EBK CHEMISTRY: AN ATOMS FIRST APPROACH
EBK CHEMISTRY: AN ATOMS FIRST APPROACH
2nd Edition
ISBN: 9780100552234
Author: ZUMDAHL
Publisher: YUZU
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Chapter 21, Problem 130AE

(a)

Interpretation Introduction

Interpretation: By using given bond energies the enthalpy ΔH for the reaction of two molecules of glycine to form a peptide linkage is to be estimated. The entropy ΔS to favor the formation of peptide linkages between two molecules of glycine is to be predicted. Also, the formation of proteins is a spontaneous process or not is to be predicted.

Concept introduction: When two or more amino acids are linked in a chain, a compound called peptide is formed in which the carboxyl group of each acid is joined to the amino group of the next forming a -OC-NH- bond, known as peptide bond. Glycine is an amino acid with a single hydrogen atom present in its side chain. The chemical formula of glycine is (C2H5NO2) . When two molecules of glycine joined through a peptide linkage a dipeptide is formed.

To determine: The ΔH for the reaction of two molecules of glycine to form a peptide linkage by using bond energies.

(a)

Expert Solution
Check Mark

Answer to Problem 130AE

Answer

The ΔH for the reaction is 1302 KJ_ . Hence, the reaction is an endothermic reaction.

Explanation of Solution

Explanation

The enthalpy ΔH for the reaction is 1302 KJ_ . Hence, the reaction is an endothermic reaction.

The reaction between two molecules of glycine to form peptide linkage is,

EBK CHEMISTRY: AN ATOMS FIRST APPROACH, Chapter 21, Problem 130AE

Figure 1

The enthalpy of the reaction is calculated by using the formula,

ΔHrxnο=D(bonds broken)-D(bonds formed)

In the given reaction shown in Figure 1 bonds are broken and formed during peptide formation and energy is absorbed and released. The bond energies are shown in (Table 3-3) respectively.

Therefore, the energy required to break the bonds is calculated as,

2 moles of 2N-H bonds = 4×391 KJ = 1564 KJ2 moles of C-N bonds = 2×305 KJ = 610 KJ2 moles of 2C-H bonds = 4×413 KJ = 1652 KJ2 moles of C-C bonds = 2×347 KJ = 694 KJ

2 moles of C=O bonds = 2×745 KJ = 1490 KJ2 moles of C-O bonds = 2×358 KJ = 716 KJ2 moles of O-H bonds = 2×467 KJ = 934 KJTotal energy required to break the bonds = 7660 KJ

Similarly, the energy released when the bonds are formed are calculated as,

3 N-H bonds = 2×392 KJ = 782 KJ3 C-N bonds = 3×305 KJ = 915 KJ4 C-H bonds = 4×413 KJ = 1652 KJ2 C-C bonds = 2×347 KJ = 694 KJ

2 C=O bonds = 2×745 KJ = 1490 KJ1 C-O bond = 1×358 KJ = 358 KJ1 O-H bond = 1×467 KJ = 467 KJTotal energy released to form the bonds = 6358 KJ

Therefore, the total enthalpy for the reaction is calculated by using the formula,

ΔHrxnο=D(bonds broken)-D(bonds formed)= 7660 KJ - 6358 KJ1302 KJ_

Hence, the total enthalpy of formation for the reaction of two molecules of glycine to form a peptide linkage is positive. It is an endothermic reaction.

(b)

Interpretation Introduction

Interpretation: By using given bond energies the enthalpy ΔH for the reaction of two molecules of glycine to form a peptide linkage is to be estimated. The entropy ΔS to favor the formation of peptide linkages between two molecules of glycine is to be predicted. Also, the formation of proteins is a spontaneous process or not is to be predicted.

Concept introduction: When two or more amino acids are linked in a chain, a compound called peptide is formed in which the carboxyl group of each acid is joined to the amino group of the next forming a -OC-NH- bond, known as peptide bond. Glycine is an amino acid with a single hydrogen atom present in its side chain. The chemical formula of glycine is (C2H5NO2) . When two molecules of glycine joined through a peptide linkage a dipeptide is formed.

To determine: The entropy ΔS to favor the formation of peptide linkages between two molecules of glycine.

(b)

Expert Solution
Check Mark

Answer to Problem 130AE

Answer

The entropy ΔS for the reaction is negative. Hence, the reaction is non-spontaneous.

Explanation of Solution

Explanation

The entropy ΔS to favor the formation of peptide linkages between two molecules of glycine is negative.

The reaction between two molecules of glycine to form a peptide linkage is found to be an endothermic process because of the positive value of enthalpy. In the system an endothermic process absorbs heat from the surrounding and therefore, decreases the entropy of the surrounding because the molecular motion in the surrounding decreases. Therefore, for the given reaction ΔS becomes negative and the reaction is non-spontaneous with respect to the entropy ΔS .

(c)

Interpretation Introduction

Interpretation: By using given bond energies the enthalpy ΔH for the reaction of two molecules of glycine to form a peptide linkage is to be estimated. The entropy ΔS to favor the formation of peptide linkages between two molecules of glycine is to be predicted. Also, the formation of proteins is a spontaneous process or not is to be predicted.

Concept introduction: When two or more amino acids are linked in a chain, a compound called peptide is formed in which the carboxyl group of each acid is joined to the amino group of the next forming a -OC-NH- bond, known as peptide bond. Glycine is an amino acid with a single hydrogen atom present in its side chain. The chemical formula of glycine is (C2H5NO2) . When two molecules of glycine joined through a peptide linkage a dipeptide is formed.

To determine: The formation of proteins is a spontaneous process or not.

(c)

Expert Solution
Check Mark

Answer to Problem 130AE

Answer

Formation of proteins is a non-spontaneous process.

Explanation of Solution

Explanation

The formation of proteins is a non-spontaneous process.

The formation of proteins occurs when large numbers of amino acids are joined through peptide linkage. The formation of peptide linkage is a non-spontaneous process as stated above due to the negative value of ΔS . Hence, formation of proteins is a non-spontaneous process.

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Chapter 21 Solutions

EBK CHEMISTRY: AN ATOMS FIRST APPROACH

Ch. 21 - Prob. 11RQCh. 21 - Prob. 12RQCh. 21 - Prob. 1QCh. 21 - Prob. 2QCh. 21 - What is wrong with the following names? Give the...Ch. 21 - Prob. 4QCh. 21 - Prob. 5QCh. 21 - Prob. 6QCh. 21 - Prob. 7QCh. 21 - Prob. 8QCh. 21 - Prob. 9QCh. 21 - Prob. 10QCh. 21 - Prob. 11QCh. 21 - Prob. 12QCh. 21 - Prob. 13ECh. 21 - Prob. 14ECh. 21 - Draw all the structural isomers for C8H18 that...Ch. 21 - Draw all the structural isomers for C8H18 that...Ch. 21 - Prob. 17ECh. 21 - Prob. 18ECh. 21 - Draw the structural formula for each of the...Ch. 21 - Prob. 20ECh. 21 - Prob. 21ECh. 21 - Prob. 22ECh. 21 - Prob. 23ECh. 21 - Prob. 24ECh. 21 - Name each of the following alkenes. a. CH2 = CH ...Ch. 21 - Name each of the following alkenes or alkynes. a....Ch. 21 - Prob. 27ECh. 21 - Prob. 28ECh. 21 - Prob. 29ECh. 21 - Prob. 30ECh. 21 - Name each of the following. a. b. CH3CH2CH2CCl3 c....Ch. 21 - Prob. 32ECh. 21 - There is only one compound that is named...Ch. 21 - Prob. 34ECh. 21 - Prob. 35ECh. 21 - Prob. 36ECh. 21 - Prob. 37ECh. 21 - Prob. 38ECh. 21 - Prob. 39ECh. 21 - Prob. 40ECh. 21 - Draw all structural and geometrical (cistrans)...Ch. 21 - Prob. 42ECh. 21 - Prob. 43ECh. 21 - Prob. 44ECh. 21 - If one hydrogen in a hydrocarbon is replaced by a...Ch. 21 - There are three isomers of dichlorobenzene, one of...Ch. 21 - Prob. 47ECh. 21 - Prob. 48ECh. 21 - Prob. 49ECh. 21 - Minoxidil (C9H15N5O) is a compound produced by...Ch. 21 - Prob. 51ECh. 21 - Prob. 52ECh. 21 - Name all the alcohols that have the formula...Ch. 21 - Prob. 54ECh. 21 - Prob. 55ECh. 21 - Prob. 56ECh. 21 - Prob. 57ECh. 21 - Prob. 58ECh. 21 - Prob. 59ECh. 21 - Prob. 60ECh. 21 - Prob. 61ECh. 21 - Prob. 62ECh. 21 - Prob. 63ECh. 21 - Prob. 64ECh. 21 - Prob. 65ECh. 21 - Prob. 66ECh. 21 - Prob. 67ECh. 21 - Prob. 68ECh. 21 - Prob. 69ECh. 21 - Complete the following reactions. a. CH3CO2H +...Ch. 21 - Prob. 71ECh. 21 - Prob. 72ECh. 21 - Prob. 73ECh. 21 - Prob. 74ECh. 21 - Prob. 75ECh. 21 - The polyester formed from lactic acid, is used for...Ch. 21 - Prob. 77ECh. 21 - Prob. 78ECh. 21 - Prob. 79ECh. 21 - Prob. 80ECh. 21 - Prob. 81ECh. 21 - Prob. 82ECh. 21 - Prob. 83ECh. 21 - Prob. 84ECh. 21 - Prob. 85ECh. 21 - Prob. 86ECh. 21 - Prob. 87ECh. 21 - Prob. 88ECh. 21 - Prob. 89ECh. 21 - Prob. 90ECh. 21 - Prob. 91ECh. 21 - Prob. 92ECh. 21 - Prob. 93ECh. 21 - Prob. 94ECh. 21 - Prob. 95ECh. 21 - Prob. 96ECh. 21 - Prob. 97ECh. 21 - Prob. 98ECh. 21 - Prob. 99ECh. 21 - Prob. 100ECh. 21 - Prob. 101ECh. 21 - Prob. 102ECh. 21 - Prob. 103ECh. 21 - Prob. 104ECh. 21 - Prob. 105ECh. 21 - Prob. 106ECh. 21 - Which base will hydrogen-bond with uracil within...Ch. 21 - Prob. 108ECh. 21 - The base sequences in mRNA that code for certain...Ch. 21 - Prob. 110ECh. 21 - Prob. 111AECh. 21 - Prob. 112AECh. 21 - Prob. 113AECh. 21 - Prob. 114AECh. 21 - Prob. 115AECh. 21 - Prob. 116AECh. 21 - Prob. 117AECh. 21 - Prob. 118AECh. 21 - Prob. 119AECh. 21 - Prob. 120AECh. 21 - Prob. 121AECh. 21 - Prob. 122AECh. 21 - Prob. 123AECh. 21 - Prob. 124AECh. 21 - Prob. 125AECh. 21 - Prob. 126AECh. 21 - Prob. 127AECh. 21 - Prob. 128AECh. 21 - Prob. 129AECh. 21 - Prob. 130AECh. 21 - Prob. 131AECh. 21 - Prob. 132AECh. 21 - Prob. 133AECh. 21 - Prob. 134AECh. 21 - When heat is added to proteins, the hydrogen...Ch. 21 - Prob. 136AECh. 21 - Prob. 137CWPCh. 21 - Prob. 138CWPCh. 21 - Prob. 139CWPCh. 21 - Name each of the following alkenes and alkynes. a....Ch. 21 - a. Name each of the following alcohols. b. Name...Ch. 21 - Prob. 142CWPCh. 21 - Prob. 143CWPCh. 21 - Prob. 144CWPCh. 21 - Prob. 145CPCh. 21 - Prob. 146CPCh. 21 - Prob. 147CPCh. 21 - Prob. 148CPCh. 21 - Prob. 149CPCh. 21 - Prob. 150CPCh. 21 - Prob. 151CPCh. 21 - Prob. 152CPCh. 21 - Prob. 153CPCh. 21 - Prob. 154CPCh. 21 - Stretch a rubber band while holding it gently to...Ch. 21 - Alcohols are very useful starting materials for...Ch. 21 - Prob. 157CPCh. 21 - Prob. 158CPCh. 21 - Prob. 159IPCh. 21 - Prob. 160IPCh. 21 - Prob. 161MPCh. 21 - Prob. 162MP
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