Problem 6 - Biochemistry

Question

For the process \(A \equiv B, K_{c g}(A B)\) is 0.02 at \(37^{\circ} \mathrm{C}\). For the process \(\mathrm{B} \rightleftharpoons \mathrm{C}, K_{\mathrm{eq}}(\mathrm{BC})=1000\) at \(37^{\circ} \mathrm{C}\) a. Determine \(K_{\mathrm{rg}}(\mathrm{AC}),\) the equilibrium constant for the overall process \(A \rightleftharpoons C,\) from \(K_{c q}(A B)\) and \(K_{c g}(B C)\) b. Determine standardstate free energy changes for all three processes, and \(\mathrm{us}=\Delta G^{\circ}(\mathrm{AC})\) to determine \(K_{\mathrm{rg}}(\mathrm{AC}) .\) Make sure that this value agrees with that determined in part a of this problem.

Step-by-Step Solution

Answer

The equilibrium constant for the overall process \(A \rightleftharpoons C\) is 20. The standard state free energy changes for the processes \(A \rightleftharpoons B\), \(B \rightleftharpoons C\), and \(A \rightleftharpoons C\) can be calculated using the expression -RTlnK. Additional analysis shows that the calculated equilibrium constant from the overall free energy change agrees with the directly calculated equilibrium constant.

Step 1: Determine the Overall Equilibrium Constant

According to the law of mass action, the equilibrium constant for a net reaction made up of two or more steps is equal to the product of the equilibrium constants for the individual steps. Therefore, we can represent the overall equilibrium constant for the process \(A \rightleftharpoons C\) as \(K_{c g}(A C)=K_{c g}(A B) \times K_{c g}(B C)\). Substituting the provided values into this expression, we find \(K_{c g}(A C)=0.02 \times 1000 = 20\)

Step 2: Calculating Free Energy Changes

The equation relating standard free energy change (\(\Delta G^{\circ}\)) and equilibrium constant (K) is \(\Delta G^{\circ}=-R T \ln K\), where R is the universal gas constant (8.314 J /(mol \cdot K)) and T is the temperature in Kelvin (310K for 37°C). Plugging in the provided values of K for each process, we can calculate \(\Delta G^{\circ}(A B)=-R T \ln K_{c g}(A B)\) and \(\Delta G^{\circ}(B C)=-R T \ln K_{c g}(B C)\). The overall free energy change \(\Delta G^{\circ}(A C)\) is the sum of these two values.

Step 3: Use Delta G AC to Confirm Overall Equilibrium Constant

The equilibrium constant derived from the calculated \(\Delta G^{\circ}(A C)\) should match the \(K_{c g}(A C)\) obtained in the first step. We can use the equation \(\Delta G^{\circ}=-R T \ln K_{c g}(A C)\) to calculate \(K_{c g}(A C)\) from the derived \(\Delta G^{\circ}(A C)\). We find that the calculated \(K_{c g}(A C)\) is indeed equal to 20, confirming accuracy.

Key Concepts

Free Energy Change

In chemical reactions, understanding the concept of free energy change, particularly the standard free energy change (abla G^{\circ}abla), is crucial. Free energy change helps us determine whether a reaction is spontaneous. A negative free energy change indicates that the reaction can occur spontaneously, while a positive one suggests it is non-spontaneous under standard conditions.

The relationship between free energy change and the equilibrium constant (K) is given by the formula:

• \( \Delta G^{\circ} = -RT \ln K \)

where \( R \) is the universal gas constant (8.314 J/mol·K), and \( T \) is the temperature in Kelvin.

This equation shows that large equilibrium constants (implying products favored at equilibrium) result in large negative \( \Delta G^{\circ} \), indicating spontaneity. Conversely, small equilibrium constants imply positive \( \Delta G^{\circ} \). Understanding this relationship is key in predicting how reactions will proceed.

Law of Mass Action

The law of mass action is fundamental in understanding how equilibrium is established in chemical reactions. It states that the rate of any chemical reaction is proportional to the product of the concentrations of the reactants, each raised to a power corresponding to the number of molecules of the substance involved in the reaction.

For a reversible reaction at equilibrium, the law helps us understand how the equilibrium constant, \( K_{eq} \), is determined. For a reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant expression is:

• \( K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

The law of mass action means that the equilibrium state is reached when the ratio of the product of concentration terms for products over reactants remains constant.

This principle is beautifully illustrated when combining reactions. The overall equilibrium constant for a series of reactions is the product of the individual equilibrium constants, as seen in the solution process for transforming \( A \rightleftharpoons C \) from \( A \rightleftharpoons B \) and \( B \rightleftharpoons C \).

Thermodynamics

Thermodynamics is the branch of physical science focused on heat and temperature and their relation to energy and work. It applies crucial concepts to chemistry, particularly to understanding how energy influences reactions.

At the core of thermodynamics are the laws governing energy conservation and entropy. These play a crucial role in determining reaction spontaneity and equilibrium. The first law of thermodynamics, the law of energy conservation, tells us that energy cannot be created or destroyed, only transformed.

When studying chemical reactions, thermodynamics helps us through related concepts like enthalpy, entropy, and free energy.

• Enthalpy (\( H \)) is the heat content of a system.
• Entropy (\( S \)) is a measure of disorder or randomness.
• Free energy combines enthalpy and entropy to predict the feasibility of a process.

These concepts work together to predict whether a reaction will proceed spontaneously, helping us understand equilibria and calculate equilibrium constants using the free energy changes calculated through given equations, as demonstrated in the solution process of reactions \( A \equiv B \) and \( B \rightleftharpoons C \).