ChemistryThe equilibrium constant (at 298 K) for the mono-phosphorylation of glucoseusing ATP is 860 (Glucose + ATP → Glucose-6-Phosphate + ADP). For thehydrolysis of Glucose-6-Phosphate (Glucose-6-Phosphate + H₂O → Glucose +Pi), it is 260. What is the standard free energy in cal/mol for the hydrolysis ofATP at 298 K (ATP + H₂O → ADP +P;)? Show your work.a.~-3300b. 3800C.- -4000d. ~ -7300e. ~-12300explain why?

To Calculate standard free energy we would use formula , formula used is : ∆G° = - RTlnK So ,first…

The equilibrium constant (at 298 K) for the mono-phosphorylation of glucose using ATP is 860 (Glucose + ATP → Glucose-6-Phosphate + ADP). For the hydrolysis of (Glucose-6-Phosphate + H₂2O → Glucose + P₁), it is 260. What is the standard free energy in cal/mol for the hydrolysis of ATP at 298 K (ATP + H₂O → ADP + Pi)? Show your work. Glucose-6-Phosphate a.~-3300 b. - 3800 C. ~ -4000 d. -7300 N e. ~-12300

The equilibrium constant (at 298 K) for the mono-phosphorylation of glucose using ATP is 860 (Glucose + ATP → Glucose-6-Phosphate + ADP). For the hydrolysis of (Glucose-6-Phosphate + H₂2O → Glucose + P₁), it is 260. What is the standard free energy in cal/mol for the hydrolysis of ATP at 298 K (ATP + H₂O → ADP + Pi)? Show your work. Glucose-6-Phosphate a.~-3300 b. - 3800 C. ~ -4000 d. -7300 N e. ~-12300

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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