Question

State the answer and show your work. State units in the answer. 10 points total [F = 96.485 kJ/V-mole; R = 8.314 J/mole-K] 1. The standard state free energy change (AG°') of hydrolysis for phosphoenolpyruvate (PEP) is -62.2 kJ/mole. The standard state free energy change of hydrolysis for ATP is -30.5 kJ/mole. What is the standard state free energy change for the following coupled reaction? PEP + ADP ---------> Pyruvate + ATP 2. In glycolysis, the standard state free energy change (AGⓇ) of the phosphofructokinase reaction is - 14.2 kJ/mole. What is the equilibrium constant at 37°C? Round the answer to one decimal place and use scientific notation, ir 545 x 10-6

Answer 1

1. Hydrolysis of PEP P o Pyruvate & Phosphate 10" - 62.2 kJmol. Hydrolysis of APP ATP ADP + Phosphate (80" - 30.5 kJ mol elhen the a reactions are coupled, PEP + ADP — Pyruvate + ATP ". Net standard free energy low -040")

Net standerd fret energy a 4-62.2 -(-30-5)} <3 col? Net standard forer energy: - 31.7 kJm01"| Given. Sola -14.2 kJ/mole T 37°C - 310k Cele know, IR Universal Gas 341 - RT ln K Constant Kay equilibrium Constant Ale know, IR 28.314 I Kn molt 89.-lok RT aft lnk a f 14.2x 103 18.314 x 310

Ink 25.5095 Ka 247.027 1 decimal As we have been asked to answer im place, ka247.0 equilibrium constanta 247.0