   Chapter 13, Problem 88AE

Chapter
Section
Textbook Problem

# Peptide decomposition is one of the key processes of digestion, where a peptide bond is broken into an acid group and an amine group. We can describe this reaction as follows: Peptide ( a q ) + H 2 O ( l ) ⇌ acid   group ( a q ) + amine     g r o u p ( a q ) If we place 1.0 mole of peptide into 1.0 L water, what will be the equilibrium concentrations of all species in this reaction? Assume the K value for this reaction is 3 .1 × 10−5.

Interpretation Introduction

Interpretation: The reaction involving the decomposition of a peptide bond into an acid group and an amine group is given. The equilibrium concentrations of all the species involved in the reaction when 1.0mole of peptide is placed in 1.0L of water are to be calculated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction.

When the equilibrium constant is expressed in terms of concentration, it is represented as K .

To determine: The equilibrium concentrations of all the species involved in the given reaction.

Explanation

Explanation

Given

The stated reaction is,

Peptide(aq)+H2O(l)acidgroup(aq)+aminegroup(aq)

The initial number of moles of peptide is 1.0mole.

The value of the equilibrium constant for the reaction is 3.1×105.

At equilibrium, the equilibrium ratio is expressed by the formula,

K=ConcentrationofproductsConcentrationofreactants

Where,

• K is the equilibrium constant.

The equilibrium constant expression for the given reaction is,

K=[acidgroup][aminegroup][Peptide] (1)

The amount of the peptide that reacted till the equilibrium stage is 0.0056M_.

The initial concentration of the given peptide is 1.0mol/L that is 1.0M.

The amount of the peptide that reacted till the equilibrium stage is assumed to be x .

The equilibrium concentrations are represented as,

Peptide(aq)+H2O(l)acidgroup(aq)+aminegroup(aq)Initialconcentration1.000Change-x+x+xEquilibriumconcentration1.0-xxx

H2O is present in the liquid state and hence it does not affect the value of the equilibrium constant.

According to the ICE table formed,

The equilibrium concentration of the peptide is (1.0x)M.

The equilibrium concentration of the acid group is xM.

The equilibrium concentration of the amine group is xM.

Substitute the equilibrium concentration values of the peptide, the acid group and the amine group in equation (1).

The equilibrium constant expression for the given reaction is,

K=[acidgroup][aminegroup][Peptide]=[x][x][1.0x]=[x]2[1.0x]

The value of the equilibrium constant (K) for the reaction is given to be 3

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