Question

Carbohydrates are metabolized according to: CH2O(aq)+O2(aq)=CO2(aq)+H2O(l) DeltaG= -509 KJ/mol How much energy is available per mole of carbohydrate at typical blood levela of the chemicals? Assume 298 K [CH2O]= 5.0*10^-3 M [CO2]= 1.7*10^-3 M [O2]= 2.5*10^-4 M

Answer 1

The reaction:

CH₂O(aq)+O₂(aq) = CO₂(aq)+H₂O(l)

The data:

• ΔG⁰ = -509 KJ/mol
• [CH₂O]= 5.0x10^-3 M
• [CO₂]= 1.7x10^-3 M
• [O₂]= 2.5x10^-4 M

1) Calculate k using the equilibrium expression:

k = [products] / [reagents]

k = [CO₂] / [CH₂O] [O₂]

k = 1.7x10^-3 M / (5.0x10^-3 M)(2.5x10^-4 M)

k = 1360 M^-1

2) Calculate ΔG (the available energy) using the formula:

ΔG = ΔG⁰ - RT*lnk

Where R = 8.314 J/molK and T = 298K

ΔG = -509 kJ/mol - (8.314 J/molK)(298K)*ln(1360)

ΔG = -509 kJ/mol - 17.9 kJ/mol

ΔG = -526.9 kJ/mol

So, the energy that is available is 526.9 kJ/mol carbohydrate.