ThermodynamicsSampleProblemsAugust2004：2004八月热力学样本问题.doc

2004 B. ChazotteBIOCHEMISTRY 304Thermodynamics Sample Problems August 20041A)What is the arithmetic change in DG that accompanies a 25-fold increase in Keq at 27 C?This is the key equation (1) needed to do the calculation:DG = -2.303 RT log Keq(1)DG= DG2 - DG1 (2) since free energies are state functions and are additive In one approach, we can assume an arbitrary equilibrium constant of Keq1= 1.000 x 10-5 (3)a 25-fold increase would be Keq2 = 2.500 x 10-4 (4)and substitute into equation (2):DG2 - DG1 = -2.303 RT log Keq2 - log Keq1(5)DG2 - DG1 = -2.303 RT log Keq2 / Keq1(6)and solve equation (6)or we can more simply write based on eq (6) using the factor of 25 increase in Keq and then solve: DG= -2.303 RT log 25(7)DG= -2.303 (1.987 cal-1 mol-1 deg-1) x (300 K) x 1.398(8)DG= -1,919 cal mol-1 = - 8,030 J mol-1 (9)b) What is the effect of a decrease of 1.36 kcal/mol in DG on the equilibrium constant of a reaction at 15 C?DG = DG2 - DG1 = -1,360 cal mol-1(2)DG = -2.303 RT log Keq2 / Keq1(6) -1,360 cal mol-1= -2.303 (1.987 cal-1 mol-1 deg-1) x (288 K) log Keq2 / Keq1 (10)-1,360 cal mol-1= -1,318 cal1 mol-1 log Keq2 / Keq1(11)log Keq2 / Keq1= 1.032(12)Keq2 / Keq1= 10.8(13)2) Determine the enthalpy for a chemical reaction that has an equilibrium constant of 1.782 x 103 at 20 C and an equilibrium constant of 2.065 x 105 at 37 C. This requires the use of the equation based on the vant Hoff isochore which assumes that the enthalpy is independent of temperature.Keq2 = DH 1 - 1(14) ln Keq1 R T1 T2 substituting in:2.065 x 105 = DH 1 - 1ln1.782 x 103 8.314 J mol-1 deg-1 293.15 310.15 (15) = DH ln 1.159 x 102 8.314 J mol-1 deg-1 x 1.870 10-4 deg 1 (16) DH 4.753 = 8.314 J mol-1 deg-1 x 1.870 10-4 deg 1(17)rearranging for DHDH = 3.952x101 J mol-1 deg-1 / 1.870 10-4 deg 1 (18)DH = 2.113x105 J mol-1 (19)or DH = 5.051 x104 cal mol-1 = 50.51 kcal mol-1 (19a)Using the incorrect R= 8.2314 the results would be: DH = 5.001 x104 cal mol-1 = 50.00 kcal mol-1 = 2.092x105 J mol-1 3.For the reaction 2A + 1 B 2C + 2D determine the equilibrium concentration of C given the following. At equilibrium A =4.000 x 10-3 M, B =1.000 x 10-3, D = 1.500 x 10-1, DG = -3,521 J at 25CFirst we need to determine Keq using the following equation:DG = -2.303 RT log Keq(20)where Keq =C2 D2A2 B1rearranging: log Keq = DG / -2.303 RT(21)substituting :log Keq = -3,521 J mol-1 / -(2.303 x 8.314 J mol-1 deg-1 x 298.15 K)(22)log Keq = -3,521 J mol-1/-5,709 J mol-1(23)log Keq = 0.617(24)Keq = 4.138 x 100 = C2 D2 = C2 1.500 x 10-1 2A2 B1 4.000 10-32 1.000 x 10-31 (25) C2 2.250x10-24.138 x 100 = 1.60 x 10-5 1.00 x 10-3 (26)rearranging 4.196 x 101 1.6x 10-5 1.00 x 10-3 C2 = 2.250x10-2(27) C2= 2.937 x 10-6 M2(28) C= 1.714 x 10-3M(29)4. Assume that you have a solution of 0.1M glucose -6-P. To this reaction you add the appropriate amount of the enzyme phosphoglucomutase which catalyzes the following reaction at 25 C:phosphoglucomutase Glucose-6-P Glucose-1-P G = 1.8 kcal mol-1Abbrev: (G-6-P) (G-1-P)a) Does this reaction proceed as written and if so what are the final concentrations of G-6-P and G-1-P?Since the reactions initial conditions are NOT equilibrium conditions (you have G-6-P and no G-1-P), the reaction will proceed until it reaches equilibrium. We will need this equation:DG = -2.303 RT log Keq (30) From the reaction we know Keq = G-1-P / G-6-P = products / reactants (31) We can write using equations 30 and 31: G-1-P 1,800 cal mol-1 = -2.303 (1.987 cal deg-1 mol-1 x 298.15 K) log G-6-P(32) G-1-P 1,800 cal mol-1 = -2.303 (592.4 cal mol-1) log G-6-P (33) G-1-P 1,800 cal mol-1 = -1364.3 cal mol-1 log G-6-P(34) G-1-P -1.32 = log G-6-P (35)Invert reactant and product for positive log. G-6-P 1.32 = log G-1-P (36) G-6-P20.86 2.1 10 1 = G-1-P = Keq(37)Thus we now know that there is one G-1-P for every 21 G-6-P molecules are equilibrium from equation (37).Remembering the initial G-6-P concentration of 0.1 M and the equilibrium ratio (Kre) in equation (37) we can set up the following considering (1+21 = 22):G-1-P = (1/22) (0.1 M) 0.0045 M (38)G-6-P = 21/22 (0.1 M) 0.096 M(39)b) Under what cellular conditions, if any, would this reaction continuously produce G-1-P at a high rate?This is an equilibrium question. We know thermodynamically that there will be much more G-6-P than G-1-P (See 4a). Thus we need coupled reactions where G-1-P must be constantly removed by other reactions so that a high ratio of G-6-P/ G-1-P is maintained to drive the reaction forward since G is slightly positive, which also requires a constant resupply of G-6-P. Only than can the reaction continue to proceed at an appreciable rate.5.For the reactions at 37 C.A + B C + D G1 = +45 kJ mol-1 C + D G + H G2 = -98 kJ mol-1 What is the overall equilibrium constant?These are coupled reaction, the first endogonic, the second exogonic. Method 1 Free energies are additive. Solve for overall G and for its Keq .Goverall = G1 + G2 = 45 kJ mol-1 - 98 kJ mol-1 = -53 kJ mol-1Need the equation to relate G to KeqGoverall = -RT ln Keq overall or Goverall = -2.303 RT log Keq overall-53 kJ mol-1 = -2.303 RT log Keq overall -53 kJ mol-1-2.303 (8.3145 J K-1mol-1 x (273.15 K + 37 C) = log keq overall -53 kJ mol-1-2.303 (8.3145 J K-1mol-1 x (273.15 K + 37 C) = log keq overall -53 kJ mol-1-2.303 (8.3145 J K-1mol-1 x (310.15 K ) = log keq overall -53,000 J mol-1-2.303 (2578.7 J mol-1) = log keq overall -53,000 J mol-15938.8 J mol-1 = log keq overall 8.924= log Keq overall 8.40 x 10 8= Keq overall METHOD 2Equilibrium constants are multiplicative. Solve for e