Unformatted text preview: 1. The equilibrium constant for some process A B is 0.5 at 20°C and 10 at 30°C. Assuming that ΔH° is independent of temperature, calculate ΔH° for this reaction. Determine ΔG° and ΔS° at 20°C and 30°C. Solution: @ 293K ΔG° = ‐RTlnKeq = ‐(8.314 x 10‐3 kJ/mol∙K)*(293)*ln(0.5) = 1.69 kJ/mol = ΔG°293 @303K ΔG° = ‐RTlnKeq = ‐(8.314 x 10‐3 kJ/mol∙K)*(303)*ln(10) = ‐5.80 kJ/mol = ΔG°303 ΔG° = ΔH° ‐ TΔS; y = ΔG° m = ‐ ΔS x = T b = ΔH° Therefore, ‐ ΔS = (change in ΔG°)/(change in T) Assume slope‐intercept form = [(‐5.80)‐(1.69)]/[303‐293] = ‐0.75 kJ/mol ΔS° = 0.75 kJ/mol = ΔS°293 = ΔS°303 @293K ΔG° = ΔH° ‐ TΔS 1.69 = ΔH° ‐ (293)*(0.75) ΔH° = 1.69 + (293)*(0.75) = 221.4 kJ/mol = ΔH°293 @303K ΔG° = ΔH° ‐ TΔS ‐5.80 = ΔH° ‐ (293)*(0.75) ΔH° = ‐5.80 + (303)*(0.75) = 221.4 kJ/mol = ΔH°303 2. An enzymatic hydrolysis of fructose‐1‐P was allowed to proceed to equilibrium at 25°C. The free energy of hydrolysis for the given reaction is ‐15.9 kJ/mol. When the system reached equilibrium, the resulting concentration of fructose‐1‐P was 65.2 µM. What was the original concentration of fructose‐1‐P prior to the hydrolysis reaction (you can assume that the original concentration is significantly larger than the resulting concentration)? What is the equilibrium constant for this reaction? Fructose‐1‐P + H2O fructose + Pi Solution: ΔG° = ‐RTlnKeq ‐15.9 kJ/mol = ‐(8.314 x 10‐3 kJ/mol ∙ K) * (298K) * ln(Keq) Keq = e‐15.9/(‐8.314x10‐3)(298) = 2.718‐15.9/(‐8.314x10‐3)(298) = 612 Keq = [products]/[reactants] 612 = [fructose]*[Pi]/[Fructose‐1‐P] = [X‐65.2x10‐6]*[X‐65.2x10‐6]/[65.2x10‐6] = X2/65.2x10‐6 {mathematically, this is ok because we take the assumption that X >>> 65.2x10‐6} X = [(612)*(65.2x10‐6)]0.5 = 0.2 ; therefore the original concentration of fructose‐1‐P is 0.2M 3. Given the following reactions, T = 40°C, and all Ks given are equilibrium constants: K1 K A B 2 C K3 A C K1 = 0.47 K3 = 400 Calculate K2 and the standard state free energies for all processes. Solution: K1 = [B]/[A] K2 = [C]/[B] K3 = [C]/[A] [C] = K3*[A] [B] = K1 * [A] K2 = [C]/[B] = {K3 * [A]}/{K1 * [A]} = K3/K1 = 400/0.47 = 851 = K2 ΔG°1 = ‐RTlnK1 = ‐(8.314x10‐3 kJ/mol ∙ K) * (313 K) * ln(0.47) = 1.96 kJ/mol = ΔG°1 ΔG°2 = ‐RTlnK2 = ‐(8.314x10‐3 kJ/mol ∙ K) * (313 K) * ln(851) = ‐17.6 kJ/mol = ΔG°2 ΔG°3 = ‐RTlnK3 = ‐(8.314x10‐3 kJ/mol ∙ K) * (313 K) * ln(400) = ‐15.6 kJ/mol = ΔG°3 4. Given the hydrolysis of ATP (ΔG° = ‐30.5 kJ/mol) occurring at 47°C and the following concentrations: [ADP] = 3.59 pM [ATP] = 20.4 µM [Pi] = 15.9 nM Calculate ΔG Solution: ATP + H2O ADP + Pi ΔG = ΔG° + RT ln [prod/react] = ‐30.5 kJ/mol + (8.314 x 10‐3 kJ/mol ∙ K) * (320 K) * ln [(3.59 x 10‐12)*(15.9 x 10‐9)/(20.4 x 10‐6)] = ‐119.7 kJ/mol = ΔG 5. If 50 mL of 0.01 M HCl is added to 100 mL of 0.05 M phosphate buffer at pH 7.2, what is the resultant pH? What are the concentrations of H2PO4‐ and HPO42‐ in the final solution? H3PO4 ↔ H2PO4‐ ↔ HPO42‐ ↔PO43‐ pKa1 = 2.15 pKa2 = 7.20 pKa3 = 12.40 Solution: 7.20 = pKa + log [A‐/HA] ; A‐ = HPO42‐ HA = H2PO4‐ 7.20 = 7.20 + log [HPO42‐/ H2PO4‐] (HPO42‐) + (H2PO4‐) = 0.05 M H2PO4‐ = 0.05 ‐ HPO42‐ Therefore 7.20 = 7.20 + log [HPO42‐/0.05 ‐ HPO42‐] [HPO42‐] = 0.025 = [H2PO4‐] 0.050 L *(0.01M HCl) * (1/0.150 L) = 3.33 x 10‐3 M HCl 0.025M HPO42‐ * (0.100 L) * (1/0.150 L) = 1.67 x 10‐2 M HPO42‐ So, when the acid reacts with the buffer, protons will get added to H2PO4‐ [HPO42‐] = 1.67 x 10‐2 – 3.33 x 10‐3 = 1.33 x 10‐2 M = [HPO42‐] [H2PO4‐] = 1.67 x 10‐2 + 3.33 x 10‐3 = 0.02 M = [H2PO4‐] pH = 7.20 + log [(1.33 x 10‐2)/(0.02)] = 7.03 = pH 6. What is the pH of a gluatamic acid solution where a. NH3+ is 3/8 dissociated? b. The side chain is 5/9 dissociated? pK1 = 2.2 pK2 = 4.3 pK3 = 9.7 Solution: a) pH = pKa + log (Glu2‐/Glu‐) = 9.7 + log [(3/8)/(1‐3/8)] = 9.48 = pH @ NH3+ is 3/8 dissociated b) pH = pKa + log (Glu‐/Glu°) = 4.3 + log [(5/9)/(1‐5/9)] = 4.40 = pH @ side chain is 5/9 dissociated ...
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This note was uploaded on 04/17/2011 for the course BIS 102 taught by Professor Hilt during the Winter '08 term at UC Davis.
 Winter '08
 Hilt

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