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Use the hydrolysis of adenosine triphosphate (ATP) as a source of energy. Symbolically, this reaction can be written as:

ATP→ADP+Pi​+Energy
Where ADP represents -30.5 kJ. Calculate the equilibrium constant (K) for this reaction.

User Juan Besa
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Final answer:

The question inquires about calculating the equilibrium constant (K) from the free energy released during ATP hydrolysis. However, additional data is needed beyond the -30.5 kJ/mol energy release to calculate K. ATP is crucial in biological systems, serving as the energy currency and facilitating energy coupling for cellular work.

Step-by-step explanation:

When considering the hydrolysis of adenosine triphosphate (ATP), the energy released can be used to calculate the equilibrium constant (K). ATP hydrolysis generates adenosine diphosphate (ADP) and an inorganic phosphate (Pi), releasing a specific amount of free energy. Here, the reaction is given as ATP + H2O → ADP + Pi + Energy, with the reaction releasing -30.5 kJ/mol under standard conditions.

However, the student's question asks for the calculation of K given only the energy released for the formation of ADP; additional information such as the concentrations of the reactants and products at equilibrium is needed to calculate K. In general, the equilibrium constant K can be calculated from the standard free energy change (ΔG°) using the formula K = e^(-ΔG°/RT), where R is the ideal gas constant and T is the temperature in Kelvin. Without all the necessary data, calculating K is not possible.

In biological systems, ATP is often referred to as the energy currency of the cell due to its role in energy transfer during cellular processes. The release of energy during the ATP hydrolysis reaction is used for energy-coupling processes, enabling the cell to perform work, like operating transmembrane ion pumps or other endergonic reactions.

It is worth noting that while -30.5 kJ/mol is the value under standard conditions, in living cells, the free energy change for ATP hydrolysis can be as high as -57 kJ/mol. This difference is due to the non-standard conditions within the cell, emphasizing that the actual operating environment can greatly impact biochemical reactions.

User WiatroBosy
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