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Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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
Transcribed Image Text:Mountain climbing. Proteins are thermodynamically
unstable. The AG of the hydrolysis of proteins is quite
negative, yet proteins can be quite stable. Explain this
apparent paradox. What does it tell you about protein
synthesis?
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- Consider the following reversible reaction at equilibrium: C6H12O6(aq) + 6 O2(g) 6 CO2(g) + 6 H2O(l). Given that this reaction is exothermic, if heat is added to the equilibrium system, how is the stress relieved?arrow_forwardGlucose-1-phosphate = glucose-6-phosphate AG'o = - 7.3 kJ mol-1 Glucose-6-phosphate = fructose-6-phosphate AG'O = + 1.7 kJ mol-1 Calculate the equilibrium constant, K'eg, for the conversion of glucose-1-phosphate to fructose-6-phosphate. You can use a value of RT = 2.48 kJ mol-1.arrow_forwardConsider the vaporization of carbon disulfide. CS₂ (1) = CS₂(g) Free energies and enthalpies of formation can be found here Calculate the standard free energy for this reaction and the standard enthalpy for this reaction. AG = K25C 2.5 Calculate the equilibrium constants for this reaction at 25 °C and at CS₂'s normal boiling point. Tbp 0.346 Incorrect Calculate the normal boiling point of CS₂. 290.322 kJ/mol Incorrect ΔΗ° = 27.7 °C Kbp 0.355 Incorrect kJ/molarrow_forward
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- What’s the answer for thisarrow_forwardA 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= kJarrow_forwardST5A.2 - Calculate the equilibrium constant given that the standard Gibbs free energy of reaction is -8.16 kJ/mol at 249 K for the reaction. Type your answer...arrow_forward
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