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Chapter 13: Problem 76

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

Expert verified
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

01

Write the equilibrium expression

using the given equilibrium constant, Kc, and the concentrations of all species in the reaction: \[ K_c = \frac{[\text{acid group}][\text{amine group}]}{[\text{peptide}][\text{H}_{2}\text{O}]}\]
02

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
03

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.
04

Insert the equilibrium concentrations into Kc expression

: \[ 3.1 \times 10^{-5} = \frac{x^2}{(1-x)^2}\]
05

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: \[ x \approx 5.57 × 10^{-3}\]
06

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

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Most popular questions from this chapter

As we have attempted to lessen our dependence on fossil fuels, the demand for biofuels, such as ethanol, which is produced by the fermentation of the sugars found in corn, has increased. Using Le Châtelier's principle, predict which way the equilibrium will shift during the fermentation of sugar for each of the following changes. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(a q) $$ a. when the concentration of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is increased b. when the concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is decreased c. when \(\mathrm{CO}_{2}\) gas is added to the solution d. when the volume of water in the solution is doubled

At \(25^{\circ} \mathrm{C}\), gaseous \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}(g)\) and \(\mathrm{Cl}_{2}(g)\) to the extent that \(12.5 \%\) of the original \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (by moles) has decomposed to reach equilibrium. The total pressure (at equilibrium) is \(0.900\) atm. Calculate the value of \(K_{\mathrm{p}}\) for this system.

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. \(2.0 \mathrm{~mol}\) pure \(\mathrm{NOCl}\) in a \(2.0\) - \(\mathrm{L}\) flask b. \(1.0 \mathrm{~mol} \mathrm{NOCl}\) and \(1.0 \mathrm{~mol} \mathrm{NO}\) in a \(1.0\) -L flask c. \(2.0 \mathrm{~mol} \mathrm{NOCl}\) and \(1.0 \mathrm{~mol} \mathrm{Cl}_{2}\) in a \(1.0\) -L flask

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=0.041 M\), \(\left[\mathrm{O}_{2}\right]=0.0078 M\), and \([\mathrm{NO}]=4.7 \times 10^{-4} M .\) Calculate the value of \(K\) for the reaction.

The formation of glucose from water and carbon dioxide is one of the more important chemical reactions in the world. Plants perform this reaction through the process of photosynthesis, creating the base of the food chain: $$ 6 \mathrm{H}_{2} \mathrm{O}(g)+6 \mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) $$ At a particular temperature, the following equilibrium concentrations were found: \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=7.91 \times 10^{-2} M,\left[\mathrm{CO}_{2}(g)\right]=\) \(9.3 \times 10^{-1} M\), and \(\left[\mathrm{O}_{2}(g)\right]=2.4 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.
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