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:

CH_{2}O(aq)+O_{2}(aq)
= CO_{2}(aq)+H_{2}O(l)

The data:

- ΔG
^{0}= -509 KJ/mol - [CH
_{2}O]= 5.0x10^{-3}M - [CO
_{2}]= 1.7x10^{-3}M - [O
_{2}]= 2.5x10^{-4}M

1) Calculate k using the equilibrium expression:

k = [products] / [reagents]

k = [CO_{2}] /
[CH_{2}O] [O_{2}]

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 = ΔG^{0} - 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.**

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