LS1a Fall 2008 Lect 3-4 Lecture Slides final

LS1a Fall 2008 Lect 3-4 Lecture Slides final - Chemical...

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Unformatted text preview: Chemical Equilibrium and Thermodynamics Life Sciences 1a Lecture Slides Set 2 Fall 2008 Prof. Daniel Kahne Lectures 3 & 4: The Chemical Foundations of Life II 1. Understanding thermodynamics and chemical equilibrium a. First law of thermodynamics b. The concept of chemical equilibrium and Keq c. Gibbs free energy, G and Gorxn d. Relationship between Keq and Gorxn e. Coupled reactions f. State of the cell 2. Understanding other forms of equilibria a. Acidity, Ka and pKa b. The Henderson-Hasselbach equation c. Understanding the Le Chatelier's principle 1 The flask vs the cell The cell The flask All living systems require energy First law of thermodynamics Energy cannot be created or destroyed 2 Cells get energy from food Favored Reaction high energy C6H12O6 , O2 ENERGY low energy CO2 , H2O Energy is released when glucose and oxygen react to give carbon dioxide and water Some reactions release energy but other reactions require energy Disfavored reaction high energy ENERGY C6H12O6 , O2 low energy CO2 , H2O In cells, reactions can go in both directions. How? 3 How do we know which direction is favored? Favored Reaction Disfavored reaction Chemical equilibrium... Equilibrium: a dynamic balancing act A H O H H O B H O H H H O O H H Liquid water H O H H H H O H O Water vapor H O H H Water, carbon dioxide H O + H O C O H O O C O H Carbonic acid At equilibrium, the concentrations of two interconverting states do not change (forward rate = reverse rate) 4 Hydration of carbon dioxide CO2 (aq.) + H2O (l) H2CO3 (aq.) Which side is favored? The equilibrium constant reflects which side of the reaction is favored CO2 (aq.) + H2O (l) Keq H2CO3 (aq.) = [products]/[reactants] = [H2CO3(aq.)]/[CO2(aq.)][H2O(l)] = 3.09 X 10-5 M-1 at 25 o C Keq < 1: favors the left side (reactants) Keq > 1: favors the right side (products) Keq = 1: favors neither side 5 The position of a chemical equilibrium is independent of the initial state CO2 (aq.) + H2O (l) 100% H2CO3 concentration 1 2 1 H2CO3 (aq.) State 1 [H2CO3]/[CO2] = 1 Keq[H2O] 100% CO2 concentration time 2 State 2 (equilibrium) [H2CO3]/[CO2] = Keq[H2O] time When the ratio of concentrations of products to reactants stop changing, you have reached equilibrium The state of 100% CO2 (aq.) is more stable than the state of 100% H2CO3 (aq.) CO2 (aq.) + H2O (l) H2CO3 (aq.) 100% H2CO3 (aq.) Gorxn 100% CO2 (aq.) Gorxn : Change in G when a reaction proceeds to completion under standard conditions ... why do we have any H2CO3 (aq.) at equilibrium? 6 All systems tend towards a state of lower Gibbs free energy CO2 (aq.) G Gorxn + H2O (l) H2CO3 (aq.) 100% CO2 (aq.) [H2CO3]/[CO2] [H2CO3]/[CO2] =0 = Keq[H2O] 100% H2CO3 (aq.) [H2CO3]/[CO2] = infinity The equilibrium state has the lowest Gibbs free energy G is a measure of how far you are away from equilibrium G G orxn Cfinal Cinitial GAiAf = GAf GAi < 0 spontaneous step GCiCf = GCf CCi > 0 disfavored step (note: reverse step (Cf -> Ci) is spontaneous, GCf->Ci < 0) 100% H2CO3 (aq.) Ainitial Afinal B Binitialfinal 100% CO2 (aq.) GBiBf = GBf GBi = 0 at equilibrium ... because all reactions tend towards a state where G = 0 7 We can calculate Grxn from Keq A + B C + D [C][D] G = Grxn + RT ln [A][B] At equilibrium, 0 = Grxn + RT ln Keq Therefore, Grxn = RT ln Keq Grxn tells us which direction is favored Grxn = RT ln Keq Keq > 1 Grxn < 0 "downhill" "favorable" "spontaneous" "uphill" "disfavored" A Energy B A A B Energy Keq < 1 Grxn > 0 B Keq = 1 Grxn = 0 Neither side is favored 8 How does Nature drive disfavored reactions? Energy A B ADP B (P) (P) Grxn > 0 Energy Grxn << 0 ATP ATP A ADP Energy Grxn < 0 Cells can enable uphill reactions to proceed efficiently by coupling them with favorable reactions (e.g., ATP hydrolysis) Chemical equilibrium and thermodynamics CO2 (aq.) + H2O (l) 100% H 2CO3 conc. 1 2 1 H2CO3 (aq.) State 1 Gorxn = -RT ln Keq [H2CO3]/[CO2] = 1 Keq[H2O] G lowest G < 0 (towards equilibrium) State 2 (equilibrium) Gorxn = -RT ln Keq [H2CO3]/[CO2] = Keq[H2O] G = lowest G = 0 (at equilibrium) time G G decreases as a system approaches equilibrium 2 G 0 time time All reactions tend towards a state of chemical equilibrium and lowest energy where G = 0 9 Cells are not at equilibrium The cell The flask Important equilibria Dissolution CO2 (g) Reaction Keq =0.8317 CO2 (aq.) H2CO3 (aq.) M-1 CO2 (aq.) + H2O (l) Acid-Base Keq = 3.09 X 10-5 H2CO3 (aq.) + H2O (l) H3O+ (aq.) + HCO3- (aq.) Keq = 4.46 x 10-6 10 The concept of acidity H2CO3 (aq.) + H2O (l) H3O+ (aq.) + HCO3- (aq.) Acidity and Ka H2CO3 (aq.) + H2O (l) Keq H3O+ (aq.) + HCO3- (aq.) [H3O+(aq.)][HCO3-(aq.)] -6 = [H2CO3(aq.)][H2O(l)] = 4.46 x 10 [H3O+(aq.)][HCO3-(aq.)] = 2.5 x 10-4 M [H2CO3(aq.)] Ka = Keq[H2O(l)] = In general, Ka = [H3O+(aq.)][A-(aq.)] [HA(aq.)] 11 pKa values of common acids acid Phosphoric acid DNA backbone O O P H O OH OH H conjugate base + Ka 10-2 10-2 pKa (= log Ka) 2 + O OH P OH -O O increasing acidity O P H O O O H + + O P O -O 2 3 5 9 OH O OH H Citric acid Acetic acid Ammonium ion O OH O OH HO O H O + + HO -O 10-3 10-5 10-9 O O H O H + + O -O H N H H H H + N H H H + + What does pH mean? AH A + pH = log [H+] (In water, pH = log [H3O+]) The lower the pH, the higher the [H+], indicating a more acidic solution Each pH unit represents a 10-fold change in [H+] H+ 12 The Henderson-Hasselbalch equation pKa = pH + log Two key implications: 1) If pH increases by 1: The ratio of [A] (deprotonated) to [HA] increases by 10-fold Conversely, if pH decreases by 1: The ratio of [HA] (protonated) to [A] increases by 10-fold 2) When pH = pKa, then [A] = [HA] [HA] [A] pH in a cell (pH ~ 7.4) In a solution with a mixture of acids, the overall pH determines the [HA]/[A-] ratios of each acid 13 Le Chatelier's principle Dissolution CO2 (g) Reaction Keq =0.8317 CO2 (aq.) H2CO3 (aq.) M-1 CO2 (aq.) + H2O (l) Acid-Base Keq = 3.09 X 10-5 H2CO3 (aq.) + H2O (l) H3O+ (aq.) + HCO3- (aq.) Keq = 4.46 x 10-6 [H2CO3]/[HCO3-] = 1:6300 (pH 7.4) Key Points to Bring Home All reactions tend towards a state of equilibrium where the ratio of [products] to [reactants] approaches Keq the Gibbs free energy (G) in the system approaches a minimum the G becomes zero Gorxn can be calculated from the value of Keq, and both of these parameters tell us which side of a reaction is favored The cell is always away from the state of equilibrium Strong acids have low pKas while weak acids have large pKas The Henderson-Hasselbalch equation tells us the ratio of concentrations of the protonated to deprotonated states of an acid at a particular pH Disfavored reactions can be driven by coupling them to strongly favored reactions like ATP hydrolysis or by using the Le Chatelier's principle 14 ...
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This note was uploaded on 09/23/2008 for the course LIFESCI 1a taught by Professor Kahne during the Fall '08 term at Harvard.

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