Question

A key step in the metabolism of glucose for energy is the isomerization of glucose-6-phosphate (G6P) to fructose6 -phosphate \((\mathrm{F} 6 \mathrm{P}): \mathrm{G} 6 \mathrm{P} \rightleftharpoons \mathrm{F} 6 \mathrm{P} ; K=0.510\) at \(298 \mathrm{~K}\) (a) Calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) when \(Q\), the [F6P]/[G6P] ratio, equals 10.0 . (c) Calculate \(\Delta G\) when \(Q=0.100\). (d) Calculate \(Q\) if \(\Delta G=-2.50 \mathrm{~kJ} / \mathrm{mol} .\)

Short answer

1. \(\Delta G^{\circ} \approx 1.676 \mathrm{kJ/mol}\).\2. \(\Delta G\ \approx 7.381 \mathrm{kJ/mol}\).\3. \(\Delta G\ \approx -4.029 \mathrm{kJ/mol}\).\4. \(Q \approx 0.185\).\

Step-by-step solution

- Calculate \(\Delta G^{\circ}\) Using the Equilibrium Constant

Use the equation \Delta G^{\circ} = -RT \ln(K)\.\Given: \K = 0.510\ and \T = 298 \mathrm{K}\, and \R = 8.314 \mathrm{J/(mol\cdot K)}\.\Calculate: \[\Delta G^{\circ} = - (8.314 \mathrm{J/(mol\cdot K)}) \times 298 \mathrm{K} \times \ln(0.510)\]\Calculate the natural logarithm: \ \ln(0.510) = -0.6733\ (approx)\When substituted: \[\Delta G^{\circ} = - (8.314 \times 298 \times -0.6733) = 1676.12 \mathrm{J/mol}\ \approx 1.676 \mathrm{kJ/mol}\]\.

- Calculate \(\Delta G\) When \(Q = 10.0\)

Use the equation \Delta G = \Delta G^{\circ} + RT \ln(Q)\.\Substitute \Delta G^{\circ} = 1676.12 \mathrm{J/mol}\, \T = 298 \mathrm{K}\, \R = 8.314 \mathrm{J/(mol\cdot K)}\, and \Q = 10.0\.\Calculate: \[\Delta G = 1676.12 + 8.314 \times 298 \times \ln(10.0)\]\Calculate the natural logarithm: \ \ln(10.0) = 2.3026\ (approx)\When substituted: \[\Delta G = 1676.12 + (8.314 \times 298 \times 2.3026) = 1676.12 + 5704.63\]\Final Calculation: \[\Delta G = 7380.75 \mathrm{J/mol} \approx 7.381 \mathrm{kJ/mol}\]\.

- Calculate \(\Delta G\) When \(Q = 0.100\)

Use the equation \Delta G = \Delta G^{\circ} + RT \ln(Q)\.\Substitute \Delta G^{\circ} = 1676.12 \mathrm{J/mol}\, \T = 298 \mathrm{K}\, \R = 8.314 \mathrm{J/(mol\cdot K)}\, and \Q = 0.100\.\Calculate: \[\Delta G = 1676.12 + 8.314 \times 298 \times \ln(0.100)\]\Calculate the natural logarithm: \ \ln(0.100) = -2.3026\ (approx)\When substituted: \[\Delta G = 1676.12 + (8.314 \times 298 \times -2.3026) = 1676.12 - 5704.63\]\Final Calculation: \[\Delta G = -4028.51 \mathrm{J/mol} \approx -4.029 \mathrm{kJ/mol}\]\.

- Calculate \(Q\) Given \(\Delta G = -2.50 \mathrm{kJ/mol}\)

Use the equation \Delta G = \Delta G^{\circ} + RT \ln(Q)\ and solve for \(Q\).\Substitute \Delta G = -2.50 \mathrm{kJ/mol} = -2500 \mathrm{J/mol}\, \Delta G^{\circ} = 1676.12 \mathrm{J/mol}\, \T = 298 \mathrm{K}\, and \R = 8.314 \mathrm{J/(mol\cdot K)}\.\Rearrange: \[\ln(Q) = \dfrac{\Delta G - \Delta G^{\circ}}{RT}\]\Calculate: \[\ln(Q) = \dfrac{-2500 - 1676.12}{8.314 \times 298} = \dfrac{-4176.12}{2476.372}\]\ \[\approx -1.686\]\Find antilog: \Q = e^{-1.686} \ \[\approx 0.185\]\.

Key concepts

Isomerization of Glucose

Isomerization is a crucial process in metabolism. Specifically, the isomerization of glucose-6-phosphate \((G6P)\) to fructose-6-phosphate \((F6P)\) is a key step in glycolysis, the pathway by which cells harvest energy.
This process involves the rearrangement of the molecular structure without altering the molecular formula.
This transformation is facilitated by enzymes, making the reaction both highly specific and efficient.
In this case, G6P and F6P are structural isomers. They have the same number of atoms but arranged differently.
Understanding this swap is fundamental because it sets the stage for subsequent steps that lead to energy production in the form of ATP.
For students, recognizing the significance of such biochemical transformations is essential for grasping metabolic dynamics.

Equilibrium Constant

The equilibrium constant, denoted as \(K\), is a crucial element in understanding chemical reactions.
It provides the ratio of products to reactants at equilibrium.
For the isomerization of glucose, the given equilibrium constant \( K = 0.510 \) implies a specific concentration ratio of G6P and F6P at 298 K.
This constant helps in determining the direction of the reaction and its efficiency.
An equilibrium constant less than 1 suggests that the reactants are favored at equilibrium, which is the case here.
Knowing \(K\) allows us to perform further calculations involving Gibbs free energy changes, giving us insight into the thermodynamics of the reaction.
Remember, changes in conditions such as temperature can alter the equilibrium constant, hence influencing the reaction's progress.

Natural Logarithm

The natural logarithm, represented as \( \ln \), is a mathematical function that plays a significant role in thermodynamics.
In the context of Gibbs free energy calculations, it helps quantify the relationship between equilibrium constants and reaction quotients.
The natural logarithm of a number \( x \) is the power to which \( e \) (approximately 2.718) must be raised to result in \( x \).
For example, \( \ln(0.510) = -0.6733 \) means that \( e \) must be raised to about -0.6733 to equal 0.510.
This function is essential in our calculations as it transforms multiplicative relationships into additive ones, simplifying complex chemical equations.
Without a good handle on natural logarithms, solving thermodynamic problems efficiently would be considerably more challenging.

Standard Gibbs Free Energy Change

The standard Gibbs free energy change \( \Delta G^{\circ} \) indicates whether a reaction can occur spontaneously under standard conditions (298 K, 1 atm, and 1 M concentration).
Calculating \( \Delta G^{\circ} \) for the isomerization of glucose helps us understand the reaction's feasibility.
It is given by the formula: \[ \Delta G^{\circ} = -RT \ln(K) \] Here, \( R \) is the gas constant (8.314 J/(mol·K)), \( T \) is the temperature in Kelvin, and \( K \) is the equilibrium constant.
For this glucose isomerization, substituting in the values gives us \( \Delta G^{\circ} \approx 1.676 \) kJ/mol, indicating the energy required or released under standard conditions.
Positive \( \Delta G^{\circ} \) values mean the reaction is not spontaneous, while negative values suggest it proceeds without external influence.
This concept is a cornerstone in thermodynamics, explaining the energy landscape of chemical reactions and their spontaneity.