Announcements. Next exam on G, Macronutrients, Kinetics and Smog Thursday.

Transcription

1 Announcements Turn on the Clicker (the red LED comes on). Push Join button followed by 20 followed by the Send button (switches to flashing green LED if successful). Next exam on G, Macronutrients, Kinetics and Smog Thursday. As usual please do not enter class room until let in. I have graded lab reports for J. Matzke, A. Graul, D. Tatro, M. Hoeft, J. Adam, A. Kaur, K. Mahar & J. Moss. If one of your group didn't get it before lecture please collect after class.

2 Review: Reaction Mechanisms Elementary Steps Unimolecular: A --> P Rate Law: -d[a]/dt=d[p]/dt = k[a] Bimolecular: A + B --> P Rate Law: -d[a]/dt=-d[b]/dt = d[p]/dt = k[a][b] Also 2 A --> P has rate law -d[a]/dt = k[a] 2 Mechanism consists of sequence of elementary steps. Rate limiting or rate determining steps (overall rate is determined by slow step) Steady state approximation (an intermediate product concentration stays the same during the reaction)

3 Review: Steady State Approximation Assume that the concentration of an intermediate does not change during the reaction. Example mechanism for the RXN: A + C > D A k1 > B B + C k2 > D assume d[b]/dt = [B]/ t = 0 = k1[a] - k2[b][c] Solve for [B]ss k1[a]/{k2[c]}, overall rate of A + C > D is d[d]/dt = k2[b][c] Substituting in [B] ss gives d[d]/dt = k 2 {k1[a]/{k 2 [C]}}[C] = k1[a].

4 Review: Preequilibria Assumes that the one reaction goes forward and reverse so fast that the concentration of the intermediate is controlled by that reaction. Example mechanism for A + C > D A < k f, k r > B B + C k 2 > D. d[b]/dt = k f [A] -k r [B] k 2 [B][C] assume k 2 <<< k f and k r. So d[b]/dt = 0 = k f [A] - k r [B] [B] k f [A]/k r d[d]/dt = k 2 [B][C] = k 2 {k f [A]/k r }[C] = {k 2 k f /k r }[A][C]. As we will see k f /k r = K eq the equilibrium constant for the first reaction.

5 T dependence of NO + O 3 > NO 2 + O 2 T (K) K (M -1 s -1 ) ln k 1/T (K -1 ) E E E E E E k(t) = Aexp(-E a /{RT}) ln(k(t)) = lna - (Ea/R) (1/T)

6 Activation Energy and Catalysis

7 Heterogeneous Catalysis (Catalytic Converter) Cartoon is for 3H 2 + N 2 > 2NH 3 Chang Figure Image at right courtesy of Dudley Metropolitan Borough Council of the UK. Chang Figure 14.21

8 Chapter 15-Chemical Equilibria Not on Exam 2 Dynamic Equilibria/What is equilibrium? Equilibrium constants, mass action expressions Reaction quotient Q and direction of change K p vs K c K and G Le Châtelier s principle (equilibrium response to change in conditions) Effect of catalysts Calculations of equilibrium concentrations/pressures Dependence of K on T Heterogeneous equilibria

9 N 2 + O 2 2 NO N 2 Concentration (M) O 2 NO Time

10 Partial Mechanism for Reverse Reaction: 2 NO > N 2 + O 2 2 NO k 1 > N 2 O 2 O 2 k 2 k-2 2 O (fast equilibrium) N 2 O 2 k 3 > N 2 + O 2 N 2 O 2 + O k 4 > NO + NO 2 Overall Rate Law: [N 2 ]/ t = k r [NO][O 2 ] -1/2

11 N 2 + O 2 2 NO N 2 Concentration (M) O 2 NO Time

12 Enthalpies of Solution Review Contributions: H ionic > 0, H H-bonds > 0, H ion-dipole < 0 H H-bonds + H ion-dipole = H hyd H soln = H hyd + H ionic, overall sign depends on balance. S quantifies the disorder of a system larger S means more disorder Spontaneous processes: S univ = S sys + S surr >0 S sys S o rxn = ΣSo prod - ΣSo reac G = H sys T S sys is easier to use than S univ G <0 = spontaneous, G > 0 non-spontaneous (exergonic) (endergonic)

13 Review G = H T S H H H H <0, S > 0 G < 0 always spontaneous <0, S < 0 G? spontaneous at low T >0, S < 0 G > 0 never spontaneous >0, S > 0 G? spontaneous at high T Water near its boiling point is example of the last case.

14 Review Calculating G From H o f and So. G = H sys T S sys From G o f. Go RXN = Σ Go f (prod) Σ Go f (reac) G < spontaneous, G > 0 nonspontaneous Carbohydrates, Proteins, Lipids Carbohydrates = starch, cellulose & sugars Proteins made of chains of α-amino acids coupled by condensation reactions to form peptide bonds:

15 Lipids Review Stereoisomerism of proteins 4 different groups on C make a chiral center. Called stereoisomers or enantiomers. Fatty acids bound to glycerol in a condensation reaction. Saturated have no double bonds in chains Unsaturated have double bonds in chains

16 Review Results of Food Value Calculations Calculation Path: G (kj/mol) --> Energy Available (kj/g) --> Cal/g Don't forget that 1 Cal = 1000 cal = 4184 J Energy Available (kj/g) glucose (carbohydrate) Alanine (amino acid) Steric Acid (common saturated fatty acid) CH 3 CH 2 OH (ethanol) Food Value (Cal/g)

17 Review (S, G & macronutrients) Catabolism Carbohydrates, Proteins, Lipids converted to pyruvate then to Acetyl CoA which passes into Krebs Cycle where CO 2 and H 2 O are produced along with ADP --> ATP Example of glycolysis (glucose -->pyruvate): steps with G >0 driven by coupling to ATP + H 2 O --> ADP + HPO 4 2-, where G < 0 Can tell whether process stores energy for organism by counting number of ATP --> ADP vs ADP --> ATP. DNA (deoxyribonucleic acid) Sugar backbone Bases on each sugar forms a double helix with bases pairing by H-bonding (A T and C G) DNA transcribed to mrna (decoded by ribosomes to make proteins)

18 Review (Kinetics) Kinetics is the study of the time variation of concentrations in systems of chemical reactions. Example: smog shows variation of which species are present versus time. Example: A ---> 2 B + C Average rate = R = (1/2) [B]/ t = [C]/ t = - [A]/ t

19 Reaction Rates Review Write rate using derivative notation (Example A + 2B --> C): R = -d[a]/dt = -(1/2)d[B]/dt = d[c]/dt. Rate laws of form: R = -d[a]/dt = k[a] a [B] b [C] c... Simple exponents (2, 1, 0, -1, -2) can easily be determined from initial rate data. Rate doubles on doubling a species exponent = 1 Rate halves on doubling a species exponent = -1 Rate quadruples on doubling a species exponent = 2 Rate unchanged on doubling a species exponent = 0

20 Simple Integrated Rate Laws for -d[a]/dt = k app [A] a 0 th order a = 0: [A] t = [A] o k app t 1 st order a = 1: [A] t = [A] o exp{-k app t} Linear: ln[a] t = ln[a] o -k app t 2 nd order a = 2: 1/[A] t = 1/[A] o + kt Alternate form: [ A ] = [ A ] 0 [ A ] 0 kt + 1

21 Review Pseudo-order (Swamping) method Uses large excess of all but one reactant, so concentration of only the limiting reactant (A) changes significantly. -d[a]/dt = (k[b] ob )[A] a k app [A] a 0 th order a = 0: [A] t = [A] o k app t 1 st order a = 1: [A] t = [A] o exp{-k app t} Linear: ln[a] t = ln[a] o -k app t 2 nd order a = 2: 1/[A] t = 1/[A] o + kt Can determine k and b by varying [B] o lnk app = lnk + bln[b] o ln[b] o ln[b] o

22 Review: Reaction Mechanisms Elementary Steps Unimolecular: A --> P Rate Law: -d[a]/dt=d[p]/dt = k[a] Bimolecular: A + B --> P Rate Law: -d[a]/dt=-d[b]/dt = d[p]/dt = k[a][b] Also 2 A --> P has rate law -d[a]/dt = k[a] 2 Mechanism consists of sequence of elementary steps. Rate limiting or rate determining steps (overall rate is determined by slow step) Steady state approximation (an intermediate product concentration stays the same during the reaction)

23 Review: Steady State Approximation Assume that the concentration of an intermediate does not change during the reaction. Example mechanism for the RXN: A + C > D A k1 > B B + C k2 > D assume d[b]/dt = [B]/ t = 0 = k1[a] - k2[b][c] Solve for [B]ss k1[a]/{k2[c]}, overall rate of A + C > D is d[d]/dt = k2[b][c] Substituting in [B] ss gives d[d]/dt = k 2 {k1[a]/{k 2 [C]}}[C] = k1[a].

24 Review: Preequilibria Assumes that the one reaction goes forward and reverse so fast that the concentration of the intermediate is controlled by that reaction. Examplemechanism for A + C > D A < k f, k r > B B + C k 2 > D. d[b]/dt = k f [A] -k r [B] k 2 [B][C] assume k 2 <<< k f and k r. So d[b]/dt = 0 = k f [A] - k r [B] [B] k f [A]/k r d[d]/dt = k 2 [B][C] = k 2 {k f [A]/k r }[C] = {k 2 k f /k r }[A][C]. As we will see k f /k r = K eq the equilibrium constant for the first reaction.

25 T dependence of NO + O 3 > NO 2 + O 2 T (K) K (M -1 s -1 ) ln k 1/T (K -1 ) E E E E E E k(t) = Aexp(-E a /{RT}) ln(k(t)) = lna - (Ea/R) (1/T)