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

Answer:

In equilibrium:

$[acid]=[amine]=5.5*10^{-3}$

tex][Pep]=0.989[/tex]

Explanation:

Hi,

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

$[Pep]= \frac{1mol}{1L}=1M$

For the given reaction the equilibrium constant will be:

$K=\frac{[acid]*[amine]}{[Pep]}$

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:

$[Pep]_{eq}=[Pep]-2*[acid]$

$K=\frac{[acid]^2}{1-2[acid]}$

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

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

$0=[acid]^2+6.2*10^{-5}[acid]-3.1*10^{-5}$

Solving:

$[acid]=[amine]=5.5*10^{-3}$

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