Δ G 0 and Δ S 0 at 20 0 C and 30 0 C and the reason for assuming Δ H 0 is independent of temperature should be discussed. Introduction: Δ G 0 = − R T ln K e q Δ G 0 = Δ H 0 − T . Δ S 0 Where K e q is equilibrium constant, Δ G 0 is change in Gibbs free energy at standard conditions, R is Gas constant, T is temperature, Δ H 0 is change in enthalpy at standard conditions, Δ S 0 is change in entropy at standard conditions.

BuyFind

Biochemistry

6th Edition
Reginald H. Garrett + 1 other
Publisher: Cengage Learning
ISBN: 9781305577206
BuyFind

Biochemistry

6th Edition
Reginald H. Garrett + 1 other
Publisher: Cengage Learning
ISBN: 9781305577206

Solutions

Chapter 3, Problem 2P
Interpretation Introduction

To calculate:

ΔG0and ΔS0at 200Cand 300Cand the reason for assuming ΔH0is independent of temperature should be discussed.

Introduction:

  ΔG0=RTlnKeq

  ΔG0=ΔH0T.ΔS0

Where Keqis equilibrium constant, ΔG0is change in Gibbs free energy at standard conditions, R is Gas constant, T is temperature, ΔH0is change in enthalpy at standard conditions, ΔS0is change in entropy at standard conditions.

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