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 \((a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) acid group \((a q)+\) amine group \((a q)\) If we place \(1.0\) mol peptide into \(1.0 \mathrm{~L}\) water, what will be the equilibrium concentrations of all species in this reaction? Assume the \(K\) value for this reaction is \(3.1 \times 10^{-5}\).

Short answer

The equilibrium concentrations for the peptide decomposition reaction are approximately as follows: Peptide: 0.994 mol/L, H₂O: 0.994 mol/L, Acid group: 5.57 × 10⁻³ mol/L, and Amine group: 5.57 × 10⁻³ mol/L.

Step-by-step solution

Write the equilibrium expression

using the given equilibrium constant, Kc, and the concentrations of all species in the reaction: \[ K_c = \frac{[\text{acid group}][\text{amine group}]}{[\text{peptide}][\text{H}_{2}\text{O}]}\]

Identify the initial concentrations

of the reactants and products which are: - Peptide: 1.0 mol/L - H₂O: 1.0 mol/L - Acid group: 0 mol/L - Amine group: 0 mol/L

Set up the ICE table

for the given reaction: ⠀⠀⠀⠀⠀⠀Peptide⠀⠀⠀⠀⠀⠀H₂O⠀⠀⠀⠀⠀⠀Acid Group⠀⠀⠀⠀⠀⠀Amine Group Initial⠀⠀1.0 mol/L⠀⠀⠀⠀1.0 mol/L⠀⠀⠀⠀⠀0 mol/L⠀⠀⠀⠀⠀⠀0 mol/L Change ⠀⠀ -x⠀⠀⠀⠀⠀⠀⠀ -x⠀⠀⠀⠀⠀⠀⠀+x⠀⠀⠀⠀⠀⠀⠀ +x Final ⠀⠀ 1.0-x mol/L⠀⠀⠀⠀1.0-x mol/L⠀⠀⠀⠀x mol/L⠀⠀⠀⠀⠀⠀x mol/L where x represents the concentration change at equilibrium.

Insert the equilibrium concentrations into Kc expression

: \[ 3.1 \times 10^{-5} = \frac{x^2}{(1-x)^2}\]

Solve for x

Since the Kc value is very small, we can assume that x will be much smaller than 1. Therefore, we can approximate \(1-x\) as 1, and the equation simplifies to: \[ x =\sqrt{3.1 \times 10^{-5}}\] Calculate the x value: \[ x \approx 5.57 × 10^{-3}\]

Find the equilibrium concentrations

Using the calculated x value, find the equilibrium concentrations for all species: - Peptide: 1.0 mol/L - x = 1.0 mol/L - 5.57 × 10⁻³ mol/L ≈ 0.994 mol/L - H₂O: 1.0 mol/L - x = 1.0 mol/L - 5.57 × 10⁻³ mol/L ≈ 0.994 mol/L - Acid group: x mol/L = 5.57 × 10⁻³ mol/L - Amine group: x mol/L = 5.57 × 10⁻³ mol/L The equilibrium concentrations are as follows: Peptide: ≈ 0.994 mol/L H₂O: ≈ 0.994 mol/L Acid group: ≈ 5.57 × 10⁻³ mol/L Amine group: ≈ 5.57 × 10⁻³ mol/L

Key concepts

Peptide Decomposition

Peptide decomposition is a chemical reaction that plays a crucial role in the digestion process. It occurs when a peptide bond, which connects amino acids in a peptide chain, is broken down by water (a process known as hydrolysis). This reaction results in the formation of an acid group and an amine group. Understanding how these organic compounds interact and decompose is important, as it mirrors processes occurring naturally in the human body to aid digestion.
In the given exercise, we examine the decomposition of a peptide in an aqueous environment, assessing how equilibrium is established in this reaction. This involves determining the concentrations of the peptide and its decomposition products—the acid and amine groups—at equilibrium.

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a fundamental principle in chemistry that provides insights into the ratio of products to reactants at equilibrium. The value of \( K_c \) reflects how far a reaction proceeds before reaching equilibrium. For a detailed reaction like peptide decomposition, the equilibrium constant helps quantify the balance between intact peptides and the decomposed acid and amine groups.
In this example, the equilibrium constant for peptide decomposition is given as \( 3.1 \times 10^{-5} \). This small value indicates the reaction favors the formation of reactants rather than products at equilibrium. Knowing \( K_c \) allows us to predict and calculate the equilibrium concentrations of all species involved.

ICE Table

An ICE table is a useful tool employed to track the changes in concentrations during a reaction until equilibrium is reached. ICE stands for Initial, Change, and Equilibrium—each representing a phase in the process of reaching equilibrium in a chemical reaction.
Let's break it down:

• Initial: Lists the starting concentrations of all reactants and products. In our case, peptide and water are 1.0 mol/L, while both the acid and amine groups are initially 0 mol/L.
• Change: Represents the change in concentrations, denoted by \( x \), as the reaction progresses. For every peptide molecule decomposed, the acid and amine group concentrations increase by \( x \).
• Equilibrium: Displays the concentrations at equilibrium, incorporating initial concentrations minus the change for reactants and plus the change for products.

Using an ICE table simplifies the process of solving equilibrium problems by organizing data and making it easier to apply the equilibrium constant expression.

Concentration Calculation

For this reaction, calculating the equilibrium concentrations requires solving the equilibrium constant expression using the information from the ICE table. By inserting our equilibrium values into the equation for \( K_c \), we obtain a mathematical representation:
\[ K_c = \frac{[\text{acid group}][\text{amine group}]}{[\text{peptide}][\text{H}_2\text{O}]} = \frac{x^2}{(1-x)^2} = 3.1 \times 10^{-5} \]
Given \( K_c \)'s small value, the assumption \( 1-x \approx 1 \) simplifies our calculation. By solving for \( x \), we find \( x \approx 5.57 \times 10^{-3} \).
The equilibrium concentrations are determined from this \( x \) value:

• Peptide: \( 1.0 - x \approx 0.994 \) mol/L
• H₂O: \( 1.0 - x \approx 0.994 \) mol/L
• Acid group: \( x \approx 5.57 \times 10^{-3} \) mol/L
• Amine group: \( x \approx 5.57 \times 10^{-3} \) mol/L

This process helps predict the stable state concentrations and balance of a system undergoing a reversible reaction.