Problem 84 - Chemistry The Molecular Science

Question

In muscle cells under the condition of vigorous exercise, glucose is converted to lactic acid ("lactate"), \(\mathrm{CH}_{3} \mathrm{CHOHCOOH},\) by the chemical reaction \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \longrightarrow 2 \mathrm{CH}_{3} \mathrm{CHOHCOOH} \quad \Delta_{\mathrm{r}} G^{\circ \prime}=-197 \mathrm{~kJ} / \mathrm{mol}\) (a) If all of the Gibbs free energy from this reaction were used to convert ADP to ATP, calculate how many moles of ATP could be produced per mole of glucose. (b) The actual reaction involves the production of \(3 \mathrm{~mol}\) ATP per mole of glucose. Calculate the \(\Delta_{\mathrm{r}} G^{\circ}\) for this overall reaction. (c) Is the overall reaction in part (b) reactant-favored or product-favored?

Step-by-Step Solution

Answer

(a) 6.46 moles of ATP can be produced. (b) \(-105.5 \text{ kJ/mol}\). (c) Product-favored.

Step 1: Gibbs Free Energy and ATP Calculation

The reaction of ADP converting to ATP has an energy requirement of approximately \( \Delta G^{\circ\prime} = +30.5 \text{ kJ/mol} \). To find how many moles of ATP can be synthesized from the given reaction's Gibbs free energy (\( \Delta_{r}G^{\circ\prime} = -197 \text{ kJ/mol} \)), divide the total energy by the energy required per mole of ATP produced. Thus, \( \frac{-197}{30.5} = 6.46 \). Therefore, theoretically, 6.46 moles of ATP could be formed per mole of glucose.

Step 2: Calculate Actual Gibbs Free Energy Change

For the production of 3 mol of ATP, the \( \Delta G^{\circ\prime} \) per ATP is 30.5 kJ/mol. Therefore, the total free energy change \( \Delta_{ATP} \) is \( 3 \times 30.5 = +91.5 \text{ kJ} \). Subtract this from the energy released by glucose breakdown (-197 kJ/mol): \( -197 + 91.5 = -105.5 \text{ kJ/mol} \). So the \( \Delta_{r} G^{\circ} \) for this overall reaction is \( -105.5 \text{ kJ/mol} \).

Step 3: Determine Reaction Favorability

A negative \( \Delta G \) indicates a spontaneous reaction, meaning it is product-favored. Since \( -105.5 \text{ kJ/mol} \) is negative, the process where 3 ATP are produced per glucose molecule in the actual reaction is indeed product-favored.

Key Concepts

Gibbs Free Energy

Gibbs Free Energy, represented as \( \Delta G \), is a concept that helps predict whether a chemical reaction will occur spontaneously under constant pressure and temperature conditions. It essentially measures the amount of usable energy obtained from a reaction. When \( \Delta G \) is negative, it means the reaction releases energy and is considered spontaneous, or product-favored. On the other hand, a positive \( \Delta G \) indicates that energy must be supplied for the reaction to occur, making it reactant-favored.
In the context of the glucose to lactic acid conversion, the \( \Delta_{r} G^{\circ\prime} = -197 \text{ kJ/mol} \) suggests that this reaction releases a significant amount of energy. This negative value indicates that, theoretically, up to 6.46 moles of ATP could be synthesized per mole of glucose.
Understanding Gibbs Free Energy helps us determine how energy can be harnessed in biological systems such as ATP synthesis, to support essential cellular functions.

ATP Synthesis

ATP, or adenosine triphosphate, is known as the energy currency of the cell. It provides energy required for various cellular processes. In muscles, ATP is crucial for contraction and relaxation cycles, especially during strenuous activities.
ATP synthesis primarily occurs through phosphorylation of ADP (adenosine diphosphate), a process requiring energy input. This energy comes from the breakdown of fuel molecules like glucose. As detailed previously, converting glucose to lactic acid releases energy, some of which is harnessed to convert ADP to ATP:

• Theoretical calculation: As \( -197 \text{ kJ/mol} \) is available, one might expect the formation of around 6.46 moles of ATP per mole of glucose.
• Actual biological conditions: The real-world production is 3 ATP per mole of glucose.

The difference arises due to other biological factors and inefficiencies, illustrating that while thermodynamic data like Gibbs Free Energy provides a guide, actual cellular processes can differ due to complex biochemical interplays.

Product-Favored Reactions

A product-favored reaction is one that naturally proceeds towards the formation of products at equilibrium. This happens when the Gibbs Free Energy \( \Delta G \) is negative, indicating the reaction's spontaneity. Being product-favored is crucial for biological reactions that need to continuously produce necessary compounds, like ATP.
In our particular scenario, the overall ATP-producing reaction, featuring a \( \Delta_{r} G^{\circ} = -105.5 \text{ kJ/mol} \), presents a clear picture of a product-favored process. This negativity tells us that once glucose is metabolized under exercise conditions, the reaction can proceed without additional inputs of energy, ensuring muscle cells receive the ATP they need.
Such reactions are a testament to the body's efficiency, adeptly converting energy into needed forms, even under demanding conditions. It also underscores the critical role of thermodynamics in understanding how organisms optimize their energy resources.