Rate Law Summary. Rate Laws vary as a function of time

Transcription

1 Rate Law Summary Measure the instantaneous rate of a reaction: this is a number with units of M/s! Measure the rate of loss of a reactant r... the rate of appearance of a product Repeat the experiment to determine the concentration dependence: btain the Rate Law Differential rate Law: gives rate as a function of concentration as t 0 C112 SU13 LRSVDS 1 Rate Laws vary as a function of time We are studying a chemical reaction; for example aa products Measure the rate as a function of concentration as t 0 rate = -d[a] = k[a] x (the differential rate law) dt This is a differential equation; Integration gives [A] as a function of t Differential Rate Law for 1 st order reaction (x = 1): Integrate to get: or: Rearrange: (Different forms of the same equation) C112 SU13 LRSVDS 2

2 FYI: Integration of the First rder Rate law (First order rate law) = k[ A] rate = Δ A Δt [ A] rearrange Δ A = kδt [ A] d A = kdt If Δ[A] d[a] as Δt 0 then [ A] t d A = k dt A [ A] 0 ln[ A] ln[ A] 0 = ln Integrate: at t = 0, [A] = [A] 0 0 [ A] A 0 = kt Integrated first order rate law: concentration as a function of time C112 SU13 LRSVDS 3 The Integrated First rder Rate Law is a Linear Equation ln[a] = k t + ln[a] o This is in the form of: y = m x + b y = b = x = m = A plot of vs. gives a straight line for first order reactions. ln[a] o! Experimental determination of k: ln[a]! slope = k! time! C112 SU13 LRSVDS 4

3 A Linear Equation Provides Important Information: C 3 #N C: C 3 # C N: [A] = [A] 0 e kt P A = P 0 e kt n [A] = n[a] 0 - k t n (P A ) = n(p 0 ) - k t C112 SU13 LRSVDS 5 2 nd order reaction (x = 2): INTEGRATED RATE LAW 2nd order reaction; A P rate = Δ A Δt = k[ A] 2 differential rate law: Integrate to get: 1 A = k t + 1 A 0 integrated 2nd order rate law [A]! t! C112 SU13 LRSVDS 6

4 FYI: Integration of the Second rder Rate Law rate = Δ A Δt rearrange If Δ[A] d[a] as Δt 0 then Integrate: at t = 0 [A] = [A] 0 = k[ A] 2 Second order rate law Δ[ A] [ A] = kδt d[ A] 2 [ A] = kdt 2 [ A] 2 t d A = k dt A [ A] 0 1 [ A] 1 A 0 0 = kt Integrated second order rate law: concentration as a function of time C112 SU13 LRSVDS 7 Example: What is the rate law for this Reaction? N 2 (g) N(g) + ½ 2 (g) Step 1: Measure [N 2 ] as a function of time Collect data Calculate these Time(s) [N 2 ](M) ln [N 2 ] 1/[N 2 ] [N 2 ] (M) TIME (SEC) C112 SU13 LRSVDS 8

5 Step 2: Determine the rder of Reaction Plot ln [N 2 ] vs. time Plot 1/[N 2 ] vs. time Is either plot linear? What is the rate law? C112 SU13 LRSVDS 9 ld Exam Question: At elevated temperatures, N 2 decomposes as below: N 2 (g) N(g) + ½ 2 (g) The reaction is 2 nd order in N 2 and has a rate constant of M -1 s 300 C. If the the initial [N 2 ] is M, how long will it take for the concentration to drop to M? C112 SU13 LRSVDS 10

6 alf life ( t 1/2 ) is the time it takes for the concentration of a reactant to drop to half of its initial value. For the reaction: A products t 1/2 is where For 1 st order reactions: What is t 1/2? alf Life is a Useful Quantity [A] = [A] o 2 n[a] = n[a] o kt First order t 1/2 does NT depend on concentration C112 SU13 LRSVDS 11 alf Life Can be Represented Graphically! What is the half life of C 3 NC according to the graph below?! C 3 #N C: C 3 # C N: C112 SU13 LRSVDS 12

7 alf Life of 2 ND rder Reactions Differs from 1 st rder! For 2 nd order reactions: 1 A = k t + 1 A 0 To get t 1/2, we let [A] = ½ [A] o t 1/2 of a Second rder Reaction depends on the initial concentration [A] 0 C112 SU13 LRSVDS 13 Second rder alf Life can also be represented graphically: N 2 (g) N(g) + 1/2 2 (g) [N 2 ] (M) Time (sec) C112 SU13 LRSVDS 14

8 alf Life Problem-Solving: Focus on Reactants It requires 100 years for a first-order reaction to go 30.0% of the way to completion. What is the half-life of the reaction? rganize Data: time [A] t " ln [A] % t $ ' = kt #[A] o & t 1 2 = k C112 SU13 LRSVDS 15 Information Regarding CATALYSIS Definition of a Catalyst: ow does it do this? Thermodynamic state functions are unaffected by catalysis (ΔE, Δ, Δ G, Δ S ) (Why??) A Catalyst:!!!! ow to identify a catalyst in a mechanism?! C112 SU13 LRSVDS 16

9 omogeneous catalysis: Catalyst is in the same phase as the reactants. Example: (l) 2 2 (g) + 2 (g) 2! +! C112 SU13 LRSVDS 17 eterogeneous catalysis: Catalyst is in a different phase than the reactants /2 2 2 This reaction requires breaking strong # and = bonds 435 kj 498 kj Rate is negligible without a catalyst The stronger the bond, the more need for a catalyst to get a useful rate gas molecules - - = Pt surface solid = adsorbed atoms diffusion on surface = = C112 SU13 LRSVDS 18

10 Examples of eterogeneous Catalysis C Reaction of ethylene with hydrogen gas 2. Catalysis of ydrogen Peroxide Decomposition by Manganese Dioxide (Mn 2 )! Similar Mechanism! 3. Catalytic Converters! 2 1) C, C x 2x+2 C 2 (g) + 2 2) N, N 2 N 2 (g) Catalysts: Cu, Cr 2 3, Pt, Pd, Rh C112 SU13 LRSVDS 19 eterogeneous Catalysis in Nitrogen Fixation N N 3 ΔG = K (spontaneous but high E a ) N N strong triple bond (D = 946 kj) Microorganisms convert waste into N 2 (g), which must be turned into a form plants can use Plants easily accomplish this at 298 K, 1 atm with the Enzyme. The best we can do is the aber Process for Nitrogen Fixation C112 SU13 LRSVDS 20

11 Enzymes are Biological Catalysts! Accelerate and control reaction rates. Enormous protein molecules or combinations of proteins with other molecules. Reactant is called a. The region where substrates bind is called the. Binding of glucose to hexokinase C112 SU13 LRSVDS More efficient Enzymes are superior to man-made catalysts 2. Absolute specificity 3. Regulated What factors can control enzyme activity? C112 SU13 LRSVDS 22

12 Lock and Key Enzyme Model enzyme binding sites k = A e -E /RT a reactant molecules enzyme-substrate complex products 1. ow does the active site lower E a? 2. Active site is very selective for a certain substrate 3. What is the effect on A? Each factor enhances rate by 10 5 C112 SU13 LRSVDS 23 Lowering E a AFFECTS RATES Dramatically! Uncatalyzed Catalase (enzyme in liver) k cat k uncat = /2 2 A cat e -E E a = 72 kj E a = 28 kj Compare k cat /k uncat at 37 C (body temp.); what is ratio? Assume A cat = A uncat (Is this a good assumption?) /RT a,cat A e -E a,uncat /RT uncat k cat k uncat = e [E a,cat E a,uncat ]/ RT ln k cat (28 72)kJ / mol = k uncat (0.0083kJ / mol K)(310K) =17.1 Catalase enzyme speeds up the decomposition rate by a factor of! Peptidase enzymes break up proteins into amino acids (in your stomach). Similar effect on E a. C112 SU13 LRSVDS 24

13 Regulation of Enzyme Activity Activity Depends on: Temperature p Concentration of substrate These conditions affect the active site by affecting the enzyme conformation ( ). (Left) Representation of an active site in an enzyme. (Right) Denatured enzyme; parts of the active site are no longer in close proximity. C112 SU13 LRSVDS 25