV) Section 14.4 The Change of Concentration with Time A) Integrated Rate Law: shows how the concentration of the reactant(s) varies with time 1) [A]0 is the initial concentration of A 2) [A]t is the concentration of A at some time, t during the course of the reaction B) First Order ln[a] t ln[a] 0 = kt 1) A plot of ln[a] vs. t will yield a straight line 2) The slope of the line will be k 3) Nuclear decay reactions are first order reactions (a) Since [A]t will always be half of [A]0, then plugging in 0.5 and 1 for those concentrations will lead to the following simplified first order integrated rate law: k = t1 2 (b) Note that half-life calculations are independent of any concentrations. C) Second Order 1 [A] t 1 [A} 0 = kt 1 1) A plot of vs. t will yield a straight line [A] 2) The slope of the line will be +k 3) A second-order half-life question can be solved by using 1 and 0.5 for initial and time t concentrations. D) Zero Order 1) [A] t = kt + [A] 0 2) A plot of [A] vs. t will yield a straight line 3) The slope of the line will be k EXAMPLE #4: Substance A decomposes following 1 st order kinetics with a rate constant of s -1. If the initial concentration of A is 1.00 M, calculate [A]t remaining after 2.5 minutes. Page 9

2 EXAMPLE #5: Substance B decomposes in a zero order reaction with a rate constant of 0.02 mol/l s. Calculate the concentration of B at 20.0s if [B]0 is 1.00 M. EXAMPLE #6: Using substance A from Example #4, how much time is required for 70% of substance A to decompose? EXAMPLE #7: Calculate the half-life of substance A. Page 10

3 V) Section 14.5 Temperature & Rate KINETICS NOTES PART D A) As temperature increases, reaction rate increases, therefore k would also increase; in general, the reaction rate doubles with each 10 C rise in temperature. B) Collision Model 1) Molecules can only react when they collide with each other. 2) Molecules must collide with: (a) proper orientation and (b) sufficient energy to break and form new bonds (or form the activated complex). 3) Activated complex: a temporary arrangement of particles that has sufficient energy to either form products or re-form reactants. 4) Activation energy (Ea): minimum amount of energy required for a reaction to occur; AKA transition state. C) Potential Energy Diagram D) Maxwell-Boltzmann Distribution Reaction Pathway 1) Temperature is a measure of average kinetic energy; thus, not all particles are moving at the exact same speed at the same time; there is a wide distribution of kinetic energies within a sample at any given temperature. Page 11

4 2) The lower temperature line is taller/steeper. This means there are less molecules with greater kinetic energies. 3) The higher temperature line is shorter/broader. This means there are more molecules with greater kinetic energies. 4) The shaded area represents molecules that would possess sufficient energy to allow the reaction to occur. (a) You can see that by shifting the curve towards the right, more molecules would fall in this shaded area (i.e. more molecules would be able to overcome the activation energy barrier), thus increasing the reaction rate. 5) There is a calculation where the fraction of molecules that would meet the activation energy barrier can be determined it uses the activation energy, temperature, and ideal gas law constant. 6) Arrhenius developed another equation relating k and Ea. This equation uses a frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. 7) While we will NOT be doing any of these calculations, we need to have a qualitative understanding with regards to a graph. 8) If k is determined experimentally at several temperatures, the Ea for a reaction can be calculated from the slope of a plot of ln k vs. 1 T. 1 T Page 12

5 KINETICS NOTES PART E VI) Section 14.6 Reaction Mechanisms A) The sequence of events that describes the actual process by which reactants becomes products is called the reaction mechanism. B) Reactions may occur all at once or through several discrete steps. C) Each of these processes is known as an elementary reaction or elementary process. D) The molecularity of a process tells how many molecules are involved in the process. 1) Unimolecular only one molecule involved; A product; rate = k[a] 2) Bimolecular involves two molecules; either A + A or A + B product; rate = k[a] 2 or rate = k[a][b] 3) Termolecular involves three molecules; very rare E) When determining reaction mechanisms, you will identify reaction intermediates and catalysts. 1) Intermediate: species that is neither a reactant nor a product in the overall reaction; species that is formed in an earlier elementary step and consumed in a later elementary step. 2) Catalyst: species that is neither a reactant nor a product in the overall reaction; species is found initially as a reactant in an earlier elementary step and is formed again as a product in a later elementary step. F) Multistep Mechanisms 1) The rate at which a reaction occurs will always be limited by the slowest elementary step in its mechanism. 2) Rate-determining step: the slowest step in a reaction mechanism; it is usually identified for you or you are provided additional data to identify the rate determining step. 3) The sum of the elementary steps must match the overall balanced equation (similar to Hess s Law). 4) Writing the rate law using the mechanism must match the experimentally determined rate law in order to be a valid mechanism. 5) You will never be asked to propose a mechanism. G) Slow Initial Step NO2(g) + CO(g) NO(g) + CO2(g) 1) The experimental rate law is determined to be rate = k[no2] 2 2) Proposed mechanism: Step 1: NO2 + NO2 NO3 + NO Step 2: NO3 + CO NO2 + CO2 (slow) (fast) Page 13

6 3) You are only allowed to write the rate law using the slow step and any steps before it. In this case, since the slow step is first, Step 1 is the only equation used to write the rate law. 4) When using mechanisms to write rate laws, exponents are determined by how many of the species appears (coefficients) as a reactant in the slow step and any steps prior. 5) Thus, the rate law of the mechanism, rate = k[no2] 2 matches the experimentally determined rate law. The proposed mechanism is a valid one. 6) NO3 is an intermediate in this mechanism. H) Fast Initial Step 2NO(g) + Br2(g) 2NOBr(g) 1) The experimental rate law is determined to be rate = k[no] 2 [Br2] 2) Proposed mechanism: Step 1: NO + Br2 NOBr2 Step 2: NOBr2 + NO 2 NOBr (fast) (slow) 3) Use all reactants from Steps 1 and 2 that do not cancel to write the rate law. 4) Rate = k[no] 2 [Br2] 5) Note: There is a much more complicated process that goes into writing rate laws for mechanisms with fast initial steps. If you re seriously interested in a 5 on the AP exam, you should refer to this section of your textbook. I) Multi-step PE Diagram Page 14

7 Example #8: A + A B + C C + D B + E Overall: KINETICS NOTES PART F Intermediate: Step 1 rate law: Rate law if step 1 is slow step: Rate law if step 2 is slow step: Step 2 rate law: Example #9: A + B C + D D + E B + F Overall: Catalyst: Intermediate: Rate law if 1 is slow step: Rate law if 2 is slow step: VII) Section 14.7 Catalysis A) Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. B) Catalysts change the mechanism by which the reaction occurs. C) Types of catalyst 1) Acid-base: reactants gain or lose a proton 2) Surface: either a new reaction intermediate is formed or the probability of successful collisions is modified; can hold reactants together to help bonds to break 3) Enzyme: Binds to reactants in a way that lowers Ea by reacting with reactants to form new intermediates (a) Found in biological systems (b) The substrate (reactant) fits into the active site of the enzyme (lock & key) Page 15

8 BEER S LAW NOTES Spectroscopic methods rely on the ability of substances (matter) to absorb or emit electromagnetic radiation. The spectrometer measures the amount of light absorbed by the sample by comparing the intensity of the light emitted from the light source with the intensity of the light that emerges from the sample. Beer s Law: A = abc A = Absorbance a = molar absorptivity b = path length c = molar concentration Beer s Law is effective only for substances that produce a color in solution. - cobalt salts = pink - copper salts = blue - nickel salts = green The more intense the color, the greater the concentration is. In Beer s Law, the ab portion of the equation is often thought of as constant since the molar absorptivity in an intensive property unique to the substance (i.e. will always be the same for a given substance) and the path length does not change given the spectrometer being used. Therefore, the absorbance (A) is directly proportional to the molar concentration (c). After a calibration curve is generated using several concentrations of the substance and distilled water as the standard (or blank), the concentration of an unknown solution can be determined by measuring its absorbance. Page 16