Question

The following reaction is catalyzed by the enzyme creatine kinase: A. Under standard conditions \((1 \mathrm{~atm}, 298 \mathrm{~K}, \mathrm{pH} 7.0 . \mathrm{pMg} 3.0\) and an ionic strength of \(0.25 \mathrm{M}\) ) with the concentrations of all reactants equal to \(10 \mathrm{mM}\), the free energy change, \(\Delta G\), for this reaction is \(13.3 \mathrm{~kJ} / \mathrm{mol}\). What is the standard free energy change, \(\Delta G^{\circ}\), for this reaction? B. What is the equilibrium constant for this reaction? C. The standard enthalpies of formation for the reactants are as follows: \(\begin{array}{ll}\text { Creatine } & -540 \mathrm{~kJ} / \mathrm{mol} \\\ \text { Creatine phosphate } & -1510 \mathrm{~kJ} / \mathrm{mol} \\ \text { ATP } & -2982 \mathrm{~kJ} / \mathrm{mol} \\ \text { ADP } & -2000 \mathrm{~kJ} / \mathrm{mol}\end{array}\) What is the standard enthalpy change, \(\Delta H^{\circ}\), for this reaction? D. What is the standard entropy change, \(\Delta S^{\circ}\), for this reaction?

Short answer

A. ΔG° = 13.3 kJ/mol B. K ≈ 0.00463 C. ΔH° = 12 kJ/mol D. ΔS° = -4.36 J/mol K

Step-by-step solution

Calculate the Standard Free Energy Change (ΔG°)

To find the standard free energy change (\( \Delta G^{\circ} \)), we use the relation: \[ \Delta G = \Delta G^{\circ} + RT \ln(Q) \]where \( \Delta G = 13.3\, \text{kJ/mol} \), \( R = 8.314 \times 10^{-3} \, \text{kJ/mol K} \),and \( T = 298 \, \text{K} \), and the reaction quotient \( Q = \frac{[ADP][Creatine \ \text{phosphate}]}{[ATP][Creatine]} = 1 \) under standard conditions (as all concentrations are equal).Thus, \( \Delta G = \Delta G^{\circ} \) leading to \( \Delta G^{\circ} = 13.3 \, \text{kJ/mol} \).Thus,\[ \Delta G^{\circ} = 13.3 \, \text{kJ/mol} \].

Determine the Equilibrium Constant (K)

We use the relation between standard free energy change (\( \Delta G^{\circ} \)) and the equilibrium constant (\( K \)):\[ \Delta G^{\circ} = -RT \ln(K) \]Rearrange to solve for \( K \):\[ \ln(K) = -\frac{\Delta G^{\circ}}{RT} \]Substitute the values:\( \ln(K) = -\frac{13.3}{8.314 \times 10^{-3} \times 298} \approx -5.37 \)Solving for \( K \),\( K = e^{-5.37} \approx 0.00463 \).

Calculate the Standard Enthalpy Change (ΔH°)

The standard enthalpy change (\( \Delta H^{\circ} \)) is given by:\[ \Delta H^{\circ} = \sum \Delta H_f^{\circ} \text{products} - \sum \Delta H_f^{\circ} \text{reactants} \]Using the standard enthalpies of formation,\[ \Delta H^{\circ} = \left(-1510 - 2000\right) - \left(-540 - 2982\right) \]Calculating,\[ \Delta H^{\circ} = -3510 - (-3522) = 12 \, \text{kJ/mol} \].

Determine the Standard Entropy Change (ΔS°)

Using the Gibbs free energy equation:\[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \]Rearrange to solve for \( \Delta S^{\circ} \):\[ \Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T} \]Substitute the values:\[ \Delta S^{\circ} = \frac{12 - 13.3}{298} \approx -0.00436 \, \text{kJ/mol K} \] or \(-4.36 \, \text{J/mol K}\).Thus,\[ \Delta S^{\circ} = -4.36 \, \text{J/mol K} \].

Key concepts

Standard Free Energy Change

Understanding the standard free energy change, denoted as \( \Delta G^{\circ}\), is pivotal in thermodynamics. It tells us whether a reaction is spontaneous under standard conditions, which involve a pressure of 1 atm, a temperature of 298 K, and each reactant and product at 1 M concentration.
If \( \Delta G^{\circ} < 0\), the reaction is spontaneous. If \( \Delta G^{\circ} > 0\), it is non-spontaneous, meaning it needs external energy to proceed.In our exercise, we determined \( \Delta G^{\circ} = 13.3\, \text{kJ/mol}\). This positive value indicates the reaction is non-spontaneous under the given conditions.
To find \( \Delta G^{\circ}\), we relate it to the free energy change \( \Delta G\) using the equation:\[ \Delta G = \Delta G^{\circ} + RT \ln(Q) \]Where \( R \) is the gas constant and \( Q \) is the reaction quotient. Under standard conditions, where concentrations are the same, \( Q = 1 \), so \( \ln(Q) = 0 \). Hence, \( \Delta G = \Delta G^{\circ} \). This shows how simple factor changes in conditions can significantly alter a reaction's free energy.

Equilibrium Constant

The equilibrium constant, \( K \), demonstrates the ratio of products to reactants at equilibrium. It is closely linked to \( \Delta G^{\circ} \) through the equation:\[ \Delta G^{\circ} = -RT \ln(K) \]This equation hints at how \( K \) signals reaction direction tendencies. When \( K > 1 \), the formation of products is favored. Conversely, \( K < 1 \) means reactants remain predominant at equilibrium.For the given reaction, we calculated \( K \) using\[ \ln(K) = -\frac{\Delta G^{\circ}}{RT} \],leading to a \( K \) of approximately 0.00463.
This small value indicates that, under standard conditions, the reaction heavily favors reactants over products.
Understanding \( K \) helps predict reaction behavior without experimenting, a crucial ability for chemists and engineers.

Standard Enthalpy Change

The standard enthalpy change \( \Delta H^{\circ} \) gives insight into the heat absorbed or released in a chemical reaction at constant pressure and standard conditions.It is computed using the enthalpies of formation \( \Delta H_f^{\circ} \) of each reactant and product:\[ \Delta H^{\circ} = \sum \Delta H_f^{\circ} \text{(products)} - \sum \Delta H_f^{\circ} \text{(reactants)} \]From our example, enthalpy values for creatine and ATP, among others, were used:\[ \Delta H^{\circ} = \left(-1510 - 2000\right) - \left(-540 - 2982\right) \]This calculation resulted in \( \Delta H^{\circ} = 12 \, \text{kJ/mol} \).
A positive \( \Delta H^{\circ} \) implies the reaction absorbs heat (endothermic), while a negative value signals heat release (exothermic).For our reaction, \( \Delta H^{\circ} = 12 \, \text{kJ/mol} \) shows it is mildly endothermic, affirming that energy input is necessary.

Standard Entropy Change

Entropy change, \( \Delta S^{\circ} \), describes the disorder increase or decrease in a reaction.
Entropy is a measure of the number of ways the energy can be distributed among the molecules. Its importance stands from Gibbs free energy equation:\[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \]Rearranging to find \( \Delta S^{\circ} \) leads to:\[ \Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T} \]For our specific reaction, using the values found:\[ \Delta S^{\circ} = \frac{12 - 13.3}{298} \approx -0.00436 \, \text{kJ/mol K} \]Which translates to an entropy change of \(-4.36 \, \text{J/mol K} \).A negative \( \Delta S^{\circ} \) suggests a decrease in disorder, or the products are more ordered than the reactants.
This understanding of entropy helps in deducing whether energy spreads out or accumulates during a reaction, crucial for ecosystem and process efficiency modeling.