CH 117 Notes Weeks 1 and 2
CH 117 Notes Weeks 1 and 2 General Chemistry II
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Date Created: 01/24/16
Week 1 Notes General Chemistry II Chemical Kinetics – the study of the speeds of chemical reactions and the factors affecting those speeds Reactions can be homogeneous (reactants and products are all in the same phase) or heterogeneous (multiple phases). Reaction Rate Catalyst – a substance that increases the rate of the chemical reaction without getting consumed. E.g. Hydrogen peroxide is unstable and decomposes slowly without a catalyst such as KI . (aq) Slow: 2H O 2 2 (aq) H O2 +(l) 2 (g) Fast: H O2 2 (aq) (aq) IO (aq) H O2 OR(l) 2 2 (aq) IO (aq) H O2 (aq) O 2 (g) (aq) Factors that control the speed of chemical reactions: 1) Properties of reactants and products (especially structure and type of bonding) 2) Concentration of reactants (and products) 3) Temperature 4) Presence of catalyst 5) Nature of the surface (heterogeneous reactions) *E.g. heterogeneous reaction=liquid and solid; homogeneous = liquid and liquid Rate – the change of some measurable quantity over time 8 E.g. Light travels at a rate of 2.998 x 10 m/s. How far will it travel in one hour? Rate = delta x / delta t delta x = rate * delta t = (2.998x10 m/s) (3600 s) = 1.079x10 m 8 Reaction Rate – change in the concentration of a reactant or product per unit of time RR= delta [reactant] / delta t = delta [product] / delta t **Reaction rate is defined as a positive quantity; [ ] indicate concentration (M). E.g. [NaCl] = concentration of NaCl. **When the reaction goes forward, the product M goes up and the reactant M goes down. To calculate: 1) Find the difference between the final and initial concentration of the substance of interest over the time interval of interest 2) Divide the difference by the time interval Practice Week 1 Notes You measure concentration of an acid over time during a room temperature reaction. What is the average rate of each interval? What happens to the rate as the reaction proceeds? Time (s) [Acid] (M) 0.00 5.00 x 10 3 4 25.0 9.85 x 10 50.0 1.94 x 10 4 6 100. 7.52 x 10 Remember, for each interval: Delta [acid] = [acid] – [acid] final initial Delta t = t final tinitial 4 3 4 Rate (025) = (9.85x10 M – 5.00x10 M) / (25.0 s – 0.00 s) = 1.61 x 10 M/s 4 4 5 Rate (2550) = (1.94x10 M – 9.85x10 M) / (50.0 s – 25.0 s) = 3.16 x 10 M/s Rate (50100) = (7.52x10 M – 1.94x10 M) / (100. S – 50.0 s) = 3.73 x 10 M/s 6 The rate is slowing down because the reactant is taken away. Note: The rate of reaction is often constantly changing. Practice The plot below shows [C H ]. 4st8mate the reaction rate from 0.00 s to 20.0 s and from 20.0 s to 60.0 s. C H 2C H 4 8 (g) 2 4 (g) 0.3 0.25 0.2 0.15 Concentration (M) 0.1 0.05 0 0 20 40 60 80 100 120 Time (s) Week 1 Notes Rate = delta [C H 4 /8delta t 0 s to 20 s 3 Rate = (0.12 M / 20.0 s) = 6.0 x 10 M/s 20 s to 60 s Rate = (0.09 M / 40.0 s) = 2.3 x 10 M/s 3 Reaction Rates and Stoichiometry Suppose you have the following generic reaction aA + bB cC + dD To account for the reaction stoichiometry we write the rate as Rate = (1/a) (delta [A] / delta t) = (1/b) (delta [B] / delta t) = (1/c) (delta [C] / delta t) = (1/d) (delta [D] / delta t) ** For reactants, delta [ ] will be negative, so the reaction rate needs (). Practice If the rate of the reaction below is 2.55 x 10 M/s at 273 K, what is the concentration of KOH after 15.0 s if the initial concentration of KOH is 4.50 x 10 M? 4 (aq) (aq) KOH (aq) HCl (aq) KCl (aq) H O2 (l) Rate = delta [KOH] / delta t (rate x delta t) = [KOH] final [KOH] initial [KOH] initial ate x delta t) = [KOH] final [KOH] final 4.50 x 10 M – (2.55 x 10 M/s) (15.0 s) 5 = 6.75 x 10 M Practice For the reaction below, the rate of loss of I is 020040 M/s. What is the rate of formation of HI? H 2 (g) I2 (g) 2HI (g) Rate = delta [I ]2/ delta t = ½ x delta [HI] / delta t = 0.0040 M/s 2 x rate (stoichiometry!) = 2 (0.0040 M/s) = 0.0080 M/s Average rate vs. Instantaneous rate Average reaction rate – a reaction rate calculated from a change in concentration divided by a change in time Instantaneous reaction rate – the reaction rate at a particular time after the reaction has begun Week 1 Notes To find the instantaneous reaction rate, find the slope of the line tangent to the plot of the concentration vs. time at the specific time point of interest. Practice Find the average reaction rate for HCl af(g) 4.0 s. Find the instantaneous reaction rate for H 2 (g) 4.0 s. H 2 (g) 2ICl (g) 2 (g) 2HCl (g) 1.6 1.4 1.2 1 Concentration (M) 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 Time (s) *Draw lines on plot Avg rate = delta [HCl] / delta t = (1.26 M – 0.00 M) / (4.0 s – 0.0 s) = 0.32 M/s Inst rate HCl at 4.0 s = slope of tan line =~ (1.39 M – 0.94 M) / (7.0 s – 1.0 s) = 0.11 M/s Inst rate H 2 4.0 s = ½ (Inst rate for HCl at 4.0 s) = 0.11 M/s / 2 = 0.055 M/s Effects of concentration on rate For many (but not all) reactions, as the [ ] of a reactant changes, the rate of reaction changes. Mathematically rate = k [reactant] Rate law – a mathematical expression that summarizes the relationship between concentrations and reaction rate. Rate constant (k) – a proportionality constant relating reaction rate and concentrations of reactants and other species that affect the rate of a reaction. Finding rate laws from initial rates (1) Initial rate method (to find order of reactant A) 1) Run the experiment with known initial concentrations (i.e. [A] , [B] 0…] 0 2) Measure the initial rate of reaction (slope of tangent line to concentration curve at t=0). Week 1 Notes 3) Change [A] but 0eep all other initial concentrations constant. 4) Measure the initial rate again. 5) Use the ratio of the two rates to find the order of [A] . 0 Practice Determine how changing the initial concentrations of the reactants affects the rate of the reaction between methyl acetate and hydroxide. CH C3OCH + OH 3 H COO + CH OH 3 3 Experiment Initial [CH C3OCH ] Init3al [OH] Initial rate (M/s) 1 0.040 0.040 2.2 x 10 4 4 2 0.040 0.080 4.5 x 10 3 0.080 0.080 9.0 x 10 4 Double concentration of OH double the rate Exp 2 and 3 – initial concentration changes can use to find how CH COOCH concentr3tion 3 affects rate. Comparing exp 1 and 2 When [OH] is doubled the rate doubles Rate directly proportional to [OH ] Comparing exp 2 and 3 Doubling [CH COOCH3] doubles 3he rate Rate directly proportional to [CH COOCH ] so the rate law should be 3 3 Rate = k [CH COO3H ] [OH]. N3te: The thing you double gets squared. Finding rate laws from initial rates (2) Once a rate law is known, k can be determined: Rate = k [CH COOCH ] [OH] 3k = 3 Rate / [CH CO3CH ] [OH] 3 Using the data from run 1: k = 2.2 x 10 M/s / (0.040 M) (0.040 M) = 0.14 M s 1 1 In general, you should either repeat this for each run, and take and average, or use the graphical method. Rate law A general reaction will have a rate law: Rate = k [A] [B] … n Where k = rate constant, m, n = order for A and B respectively, and m + n +… = the overall order of the reaction. Order of reaction A general reaction will have a rate law: Rate = k [A] [B] … m n Week 1 Notes If the exponent of a given substance in the rate law is 1, then the reaction is said to be 1 order in that substance. If it is 2, then the reaction is said to be secondorder in that substance. The overall order of the reaction comes from the sum of all the exponents in the rate law. Practice What are the reaction orders and overall order of the reaction between methyl acetate and hydroxide? (equation and table in previous example) Dividing the first two data sets gives: 4.5 x 10 M/s / 2.2 x 10 M/s = k (0.040 M) (0.080 M) / k (0.040 M) (0.040 M) n = 2.05 = 1 (2.00) n = log (2.05) = n x log (2.00) n =~ 1 Using exp 2 and 3 gives: 4 4 m n m n 9.0 x 10 M/s / 4.5 x 10 M/s = k (0.080 M) (0.080 M) / k (0.040 M) (0.080 M) 2.00 = (2.00) (1) n 2.00 = (2.00) m Log (2.00) = m log (2.00) m = 1 Practice Use the initial rate data for the following reaction at 273 K to… 2 NO (g)r 2 (g) NOBr (g) Determine the order of the reaction in [NO] and [Br ], t2e overall order of the reaction, and the rate law expression. Calculate the rate constant for the reaction at 273 K. Note: Refer to the same table as before and the following: Experiment [NO] (M) [Br2] (M) Initial rate of NOBr formation (M/s) 1 0.10 0.10 12 2 0.10 0.20 24 3 0.20 0.10 48 4 0.30 0.20 216 Using exp 1 and 2 we find: Week 1 Notes m n Rate = k [NO] [Br ] 2 24 M/s / 12 M/s = k [0.10 M] [0.20 M] / k [0.10 M] [0.10 M] 2.0 = (2.0) n=1 n Using exp 2 and 3 we find: 48 M/s / 12 M/s = k [0.20 M] [0.10 M] / k [0.10 M] [0.10 M] 4.0 = (2.0) m=2 m 2 So the rate law is: rate = k [NO] [Br ] 2 The reaction is 2nd –order in NO and 1st order in Br It is 3rd 2. er overall. Practice Using exp 1 and the rate law, we can find the rate constant. 12 M/s = k [0.10 M] [0.10 M] k = 12 M/s / [0.10 M] [0.10 M] = 1.2 x 10 M s 4 2 1 For confirmation, use the other experiment to calculate the rate constant. Clearly, k=1.2x10 M s4 2 1 Practice + Use initial rate data for 2H O 2 2 (aq) I (aq) 2H (aq) I3 (aq) H O2 to(l) Determine the order of reaction in [H O ] and2[I2, the overall order of reaction, and rate law expression. Calculate the rate constant for reaction at 273 K. Experiment [H 2 ]2(M) [I] (M) Initial rate of I3 formation (M/s) 4 1 0.100 0.100 1.15 x 10 4 2 0.200 0.100 2.30 x 10 3 0.100 0.200 2.30 x 10 4 4 4 0.200 0.200 4.60 x 10 Exp 1 and 2 we find: Rate = k [H O 2 2I] n 4 4 m n m n 2.30 x 10 M/s / 1.15 x 10 M/s = k [0.200 M] [0.100 M] / k [0.100 M] [0.100 M] 2.0 = (2.0) n M=1 Exp 1 and 3… 2.30 x 10 M/s / 1.15 x 10 M/s = k [0.100 M] [0.200 M] / k [0.100 M] [0.100 M] m n n 2.0 = (2.0) Week 1 Notes N=1 Experiment 4 and rate law to find rate constant. 4.6 x 10 M/s = k [0.200 M] [0.200 M] K = 4.60 x 10 M/s / [0.200 M] [0.200 M] = 1.15 x 10 1/ M s 2 From here on: E.g. dA means delta A Integrated rate law st Consider a 1 order reaction A products Rate = d [A] / dt = k [A] This integrates to: ln [A] =tkt + ln [A] 0 Straight line equation; y=mx+b This equation is similar to the integrated rate law equation. If a reaction is 1 order, a plot of ln [A] vs. t is linear. Common integrated rate laws: 1) Order: 0; Rate law: rate=k; Integrated rate law: [A]=kt+[At 0 2) Order: 1; Rate law: rate=k[A]; Integrated: ln[A]=kt+ln[t] 0 3) Order: 2; Rate law: rate=k[A] ; Integrated: 1/[A]=kt+1/[t] 0 The slope of a plot yields the most accurate k. For 0 order, expect a horizontal line as your graph. For 1 order, expect a steadily increasing line. For 2 order, expect an exponentially increasing curve. Practice Use the data in the table to determine the order of the reaction and rate constant: Aproducts Time (s) [A] (M) Ln[A] 1/[A] 2 0.000 2.000x10 3.912 50.00 50.00 1.898x10 2 3.964 52.68 2 100.0 1.801x10 4.017 55.51 250.0 1.540x10 2 4.174 64.94 500.0 1.185x10 2 4.435 84.35 3 1000. 7.027x10 4.958 142.3 [ ] is decreasing, ln [A] is decreasing, 1/[A] is increasing Make the 3 graphs (time – [A], time – Ln [A], and time 1/[A]) Reaction 1 Order. K=slope. Pick the most extreme values. K=m=(4.96+3.91 / 10000) = 3 1.05x10 1/s. Find k by taking two points on plot and find the slope. Week 1 Notes We could have found the reaction order by calculating the slope directly from the values in the table. The one with the constant slope is the correct order. Practice st 3 You have a reaction that is 1 order in reactant A. If [A] = 0.500 M an0 k=1.733x10 1/s, what is [A] after 10.0 minutes? Ln [A] = kt + ln [A] 0 Ln [A] – Ln [A] = k0 Ln ([A]/[A] ) =0kt kt [A]/[A] = 0 [A] = [A] e [A] = (0.5 M) (e (1.733x10^3 1/s) () [A] = 0.177 M 0 Practice Given a reaction that is 1 order in reactant A. If [A] = 0.500 M 0nd k = 1.733x10 1/s, how 3 many seconds until [A] = 0.0500 M? Ln [A] = kt + ln [A] 0 Ln [A] – ln [A] = k0 Ln ([A] / [A] ) =0kt T = ln ([A] / [A] ) /0k T = ln ([0.0500 M]/0.5 M) / 1.733x10 1/s 3 T = ln (.1) / 1.733x10 1/s = 1330 s *Kinetics can tell you the speed at which reactions occur, so it can tell which reactions finishes first. *Chemotherapy is basically poisoning everything. Kinetics makes it possible to cause a reaction with cancer cells that will kill that cancer cells faster than it will kill you. Halflife Halflife (t )1/2the time it takes for one half of the reactant to be converted into product (s) For a 1 order reaction: ln ([A] / [A] )t kt 0 When t=t [A1/2 ½ [At ½ 0 Week 1 Notes Then ln ((1/2 [A] )/ 0 ] ) = l0 (1/2) = kt 1/2 Ln (1/2) = ln(2) = kt 1/2= ln1/2 / k Zeroth order: t = [A] / 2k 1/2 0 First order: t 1/2 (2)/k the only order that’s independent of initial concentration Second order: t = 1/2[A] 0 Practice nd 2 1 1 The reaction below is 2 order and k=2.0x10 M hr at 527 C (degrees C). Find the halflife if 0.525 mol of CH CHO i3 injected into a 3.50 L reaction vessel at 527 C. CH C3O (g) CH (g) + C4 (g) M = 0.525 mol / 3.5 L = 0.15 M 2 T 1/21/ (2.0x10 1/M hr) (0.15 M) T 1/20333 1/hr Practice The halflife for the following reaction is 688 hr at 1000 C. Find the rate constant if the reaction is first order. CS (2 CS (g) + S (g) *First order rate constant must be 1/time T = ln (2) / k 1/2 K = ln (2) / t 1/2 K = 0.693/688 hr = 1.01 x 10 hr 3 1 Elementary Reactions (intro to mechanisms – answers how fast and how reactions occur – part of kinetics) Elementary reactions – are the reactions that occur at the molecular level. Unimolecular reactions – reactions that involve a single molecule (or particle) that rearranges into one or more different particles. Bimolecular reactions – reactions involving two molecules (or particles) that collide and rearrange. **Important definitions Transition state theory Activation energy (E ) – tae minimum energy that reactant molecules must have to be converted to product molecules. Transition state (activated complex) – a molecular structure corresponding to the maximum of a plot of energy vs. reaction progress. Week 1 Notes Intermediate – an atom, molecule or ion that is produced in one step of a reaction mechanism and used up in a later step. Unimolecular reactions The isomerization between cis2butene (cis=same side) to trans2butene is unimolecular (trans=opposite). The bond must be broken in order to rotate, which takes energy. Cistrans conversion breaks the pibond in the C=C bond. Parallel, overlapping p orbitals make a pi bond, which resists rotation. Rotating 90 degrees makes the p orbitals perpendicular, the energy required is the pibond energy. Remember double bonds consist of one sigma bond and one pi bond! To break bonds is always endothermic; to make a bond is always exothermic. The reactants must have sufficient energy. It is always harder to go from trans to cis because cis is less stable. Bimolecular reactions Iodide ions reacting with methyl bromide are bimolecular. I (aq) + CH B3 (aq) ICH (aq3 + Br (aq) The reactants must have sufficient energy AND the right orientation. There is activation energy for these reactions as well. Temperature and reaction rate Increasing temperature speeds up most reactions (direct relationship) Molecules with enough energy to overcome activation energy WILL react!
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