Problem 16 - Cell and Molecular Biology: Concepts and Experiments

Question

Consider the reaction: \\[ \begin{aligned} \text { Glucose }+\mathrm{P}_{\mathrm{i}} \rightleftharpoons_{8} \text { lucose } 6 \text { -phosphate }+\mathrm{H}_{2} \mathrm{O} \\ & \Delta G^{* \prime}=+3 \mathrm{kcal} / \mathrm{mol} \end{aligned} \\] What is the equilibrium constant, \(K_{\mathrm{eq}}^{\prime},\) for this reaction? (Note: the concentration of water is ignored.) Does the positive \(\Delta G^{0 \prime}\) in the above reaction mean that the reaction can never go spontaneously from left to right?

Step-by-Step Solution

Answer

The equilibrium constant is 0.0064. Positive \( \Delta G^{0'} \) means non-spontaneous under standard conditions, but not impossible otherwise.

Step 1: Understanding the Relationship

The reaction involves glucose and inorganic phosphate turning into glucose-6-phosphate and water. To find the equilibrium constant \( K_{eq}' \), we need the relationship between standard free energy change and \( K_{eq}' \), which is given by the formula: \[ \Delta G^{*'} = -RT \ln K_{eq}' \] where \( R \) is the gas constant \( 1.987 \text{ cal/mol K} \) and \( T \) is the temperature in Kelvin (usually 298 K).

Step 2: Plug in the Known Values

Using the formula from Step 1, and given \( \Delta G^{*'} = +3 \text{ kcal/mol} = 3000 \text{ cal/mol} \) (converted to cal), we have:\[ 3000 = -1.987 \times 298 \times \ln K_{eq}' \]

Step 3: Solve for \( \ln K_{eq}' \)

Rearrange the formula to solve for \( \ln K_{eq}' \):\[ \ln K_{eq}' = -\frac{3000}{1.987 \times 298} \]Calculate the value:\[ \ln K_{eq}' = -5.043 \]

Step 4: Calculate \( K_{eq}' \)

To find \( K_{eq}' \), take the exponential of both sides:\[ K_{eq}' = e^{-5.043} \]Calculating this gives:\[ K_{eq}' = 0.0064 \]

Step 5: Interpret the Result

The equilibrium constant \( K_{eq}' = 0.0064 \) indicates that at equilibrium, the concentration of products is much lower than the concentration of reactants, which aligns with the positive \( \Delta G^{*'} \). This does not mean the reaction cannot go spontaneously; the actual direction depends on conditions like reactant and product concentrations.

Key Concepts

Equilibrium Constant

The equilibrium constant, often symbolized as \( K_{eq} \), is a key concept in chemistry that helps describe a chemical reaction’s behavior when it reaches a state of balance. At equilibrium, the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of the products and reactants no longer change.

For the given reaction involving glucose, the equilibrium constant is calculated using the relationship between the standard free energy change \( \Delta G^{*'} \) and \( K_{eq} \). The formula used is \( \Delta G^{*'} = -RT \ln K_{eq}' \). Here, \( R \) is the universal gas constant, typically given as \( 1.987 \text{ cal/mol K} \), and \( T \) is the temperature in Kelvin, often 298 K in standard conditions.

By substituting the known values of \( \Delta G^{*'} = +3000 \text{ cal/mol} \) into the formula, we solve for \( \ln K_{eq}' \) and then determine \( K_{eq}' = 0.0064 \). This means that at equilibrium, the concentration of glucose and inorganic phosphate is significantly higher than that of glucose-6-phosphate and water.

Essentially, a \( K_{eq}' \) value less than 1 indicates that the formation of products is less favored compared to the reactants under standard conditions, reflecting a state where reactants are more prevalent.

Standard Free Energy Change

The standard free energy change, denoted as \( \Delta G^{0'} \), is crucial in predicting how a reaction behaves under standard conditions. It helps understand whether a reaction at a given temperature is spontaneous or requires energy input.

In our example, the \( \Delta G^{0'} = +3 \text{ kcal/mol} \) signifies the amount of energy needed to transform the reactants into products at equilibrium under standard conditions. A positive \( \Delta G^{0'} \) implies that the reaction is not spontaneous in the forward direction.

This measurement ties back to the equilibrium constant. If \( \Delta G^{0'} \) is positive, as it is in this glucose-phosphate reaction, then \( K_{eq}' \) will be less than 1, indicating that the reaction does not favor the formation of products at equilibrium. Instead, it favors the reactants, meaning energy must be added to drive the reaction forward.

Understanding \( \Delta G^{0'} \) helps because it provides a thermodynamic snapshot of the reaction's potential under a set of defined conditions. Altering conditions such as temperature or concentrations of reactants/products can change whether or not a reaction seems spontaneous or not.

Reaction Spontaneity

Reaction spontaneity is a concept that explores whether a reaction occurs on its own under specified conditions without any external input of energy.

Often, the sign of the standard free energy change \( \Delta G^{0'} \) helps dictate spontaneity. If \( \Delta G^{0'} \) is negative, a reaction is considered spontaneous, as it can move towards equilibrium naturally. Conversely, a positive \( \Delta G^{0'} \) suggests a non-spontaneous reaction, as additional energy is needed.

In our reaction converting glucose to glucose-6-phosphate, \( \Delta G^{0'} \) is positive, which indicates it isn't spontaneous under standard conditions. Yet, this doesn't mean the reaction cannot ever occur spontaneously. Real-world conditions such as concentration changes or coupling with other energy-releasing reactions can shift the balance, making the reaction appear spontaneous.

Thus, while standard calculations provide baseline expectations, external factors often channel real-world spontaneity, making chemistry dynamic and context-dependent. Understanding this principle highlights the importance of environmental and conditional adjustments in predicting reaction behavior.