Can\'t figure out part B and C. Thank you! C6H12O6(s) + 6O2 -------> 6CO2(g) + 6

Question

Can't figure out part B and C. Thank you!

C6H12O6(s) + 6O2 -------> 6CO2(g) + 6H20(l)

DH=-2816 kj/mol, DS=+181 J/Kmol

1. The combustion of glucose to CO2 and water is a major source of energy in aerobic organisms. It is a reaction favored mainly by a large negative enthalpy change.

a) What is value for DGo at 37 oC?

Here I used DG=DH-TDS

and found DGnaught to be -2872.11

b) In the overall reaction of aerobic metabolism of glucose, 38 mol of ATP is produced from ADP for every mole of glucose oxidized. Calculate the standard state energy change for the overall reaction when glucose oxidation is coupled to the formation of ATP.

c) What is the efficiency of the process in terms of the percentage of the available free energy changed captured in ATP?

Explanation / Answer

b) Use this formula

   DG1 = -nRT lnK

Where n =38 mol

     R = gas constant

     K = equilibrium constant

   T = Absolute temperature = 37 +273 = 310

Find the K value and then use this K value for finding DG2 by using same equation but n value will be different as glucose oxidation is coupled to the formation of ATP

c) Efficiency can find out from (1- DG1/DG2 ) x 100 % (use here numerical value not sign)

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