Question

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(aq) + H20(l) = acid group(aq) + amine group(aq)

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 3.1x10-5.

Answer 1

In equilibrium:

`[acid]=[amine]=5.5*10^(-3)`

tex][Pep]=0.989[/tex]

Step-by-step explanation:

Hi,

Placing 1 mol of peptide into 1 L of water gives you a concentration of:

`[Pep]= (1mol)/(1L)=1M`

For the given reaction the equilibrium constant will be:

`K=([acid]*[amine])/([Pep])`

Assuming that the initial concentrations of the acid and amine groups are 0 (pure water) and given that the production rate between this two products is 1:1, we can say that:

`[acid]=[amine]`

and by mass balance:

`[Pep]_(eq)=[Pep]-2*[acid]`

`K=([acid]^2)/(1-2[acid])`

`K*(1-2[acid])=[acid]^2`

`0=[acid]^2+2K[acid]-K`

`0=[acid]^2+6.2*10^(-5)[acid]-3.1*10^(-5)`

Solving:

`[acid]=[amine]=5.5*10^(-3)`

tex][Pep]_{eq}=1-2*5.5*10^{-3}=0.989[/tex]