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.

In equilibrium:

tex][Pep]=0.989[/tex]

Explanation:

Hi,

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

For the given reaction the equilibrium constant will be:

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:

and by mass balance:

Solving:

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

i'm assuming that the correct answer is indeed the first one/ "a"..