What Is the Delta G for Creatine Phosphate Hydrolysis at 1 mM Concentration?

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SUMMARY

The delta G for the hydrolysis of creatine phosphate at a concentration of 1 mM for creatine phosphate, creatine, and phosphate at 25 degrees Celsius is determined to be between -43 kJ/mol and 0 kJ/mol. This conclusion is based on the standard delta G naught value of -43 kJ/mol for the reaction. Under these specific cellular conditions, the actual delta G will be influenced by the concentrations of the reactants and products, leading to the conclusion that -43 kJ/mol < delta G < 0 kJ/mol is the correct answer.

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  • Understanding of Gibbs free energy and its significance in biochemical reactions
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  • Basic principles of thermodynamics as they apply to biological systems
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Question:

Which best describes the delta G for hydrolysis of creatine phosphate under cellular conditions in which the concentration of creatine phosphate, creatine, and phosphate all equal 1 mM at 25 degrees C. The delta G naught for the hydrolysis of creatine phosphate at 25 degrees C is -43kJ/mol

a) delta G < -43kJ/mol
b) delta G = -43kJ/mol
c) delta G > 0kJ/mol
d) -43kJ/mol < delta G < 0 kJ/mol



This was a question on my last biochem exam. I was so confused by this question, as were many of my classmates. My professor is quite possibly the worst I have encountered, and left us further confused when she tried to explain it. I got a high A on the exam, but I would be grateful if someone could explain this :)
 
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