Question

The formation of peptide bonds is an important area of chemistry. The following reaction has an equilibrium constant \((K)\) of \(3.2 \times 10^{2}\) at some temperature: Alanine \((a q)+\) leucine \((a q) \rightleftharpoons\) alanine-leucine dipeptide \((a q)+\mathrm{H}_{2} \mathrm{O}(l)\) Which direction will this reaction need to shift to reach equilibrium under the following conditions? a. \([\) alanine \(]=0.60 M,[\) leucine \(]=0.40 M,[\) dipeptide \(]=0.20 M\) b. \([\) alanine \(]=3.5 \times 10^{-4} M,[\) leucine \(]=3.6 M,[\) dipeptide \(]=\) \(0.40 M\) c. \([\) alanine \(]=6.0 \times 10^{-3} M,[\) leucine \(]=9.0 \times 10^{-3} M\), \([\) dipeptide \(]=0.30 M\)

Short answer

a. The reaction will shift to the right to reach equilibrium. b. The reaction is already at equilibrium. c. The reaction will shift to the left to reach equilibrium.

Step-by-step solution

Calculate Qtag_content# For the given initial concentrations, calculate the reaction quotient \(Q=[dipeptide]/([alanine] \cdot [leucine])\): \(Q=\dfrac{0.20}{(0.60)(0.40)} = 0.83\)

Step 2: Compare Q with Ktag_content# Given that \(Q=0.83\) and \(K=3.2 \times 10^{2}\), we have that \(Q < K\). So, the reaction will shift to the right to reach equilibrium. b. \([alanine]=3.5 \times 10^{-4} M, [leucine]=3.6 M, [dipeptide]=0.40 M\)

Calculate Qtag_content# For the given initial concentrations, calculate the reaction quotient \(Q=[dipeptide]/([alanine] \cdot [leucine])\): \(Q=\dfrac{0.40}{(3.5 \times 10^{-4})(3.6)} = 3.2 \times 10^{2}\)

Step 2: Compare Q with Ktag_content# Given that \(Q=3.2 \times 10^{2}\) and \(K=3.2 \times 10^{2}\), we have that \(Q = K\). So, the reaction is already at equilibrium. c. \([alanine]=6.0 \times 10^{-3} M, [leucine]=9.0 \times 10^{-3} M, [dipeptide]=0.30 M\)

Calculate Qtag_content# For the given initial concentrations, calculate the reaction quotient \(Q=[dipeptide]/([alanine] \cdot [leucine])\): \(Q=\dfrac{0.30}{(6.0 \times 10^{-3})(9.0 \times 10^{-3})} = 5.56 \times 10^{3}\)

Step 2: Compare Q with Ktag_content# Given that \(Q=5.56 \times 10^{3}\) and \(K=3.2 \times 10^{2}\), we have that \(Q > K\). So, the reaction will shift to the left to reach equilibrium.

Key concepts

Peptide Bond Formation

The creation of peptide bonds is a key process in biochemistry, involving the linking of amino acids to form proteins. During this process, the amino group of one amino acid reacts with the carboxyl group of another, releasing a molecule of water and forming a peptide bond.

For instance, when alanine (aq) combines with leucine (aq), they form an alanine-leucine dipeptide (aq) and water (l). This reaction can proceed in either direction—towards the formation of the dipeptide or its breakdown—depending on various conditions such as concentration and temperature.

Understanding the details of peptide bond formation is essential for fields like biochemistry and pharmacology. It's the foundation of protein structure and function, and disruptions in this process can lead to significant biochemical issues.

Equilibrium Constant (K)

The equilibrium constant (K) expresses the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to the power of its stoichiometric coefficient. The value of K indicates the extent to which a reaction will proceed. Simply put, a higher K value means a reaction favors the formation of products.

In our textbook exercise example, the equilibrium constant \(K\) for the formation of the alanine-leucine dipeptide is \(3.2 \times 10^{2}\). This relatively large value suggests that at equilibrium, the concentration of the dipeptide is favored over that of the individual amino acids alanine and leucine. Remember that K is temperature-dependent and provides no information about the speed of the reaction.

Reaction Quotient (Q)

The reaction quotient (Q) is a measure that tells us the direction in which a reaction must proceed to reach equilibrium from a given set of initial conditions. Like K, Q is calculated using the same expression, but unlike K, it uses initial concentrations instead of equilibrium concentrations.

By comparing Q to K, we can predict the reaction’s direction: If \( Q < K \) the reaction shifts to the right (towards products), indicating the system has not yet reached equilibrium. If \( Q > K \) the reaction shifts to the left (towards reactants), meaning the system has exceeded equilibrium and must revert back. Finally, if \( Q = K \) the system is at equilibrium, and no shift is needed.

This concept is crucial in the textbook exercise for analyzing the state of the peptide bond-forming reaction under various conditions and determining which direction it will shift to achieve equilibrium.