Question

In the human body, the enzyme phosphoglucomutase catalyzes the conversion of glucose-1-phosphate into glucose \(-6-\) phosphate: glucose-1-phosphate \(\rightleftarrows\) glucose-6-phosphate. At \(38^{\circ} \mathrm{C}\), the equilibrium constant, \(\mathrm{k}\), for this reaction is approximately 20. Calculate the free energy change, \(\Delta \mathrm{G}^{\circ}\), for the equilibrium conversion. Calculate the free energy change \(\Delta \mathrm{G}\) for the nonequilibrium situation in which [glucose-1phosphate \(]=0.001 \mathrm{M}\) and [glucose \(-6-\) phosphate \(]\) \(=0.050 \mathrm{M}\)

Short answer

The equilibrium free energy change (∆G°) is approximately -5.74 kJ/mol, and the non-equilibrium free energy change (∆G) is approximately 9.82 kJ/mol.

Step-by-step solution

Convert temperature to Kelvin

To use the given temperature in our calculations, we need to convert it from Celsius to Kelvin. To do this, we add 273.15. \( 38^{\circ} \mathrm{C} + 273.15 = 311.15 \mathrm{K} \)

Calculate ∆G° using the equilibrium constant

We can find the equilibrium free energy change (∆G°) using the equation: \( \Delta G^{\circ} = -RT \ln K \) Where R is the ideal gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant. Plug in the values we know: \( \Delta G^{\circ} = -(8.314 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}})(311.15 \mathrm{K}) \ln(20) \) Calculate ∆G°: \( \Delta G^{\circ} \approx -5.74 \frac{\mathrm{kJ}}{\mathrm{mol}} \)

Calculate the reaction quotient (Q)

To find the non-equilibrium free energy change (∆G), we first need to determine the reaction quotient (Q) using the given concentrations of glucose-1-phosphate and glucose-6-phosphate: \( Q = \frac{\textrm{[glucose-6-phosphate]}}{\textrm{[glucose-1-phosphate]}} \) Plug in the given concentrations: \( Q = \frac{0.050 \mathrm{M}}{0.001 \mathrm{M}} \) Calculate Q: \( Q = 50 \)

Calculate ∆G under non-equilibrium conditions

Now we can find the non-equilibrium free energy change (∆G) using the equation: \( \Delta G = \Delta G^{\circ} + RT \ln Q \) Plug in the known values: \( \Delta G = (-5.74 \frac{\mathrm{kJ}}{\mathrm{mol}}) + (8.314 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}})(311.15 \mathrm{K}) \ln(50) \) Calculate ∆G: \( \Delta G \approx 9.82 \frac{\mathrm{kJ}}{\mathrm{mol}} \) #Summary#: The equilibrium free energy change (∆G°) is approximately -5.74 kJ/mol, and the non-equilibrium free energy change (∆G) is approximately 9.82 kJ/mol.

Key concepts

Equilibrium Constant

The equilibrium constant, often represented by the symbol \( K \), is a critical factor in chemical reactions. It tells us about the balance between products and reactants at equilibrium. For the reaction: glucose-1-phosphate \( \rightleftarrows \) glucose-6-phosphate, \( K \) is valued at 20 at \( 38^{\circ} \text{C} \). This means that at equilibrium, the concentration ratio of products to reactants is 20 to 1.
Understanding \( K \):

• A high \( K \) value (greater than 1) indicates the reaction favors the formation of products more than reactants.
• A low \( K \) (less than 1) indicates that reactants are favored over products.
• In this case, \( K = 20 \) means there is a strong tendency for glucose-1-phosphate to convert into glucose-6-phosphate when equilibrium is reached.

The equilibrium position of a reaction provides valuable insight into how far a reaction proceeds. It's important to remember that \( K \) is specific to a particular temperature.

Free Energy Change

Free energy change, represented as \( \Delta G \), is a key concept in understanding the spontaneity of a reaction. There are two types of free energy changes to consider: standard free energy change \( \Delta G^{\circ} \) and the free energy change \( \Delta G \) under any condition.
Difference Between \( \Delta G^{\circ} \) and \( \Delta G \):

• \( \Delta G^{\circ} \) is the change in free energy when a reaction happens at the standard state, meaning 1 M concentration, 1 atm pressure, and a specified temperature.
• \( \Delta G \) can be used when the reaction is not at standard states or at equilibrium.

Calculating \( \Delta G^{\circ} \): For this exercise, it is calculated using the formula: \( \Delta G^{\circ} = -RT \ln K \)
Where \( R \) is the gas constant and \( T \) is the temperature in Kelvin.
A negative \( \Delta G^{\circ} \) (like \(-5.74 \text{kJ/mol}\) here) implies that under standard conditions, the reaction proceeds spontaneously.
On the other hand, to compute \( \Delta G \) in non-equilibrium conditions, we use:\( \Delta G = \Delta G^{\circ} + RT \ln Q \). A positive \( \Delta G \) suggests the reaction is non-spontaneous as written, under the given conditions.

Reaction Quotient (Q)

The reaction quotient \( Q \) lets us assess the current state of a reaction as compared to its equilibrium state. It is calculated similarly to the equilibrium constant \( K \), but it uses current, not equilibrium, concentrations.
How to calculate \( Q \):

• For any reaction \( aA + bB \rightleftharpoons cC + dD \), \( Q \) is calculated as \( Q = \frac{[C]^c [D]^d}{[A]^a [B]^b} \).
• In our problem, \( Q \) is \( \frac{\text{[glucose-6-phosphate]}}{\text{[glucose-1-phosphate]}} \).

Using the given concentrations, \( Q = 50 \), which is much higher than \( K = 20 \).
Significance of \( Q \) vs. \( K \):

• If \( Q < K \), the reaction will move forward to produce more products to reach equilibrium.
• If \( Q > K \), the reaction will proceed in reverse to produce more reactants.
• If \( Q = K \), the system is at equilibrium, and no net change occurs.

In this case, since \( Q > K \), the reaction tends to shift towards forming more reactants (glucose-1-phosphate). This ties directly into the calculated positive \( \Delta G \), indicating non-spontaneity for the forward reaction under these conditions.