Question

Consider the following

interconversion, which occurs in glycolysis (Chapter 14):
Fructose 6-phosphate 3:::::::4 glucose 6-phosphate
Keq 1.97
(a) What is Delta G for the reaction (Keq measured at 25 C)?
(b) If the concentration of fructose 6-phosphate is adjusted
to 1.5 M and that of glucose 6-phosphate is adjusted to
0.50 M, what is Delta G?
(c) Why are Delta G and Delta G different?

Answer 1

(a) ΔG for the reaction at 25°C is +1.47 kJ/mol.

(b) Adjusting the concentrations to 1.5 M for fructose 6-phosphate and 0.50 M for glucose 6-phosphate results in ΔG of +0.74 kJ/mol.

(c) ΔG and ΔG' differ due to the influence of concentrations on the standard free energy change, altering the non-standard Gibbs free energy of the reaction.

Step-by-step explanation:

(a) ΔG can be calculated using the formula ΔG = -RT ln(Keq), where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (25°C = 298 K), and ln denotes the natural logarithm. Plugging in the values gives ΔG = -8.314 J/(mol·K) × 298 K × ln(1.97), which results in ΔG = +1.47 kJ/mol.

(b) With concentrations given, ΔG = ΔG' + RT ln(Q), where Q is the reaction quotient. First, calculate Q = ([glucose 6-phosphate] 4) / ([fructose 6-phosphate]3) = (0.50⁴) / (1.5³) = 0.0741. Then, ΔG = ΔG' + 8.314 J/(mol·K) × 298 K × ln(0.0741), yielding ΔG = +0.74 kJ/mol.

(c) ΔG and ΔG' differ due to the concentrations altering the equilibrium constant, impacting the non-standard Gibbs free energy of the reaction. ΔG' represents the standard free energy change at equilibrium, whereas ΔG considers the non-equilibrium conditions, influenced by concentration changes as seen in the reaction quotient (Q). This deviation occurs because ΔG accounts for the actual concentrations present, reflecting the system's real-time state, while ΔG' pertains solely to standard conditions.

Answer 2

The Gibbs free energy change ΔG can be calculated using the standard free energy change ΔG° and the equilibrium constant Keq. ΔG can be different from ΔG° due to changes in concentrations of reactants and products. ΔG° represents the standard free energy change at equilibrium conditions, while ΔG is the actual free energy change under non-equilibrium conditions.

Step-by-step explanation:

The Gibbs free energy change (ΔG) for a reaction can be determined using the standard free energy change (ΔG°) and the equilibrium constant (Keq) for the reaction. The formula to calculate ΔG is as follows: ΔG = -RTln(Keq), where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and ln represents the natural logarithm.

(a) To find ΔG for the reaction, we need to use the value of Keq. The equation gives the value of Keq as 1.97. By substituting this value into the formula, we can calculate ΔG at the given temperature of 25°C.

(b) If the concentrations of fructose 6-phosphate and glucose 6-phosphate are adjusted to 1.5 M and 0.50 M respectively, we can calculate ΔG using the formula ΔG = ΔG° + RTln(Q), where Q is the reaction quotient. Q can be calculated using the given concentrations of the reactants and products.

(c) ΔG° and ΔG can be different due to changes in the concentrations of reactants and products. ΔG° represents the standard free energy change at equilibrium conditions, while ΔG is the actual free energy change occurring under non-equilibrium conditions.