Chapter 6, Problem 6.8NP

(a)

Interpretation Introduction

Interpretation:

The Gibbs energy change for the denaturation of the protein at $\text{pH = 2 and T = 310 K}$ should be calculated.

Concept Introduction :

The Gibbs free energy change can be calculated as follows:

  $\Delta G^0=\Delta H^0-T.\Delta S^0$

Here,

  $\begin{array}{l}\Delta G^0\text{ - Gibbs free energy change}\\ \Delta H^0\text{ - enthalpy change}\\ T\text{ - temperature}\\ \Delta S^0\text{- entropy change}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant for denaturation of the protein at $\text{pH = 2 and T = 310 K}$ should be calculated.

Concept Introduction :

The Gibbs free energy change is related to equilibrium constant as follows:

  $\Delta G=-RT\ln K_{eq}$

Here,

  $\begin{array}{l}\Delta G\text{ - Gibbs free energy change}\\ R\text{ - universal gas constant}\\ T\text{ - temperature}\\ K_{eq}\text{ - equilibrium constant}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Whether the protein is structurally stable at $\text{pH = 2 and T = 310 K}$ should be calculated.

Concept Introduction :

If the equilibrium constant for the forward reaction is greater than 1, forward reaction is favorable.