6) If we represent protein folding with the following equilibrium expression: Unfolded protein= folded protein The values of AH and AS° for the folding of lysozyme protein are -280 kJ/mol and -790 J/mol-K, respectively. a) Based on these values, calculate AG° for the folding process at 25 °C. b) At what temperature would you expect the unfolding process to become unfavorable?

6) If we represent protein folding with the following equilibrium expression: Unfolded protein= folded protein The values of AH and AS° for the folding of lysozyme protein are -280 kJ/mol and -790 J/mol-K, respectively. a) Based on these values, calculate AG° for the folding process at 25 °C. b) At what temperature would you expect the unfolding process to become unfavorable?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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1. Calculate AHrxn° for the reaction: 2 N₂(g) + 5 O₂(g) → 2 N₂O5(g) from the following data: N₂O5(g) + H₂O(l) → 2 HNO3(1) N₂(g) + 3 O₂(g) + H₂(g) → 2 HNO3(1) 2 H₂(g) + O₂(g) → 2 H₂O(l) AHrxn° = -76.6 kJ AHrxn° = -348.2 kJ AHrxn° = -671.6 kJ
A student determines the value of the equilibrium constant to be 6.29×104 for the following reaction.S(s,rhombic) + 2CO(g)SO2(g) + 2C(s,graphite)Based on this value of Keq:G° for this reaction is expected to be (greater, less) fill in the blank than zero.Calculate the free energy change for the reaction of 1.85 moles of S(s,rhombic) at standard conditions at 298K. G°rxn = _____ kJ
A student determines the value of the equilibrium constant to be 2.30 x 10¹6 for the following reaction. Ca(OH)2 (aq) + 2HCl(aq) → CaCl₂ (s) + 2H₂O(1) Based on this value of Keq AGO for this reaction is expected to be than zero. Calculate the free energy change for the reaction of 1.57 moles of Ca(OH)2 (aq) at standard conditions at 298 K. AGO rxn kJ
a) As stated in question 5a, glucose (C6H12O6(s)) is a source of cellular energy. Calculate the standard free energy for the metabolism of glucose: Given ΔG°C6H12O6(s) = -911kJ/mol; ΔG°O2(g) = 0kJ/mol; ΔG°CO2 = -394kJ/mol; ΔG°H2O(l) = -237kJ/molC6H12O6(s) + 6O2(g) ➝ 6CO2(g) + 6H2O(l) b) Cells couple the hydrolysis of adenosine triphosphate (ATP) into adenosine diphosphate (ADP) andinorganic phosphate (PO43–) to drive chemical reactions (i.e. as a source of chemical energy). The reaction is: ATP(aq) + H2O(l) ➝ ADP(aq) + PO43–(aq) Calculate K for this reaction if ΔG° = –30.5 kJ/mole. c) If all of the energy from glucose metabolism went into ATP synthesis from ADP and inorganicphosphate, how many molecules of ATP could be generated from each molecule of glucose?
Actual free-energy change is a function of the reactant and product concentrations and the temperature at which the reaction occurs. AG = Go + RT In Use the equation for the actual free-energy change to plot AG against In Q at 25 °C for the five concentrations of ATP, ADP, and P₁ shown in the table. The AG' for the reaction is -30.5 kJ/mol. Concentration (mm) 0.2 5.0 14.9 O O AG (kJ/mol) Metabolite ATP ADP P₁ Select the graph that plots the data. AG (kJ/mol) 0 -20 -40 -~ 0 200 0 -200 [C]c[D]d [A]a[B]b -400 T 2 5 0.2 10 2 T 4 3 2.2 12.1 InQ 6 4 InQ 6 8 1 4.2 14.1 10 8 5 25 10 AG (kJ/mol) AG (kJ/mol) 0 -20 -40 -10 -8 0 -20 -40 -6 -4 InQ -2 0
Use standard thermodynamic data to calculate the value of Keq at 298 K for the Haber reaction: N2 (g)+ 3 H2 (g) ⇌ 2NH3 (g) ΔGfº (KJ/mol): N2 (g) = 0H2 (g) = 0NH3 (g) = -16.4 kJ/mol
A student determines the value of the equilibrium constant to be 4.09×10-18 for the following reaction. Fe;O4(s) + 4H2(g)→3FE(s) + 4H,O(g) Based on this value of Kegi AG° for this reaction is expected to be (greater, less) greater than zero. Calculate the free energy change for the reaction of 1.76 moles of Fe;0,(s) at standard conditions at 298K. AG°. rxn = 190.5 kJ
A reaction A(aq)+B(aq)↽−−⇀C(aq)has a standard free‑energy change of −3.21 kJ/mol at 25 °C. What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively?
A chemist fills a reaction vessel with 0.137 g calcium phosphate (Ca, (PO,) solid, 0.349 M calcium (Ca ) aqueous solution, and 0.240 M phosphate (Po,") 3- POA aqueous solution at a temperature of 25.0°C. Under these conditions, calculate the reaction free energy AG for the following chemical reaction: Ca3 (PO4),(s) → 3Ca (aq) + 2PO, (aq) 4. Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule. kJ
A student determines the value of the equilibrium constant to be 8.18x10-37 for the following reaction. H₂S(g) + 2H₂O(l)—3H₂(g) + SO₂(g) Based on this value of Keq: AG° for this reaction is expected to be (greater, less) zero. than Calculate the free energy change for the reaction of 1.68 moles of H₂S(g) at standard conditions at 298K. AG rxn= kJ