PHYSICAL CHEMISTRY. VOL.1+2 (LL)(11TH)
PHYSICAL CHEMISTRY. VOL.1+2 (LL)(11TH)
11th Edition
ISBN: 9780198826910
Author: ATKINS
Publisher: Oxford University Press
Question
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Chapter 4, Problem 4.3IA

(a)

Interpretation Introduction

Interpretation: The form of the equation for Gibbs free energy of unfolding for the given helical polypeptide has to be justified.

Concept introduction: The Gibbs free energy is a thermodynamic quantity which allows the maximum amount of reversible work that can be performed.  It is given by the product of the temperature and entropy subtracted from the enthalpy of the system.

(a)

Expert Solution
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Answer to Problem 4.3IA

The form of the equation for Gibbs free energy of unfolding for the given helical polypeptide is justified below.

Explanation of Solution

The Gibbs free energy is calculated by the formula,

  ΔG=ΔHTΔS                                                                                             (1)

Where,

  • ΔG is the Gibbs free energy.
  • ΔH is the molar enthalpy.
  • ΔS is the molar entropy.
  • T is the temperature.

The polypeptide has N amino acid residues and N4 hydrogen bonds are formed to form an α helix.

Let ΔhbH be the molar dissociation enthalpy of 1 hydrogen bond.

Thus, the molar dissociation enthalpy of N4 hydrogen bonds is (N4)ΔhbH.

Since, N2 residues in the polypeptide have restricted motion.

Thus, the molar entropy for N2 residues is (N2)ΔhbS.

Substitute the value of molar dissociation enthalpy and molar entropy in equation (1).

    ΔunfoldG=(N4)ΔhbH(N2)TΔhbS

(b)

Interpretation Introduction

Interpretation: The melting temperature Tm may be written as Tm=(N4)ΔhbH(N2)ΔhbS is to be shown.

Concept introduction: The Gibbs free energy at phase transition is zero.  Hence, the value of enthalpy becomes equal to the product of the temperature and molar entropy.

(b)

Expert Solution
Check Mark

Answer to Problem 4.3IA

The melting temperature Tm may be written as Tm=(N4)ΔhbH(N2)ΔhbS.

Explanation of Solution

The Gibbs free energy for the unfolding of the given protein is given as,

    ΔunfoldG=(N4)ΔhbH(N2)TΔhbS                                                        (2)

Where,

  • ΔunfoldG is the Gibbs free energy of unfolding.
  • ΔhbH is the dissociation enthalpy of hydrogen bonds present.
  • ΔhbS is the molar entropy of the hydrogen bonds.
  • Tm is the melting temperature of the polypeptide at which it undergoes phase. transition.

At phase transition, Gibbs free energy for the unfolding (ΔunfoldG) is 0.

Substitute the value of ΔunfoldG in equation (2).

    0=(N4)ΔhbH(N2)TmΔhbS

Hence,

  (N4)ΔhbH=(N2)TmΔhbSTm=(N4)ΔhbH(N2)TΔhbS

Hence, the melting temperature Tm may be written as Tm=(N4)ΔhbH(N2)ΔhbS is justified.

(c)

Interpretation Introduction

Interpretation: The graph of Tm/(ΔhbHm/ΔhbSm) has to be plotted for 5N20.  The value of N for which Tm changes by less than 1% when N increases by 1 has to be determined.

Concept introduction: The Gibbs free energy is a thermodynamic quantity which allows the maximum amount of reversible work that can be performed and is given by the product of the temperature and entropy subtracted from the enthalpy of the system.

(c)

Expert Solution
Check Mark

Answer to Problem 4.3IA

The graphs of Tm/(ΔhbHm/ΔhbSm) for 5N20 is plotted below.  The value of N for which Tm changes by less than 1% when N increases by 1 is greater than 16_

Explanation of Solution

The melting temperature is given by the equation,

  Tm=(N4)ΔhbHm(N2)ΔhbSm                                                                                       (3)

Equation (3) follows the equation of a line, y=mx+c.

Where

  • Tm lies on the x axis.
  • (ΔhbHmΔhbSm) lies on the y axis.

Thus, the graph will be linear.

Slope of equation (3) is,

  m=(N4)(N2)

For N=5,

    m=(54)(52)=0.33

For N=20,

    m=(204)(202)=0.88

Hence the graph is represented as

PHYSICAL CHEMISTRY. VOL.1+2 (LL)(11TH), Chapter 4, Problem 4.3IA

The value of Tm changes by less than 1% when N increases by 1.

  Tm,fTm,i<0.01

Let,

  Tm,f<1.01Tm,i

Or,

  Tm,fTm,i<1.01                                                                                                      (4)

Therefore,

    Tm,fTm,i<0.01

From equation (3),

  Tm,i=(N4)ΔhbHm(N2)ΔhbSm                                                                                   (5)

When Tm,i changes by less than 1%, N increases by 1.

Hence,

  Tm,f=(N4+1)ΔhbH(N2+1)ΔhbS=(N3)ΔhbH(N1)ΔhbS                                                                                  (6)

Divide equation (6) by (5).

  Tm,fTm,i=(N3)ΔhbH(N1)ΔhbS(N4)ΔhbHm(N2)ΔhbSm

Substitute equation (4) in above equation.

    . (N3)ΔhbHm(N1)ΔhbSm×(N2)ΔhbHm(N4)ΔhbSm<1.01(N3)×(N2)<1.01×(N1)(N4)0.01N20.05N1.96>0

Hence,

  N>16_

Hence, the value of N for which Tm changes by less than 1% when N increases by 1 is >16_.

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