Question

The average adult consumes approximately 11,700 kJ per day. Assuming that the metabolic pathways leading to ATP synthesis operate at 50% thermodynamic efficiency, about 5850 kJ ends up in the form of synthesized ATP.

Imagine that creatine phosphate, rather than ATP, is the universal energy carrier molecule in the human body. Assume that the cellular concentrations of creatine phosphate, creatine, and phosphate are 19.7 mM, 1.97×10^-3 mM, and 5.80 mM, respectively. Calculate the weight of creatine phosphate that would need to be consumed each day by a typical adult human if creatine phosphate could not be recycled.

(Estimate the free energy of hydrolysis of creatine phosphate under cellular conditions to determine how many moles required. Use the standard free energy ΔG° = -43.3 kJ/mol, and take the temperature to be 37 °C.)

ΔG =  kJ/mol

Weight of creatine phosphate consumed =  g

Answer 1

Hydyrolysis reaction

phosphocreatine + h20 ---> phospahte + creatine delta G ^o= -43 kJ/mol

now delta G_rxn = G^o + RT ln [phosphate] *[creatine]/[phosphocreatine] = -43 kJ mol^-1+8.314 J K^-1 mol^-1*298K* ln (1.97×10^-3 mM * 5.80 mM / 19.7 mM) = -43 -35.578 kJ/mol = -78.578 kJ/mol

Assuming 100 % efficiency , moles requried for human body = 11700kJ / -78.578J mol^-1 = 148.89 moles required

Weight of creatine phosphate consumed = (Molar mass)* moles =  211.11 g·mol^{−1 *}148.89 mol = 31433.56 grams