Oxygen is of vital importance for all of us . Oxygen enters the body via the lungs and is transported to the tissues in our body by blood . There it can deliver energy by the oxidation of sugars.
`C_(6)H_(12)O_(6) + 6O_(2) rarr 6CO_(2) + 6H_(2)O`
This reaction releases 400 KJ of energy per mole of oxygen `O_(2)` uptake by blood is at four heme (Hm) group in this protein hemoglobin (Hb).
Free Hm consists of an `Fe^(2+)` giving `HmO_(2)` complex. Carbon monoxides can be complexed similarily giving a Hm CO complex . CO is poison as it bonds more strongly to Hm than `O_(2)` does. The equilibrium constant `K_(f)` for the reaction:
`Hm+ CO hArr HCO " "........(i)`
is 1000 times larger than the equilibrium constant `K_(2)` for the reaction:
`Hm + CO_(2) hArr HmO_(2)" " ........(ii)`
Each Hb molecules can take up four molecules of `O_(2)` absorbs a fraction of this amount, depending on the oxygen pressure , as shown in figure1 (curve 1) . Also shown are the curve (2) and (3) for blood with two kinds of dificient Hb . These occur in patients with certain hereditary diseases.

Relevant data , `O_(2)` pressure in lungs is 15 KPa , in the muscles it is 2KPa . The maximum flow of blood through heart and lungs is `4 xx 10^(-4)m^(-3)s^(-1)` . The red cells in blood occupy 40% of the volume, inside the cells the concentration of Hb has a molar mass of 64 kg `"mol"^(-1)` R=8.314 ` J "mol"^(-1) K^(-1)` , T=298k . Using the relation between K and the standard Gibbs energy `DeltaG^(@)` for a reaction, calculated the difference between the `DeltaG^(@)` values for the home reactions (i) and (ii).

Answer

`DeltaG_(1)^(@) =- RT "In" K_(1)" "DeltaG_(2)^(@) =-RT "In"K_(2)`
`DeltaG_(2)^(@) - DeltaG_(1)^(@) = DeltaG_(1)^(@) = RT "In" (K_(1))/(K_(2))`
`DeltaG_(2)^(@) - DeltaG_(1)^(@) = (8.314 J "mol"^(-1) K^(-1) xx 298k xx "In" 10000) J =23 KJ "mol^(-1)`