Chapter 21, Problem 148AE

(a)

Interpretation Introduction

Interpretation:

The value of $\Delta H$ for the reaction of two molecules of lysine which will form peptide linkage is to be estimated by using the bond energies.

Concept Introduction :

The change in enthalpy $\Delta H$ can be defined as the difference between the energy used to break the bonds and the energy gained to form the bonds. This can be represented as −

  $\Delta H={\displaystyle \sum D(bondsbroken)-{\displaystyle \sum D(bondsformed)}}$

Answer to Problem 148AE

The value of $\Delta H$ for the reaction is $1302kJ$ and the reaction is endothermic process.

Explanation of Solution

The lysine molecules with peptide linkage will be represented as −

  

The value of $\Delta H$ for the reaction will be calculated as −

  $\Delta H{°}_{r\times n}={\displaystyle \sum D(bondsbroken)-{\displaystyle \sum D(bondsformed)}}$

The values for the bonds which are broken −

Two moles of $2N-H$ bonds = $4\times 391kJ=1564kJ$

Two moles of $C-N$ bonds = $2\times 305kJ=610kJ$

Two moles of $2C-H$ bonds = $4\times 413kJ=1652kJ$

Two moles of $C-C$ bonds = $2\times 347kJ=694kJ$

Two moles of $C=O$ bonds = $2\times 745kJ=1490kJ$

Two moles of $C-O$ bonds = $2\times 358kJ=716kJ$

Two moles of $O-H$ bonds = $2\times 467kJ=934kJ$

Now, the summation of all the above energy will result in the total energy required to break the bonds = $7660kJ$ .

Now, the total energy released when bonds are formed −

  $2N-H$ Bonds = $2\times 391kJ=782kJ$

  $3C-N$ Bonds = $3\times 305kJ=915kJ$

  $4C-H$ Bonds = $4\times 413kJ=1652kJ$

  $2C-C$ Bonds = $2\times 347kJ=694kJ$

  $2C=O$ Bonds = $2\times 745kJ=1490kJ$

  $1C-O$ Bonds = $1\times 358kJ=358kJ$

  $1O-H$ Bonds = $1\times 467kJ=467kJ$

Now, the summation of all the above energy will result in the total energy released during the formation of bond = $6358kJ$ .

  $\Delta H{°}_{r\times n}=7660kJ-6358kJ=1302kJ$

The value of the energy is positive. Hence, the given reaction is endothermic reaction.

(b)

Interpretation Introduction

Interpretation:

The value of $\Delta S$ which will favor the formation of peptide linkages between two molecules of lysine is to be predicted.

Concept Introduction :

The change in enthalpy $\Delta H$ can be defined as the difference between the energy used to break the bonds and the energy gained to form the bonds. This can be represented as −

  $\Delta H={\displaystyle \sum D(bondsbroken)-{\displaystyle \sum D(bondsformed)}}$

Answer to Problem 148AE

The value of $\Delta S$ is negative and will non-spontaneous process.

Explanation of Solution

The endothermic process can be defined as the process in which heat is absorbed from the surroundings. This process causes the decrement in the entropy of surroundings and it occurs because the molecular motion is also decreased.

Hence, the value of $\Delta S_{suff}$ of the reaction will be negative and the reaction will also be non-spontaneous with respect to $\Delta S$ .

(c)

Interpretation Introduction

Interpretation:

The formation of protein is a spontaneous process or non-spontaneous process is to be predicted.

Concept Introduction :

The change in Gibbs free energy is represented as follows:

  $\Delta G=\Delta H-T\Delta S$

Here, $\Delta H$ is change in enthalpy, T is temperature and $\Delta S$ is change in entropy.

Answer to Problem 148AE

The synthesis of protein is non-spontaneous process.

Explanation of Solution

The basic equation of the protein formation is given as −

  $naminoacids\to polypeptide+\left(n-1\right)H_{2}O$

Now, relation between change in Gibbs free energy, change in enthalpy and entropy is given below:

  $\Delta G=\Delta H-T\Delta S$

Here, if change in entropy is negative and change in enthalpy is positive thus, change in Gibbs free energy will be positive at all the temperature values.

For positive Gibbs free energy value, the reaction is non-spontaneous.