The standard free energy change(∆Go) per mole for the reaction A↔ B at 30oC in an open system is -1000 cal/mole. What is the approximate free energy change (∆G) when the concentration of A and B are 100 micromolar and 100 millimolar, respectively?
Answers
Answer:
ATP (adenosine triphosphate) is a key reaction for metabolism. Tools from systems biology require standard reaction data in order to predict metabolic pathways accurately. However, literature values for standard Gibbs energy of ATP hydrolysis are highly uncertain and differ strongly from each other. Further, such data usually neglect the activity coefficients of reacting agents, and published data like this is apparent (condition-dependent) data instead of activity-based standard data. In this work a consistent value for the standard Gibbs energy of ATP hydrolysis was determined. The activity coefficients of reacting agents were modeled with electrolyte Perturbed-Chain Statistical Associating Fluid Theory (ePC-SAFT). The Gibbs energy of ATP hydrolysis was calculated by combining the standard Gibbs energies of hexokinase reaction and of glucose-6-phosphate hydrolysis. While the standard Gibbs energy of hexokinase reaction was taken from previous work, standard Gibbs energy of glucose-6-phosphate hydrolysis reaction was determined in this work. For this purpose, reaction equilibrium molalities of reacting agents were measured at pH7 and pH8 at 298.15K at varying initial reacting agent molalities. The corresponding activity coefficients at experimental equilibrium molalities were predicted with ePC-SAFT yielding the Gibbs energy of glucose-6-phosphate hydrolysis of -13.72±0.75kJ·mol -1 . Combined with the value for hexokinase, the standard Gibbs energy of ATP hydrolysis was finally found to be -31.55±1.27kJ·mol -1 . For both, ATP hydrolysis and glucose-6-phosphate hydrolysis, a good agreement with own and literature values were obtained when influences of pH, temperature, and activity coefficients were explicitly taken into account in order to calculate standard Gibbs energy at pH7, 298.15K and standard state. Copyright © 2017 Elsevier B.V. All rights reserved.
Answer:
∆G = 3.16 kCalorie/mole
Explanation:
Use this formula:
Delta G = Delta G° + 2.303RTlogKeq.
Delta G° = -1000
R = 1.987
T = 30 °C = 30 +273 = 303 K
Keq. = [B]/[A]
= 100 mM / 100 uM
= 100 × 10^-3 / 100 ×10^-6
= 10^-3 / 10^-6
= 10^3
Delta G = Delta G° + 2.303RTlogKeq.
= -1000 + 2.303×1.987×303log10^3
= -1000 + 1386.54648[3log10]
= -1000 + 1386.54648[3×1]
= -1000 + 1386.54648[3]
= -1000 + 4,159.63944
= 3,159.63944
= 3160 Calorie/mole
= 3.16 KCalorie/mole.
~PURUSHOTAM SINGH THAKUR