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

Peptide(aq) + H20(l) -----------------------> acid group(aq) + amine group(aq)

1.0 M                                                                 0                           0

1.0 - x                                                                x                            x

K = [acid][amine] / [peptide]

3.1 x 10^-5 = x^2 / 1 - x

x = 0.0056

[peptide] = 1 -x = 1 - 0.0056 = 0.99 M

[acid] = [amine] = x = 0.0056 M