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Consider the binding of AMP to phosphorylase b: AMP + Protein AMP-Protein at 25°C the equilibrium constant is 23809 M-1 and at 38 °C the equilibrium constant is 16949 M-1, if we assume that AH does not change with temperature, calculate AGº, AH° and ASº at 25°C. If the concentrations of all species are at 10 micromolar what is the value of AG at 25 C? As the reaction goes towards equilibrium will more AMP-protein be formed or will the complex dissociate

Consider the binding of AMP to phosphorylase b: AMP + Protein AMP-Protein at 25°C the equilibrium constant is 23809 M-1 and at 38 °C the equilibrium constant is 16949 M-1, if we assume that AH does not change with temperature, calculate AGº, AH° and ASº at 25°C. If the concentrations of all species are at 10 micromolar what is the value of AG at 25 C? As the reaction goes towards equilibrium will more AMP-protein be formed or will the complex dissociate

Biochemistry
Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
Section: Chapter Questions
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Calculation about delta standard G, delta H, detla S. Question attached as photo below. And my answer attempted. Need my answer verified and corrected if neccesary. Please let me know where I got wrong and what key ideas I had miss. Thanks.

n(23809) - In(16949) 8.314.Jmol!K". ( 25+298K 38+298K) 2.82555 = of1°rxn 0. = 2358§. $61 J/mol. AHim = 23.589 kJ/mol. @ equilibrium. at equilibrium AG=0 and therefore AG295 =-RTIN K whereK is the equilibrium constant of the reaction AG 248 : - (8.314JmolK?)(323 K)ln (23809 M) 27063.2. KJ/ml| AG TAS°. = aH° - oG? 23.589 kJ/mol - (-27063.2 KI/mal). 323 K 5° = 8 3.86 kJ/mwl UM= (0" mol/L Initial CJ.all C 1 M. T= 25°C goos baik to.veatant ,move. Odissociati on Ge? | trice for any côndition NjeT e equilbrium "Catculite G2ys ät iny oG.298 + RI m (K)J
Transcribed Image Text:n(23809) - In(16949) 8.314.Jmol!K". ( 25+298K 38+298K) 2.82555 = of1°rxn 0. = 2358§. $61 J/mol. AHim = 23.589 kJ/mol. @ equilibrium. at equilibrium AG=0 and therefore AG295 =-RTIN K whereK is the equilibrium constant of the reaction AG 248 : - (8.314JmolK?)(323 K)ln (23809 M) 27063.2. KJ/ml| AG TAS°. = aH° - oG? 23.589 kJ/mol - (-27063.2 KI/mal). 323 K 5° = 8 3.86 kJ/mwl UM= (0" mol/L Initial CJ.all C 1 M. T= 25°C goos baik to.veatant ,move. Odissociati on Ge? | trice for any côndition NjeT e equilbrium "Catculite G2ys ät iny oG.298 + RI m (K)J
Consider the binding of AMP to phosphorylase b: AMP + Protein AMP-Protein at 25°C the equilibrium constant is 23809 M-1 and at 38 °C the equilibrium constant is 16949 M-1, if we assume that AH does not change with temperature, calculate AG°, AH° and ASº at 25°C. If the concentrations of all species are at 10 micromolar what is the value of AG at 25 C? As the reaction goes towards equilibrium will more AMP-protein be formed or will the complex dissociate
Transcribed Image Text:Consider the binding of AMP to phosphorylase b: AMP + Protein AMP-Protein at 25°C the equilibrium constant is 23809 M-1 and at 38 °C the equilibrium constant is 16949 M-1, if we assume that AH does not change with temperature, calculate AG°, AH° and ASº at 25°C. If the concentrations of all species are at 10 micromolar what is the value of AG at 25 C? As the reaction goes towards equilibrium will more AMP-protein be formed or will the complex dissociate
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