An enzymatic hydrolysis of fructose- \(1-P\) \\[\text { Fructose- } 1-\mathrm{P}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \text { fructose }+\mathrm{P}_{1}\\] was allowed to proceed to equilibrium at \(25^{\circ} \mathrm{C}\). The original concentration of fructose-1-P was \(0.2 \mathrm{M}\), but when the system had reached equilibrium the concentration of fructose-1-P was only \(6.52 \times 10^{-5} \mathrm{M}\). Calculate the equilibrium constant for this reaction and the free energy of hydrolysis of fructose- \(1-P\).

Understanding the equilibrium constant is essential in the study of chemical reactions, especially in biochemical processes like the enzymatic hydrolysis of fructose-1-phosphate. It quantifies the relative concentrations of reactants and products in a reaction at equilibrium. In a general sense, if the equilibrium constant, denoted as \(K_{c}\), is high, the reaction favors the formation of products, while a low \(K_{c}\) indicates that the reactants are more favored.

When calculating \(K_{c}\), it is important to note that the concentrations of solids and pure liquids don't change during a reaction and thus are often omitted from the equation. In the case of fructose-1-phosphate hydrolysis, the water concentration remains relatively constant due to its large excess, allowing us to exclude it from the calculations. The change in concentration of fructose-1-P from its initial state to equilibrium was used to deduce the concentrations of the resulting fructose and phosphate. The equilibrium constant can then be calculated using these concentrations.

Gibbs free energy, represented as \(\Delta G\), is a central concept in biochemical thermodynamics. It provides invaluable insights into the spontaneity and energy changes of a chemical reaction. A negative \(\Delta G\) value suggests the reaction can occur spontaneously under constant pressure and temperature.

By using the calculated equilibrium constant, we can determine the Gibbs free energy of the hydrolysis reaction. The relationship between \(\Delta G\) and \( K_{c} \) is expressed using an equation that incorporates the gas constant \(R\) and temperature \(T\) in Kelvin. For the hydrolysis of fructose-1-phosphate, the \(\Delta G\) was found to be negative, indicating that the reaction is exergonic, thus capable of proceeding spontaneously under standard conditions.

Biochemical thermodynamics allows us to predict the direction and extent of biochemical reactions. It combines principles from both biology and physical chemistry to explain how biological systems adhere to the laws of energy conservation and transformation.

In this context, principles like the equilibrium constant and Gibbs free energy are crucial. Both these quantities reflect how biological molecules like fructose-1-phosphate behave in aqueous environments, such as within our cells. Understanding these principles, any student can gain a deeper comprehension of how energy is channeled through metabolic pathways, like the hydrolysis of fructose-1-phosphate.

Enzyme kinetics is the study of the rates at which enzymes catalyze reactions. It seems tangential but is intimately related to biochemical thermodynamics. While our focus might be on the equilibrium state, understanding enzyme kinetics provides us with insights into how quickly that state is reached.

An enzyme's efficacy is characterized by parameters such as the Michaelis constant \(K_m\) and maximum rate \(V_{max}\). These kinetic values can also influence the pathway and rate at which a reaction, like the hydrolysis of fructose-1-phosphate, approaches equilibrium. Although kinetics were not directly calculated in this exercise, the underlying principles remain important for students studying enzymatic reactions and their efficiency in biological systems.