Glucose, C6H12O6, is readily metabolized in the body. — Thermodynamics Chemistry Question
Problem Context
Glucose, C6H12O6, is readily metabolized in the body.
Write a balanced equation for the metabolism of C6H12O6 to CO2 and H2O.
Model Answer
C6H12O6 + 6O2 → 6CO2 + 6H2O
Calculate ∆Go metabolism for glucose. Given: The free energy of formation, ∆Gfo , is –917 kJ·mol–1 for C6H12O6(s); –394.4 kJ·mol–1 for CO2(g); –237.2 kJ·mol–1 for H2O(l).
Model Answer
∆Gmetabolism o = 6∆GCO2 o + 6∆GH2 O o – ∆GC6 H12 O6 o
= 6 mol(−394.4 kJ ⋅mol –1 )+ 6 mol(–237.2 kJ ⋅ mol–1) – (–917kJ ⋅ mol–1)
= –2366.4 k J – 1 4 2 3 . 2 k J+ 9 1 7 k J
= –2873kJ
If ∆Ho for this process is –2801.3 kJ, calculate ∆So at 25 °C.
Model Answer
∆Go = ∆Ho – T∆So
–2873kJ= –2801.3 k J – 2 9 8K∆So
( 2 9 8 K )∆So = 72 kJ
∆So = 0.24 kJ / K or 240 J/K
One step in the utilization of energy in cells is the synthesis of ATP4– from ADP3– and H2PO4–, according to this equation.
ADP3– + H2PO4– → ATP4– ∆Go = 30.5 kJ·mol–1
i. Calculate the number of moles of ATP4– formed by the metabolism of 1.0 g of glucose.
Model Answer
1.0 g C6H12O6 × 1 mol / 180 g = 5.6× 10–3 m o l C6H12O6
5.6 ×10–3 mol C6H12O6 × 2872.6 kJ / mol = 16 kJ
16 kJ × 1 mol ATP / 30.5 kJ = 0.52 mol ATP formed
ii. Calculate the equilibrium constant, K, for the formation of ATP4– at 25 ˚C.
Model Answer
∆Go = –RT ln K
30.5 ×103J / mol = (–8.314 J / mol ⋅K) (298 K)ln K
ln K = –12.31
K = 4.5 ×10–6