Question

Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction: HbO2(aq) + CO(aq) ∆ HbCO(aq) + O2(aq) a. Use the reactions and associated equilibrium constants at body temperature to find the equilibrium constant for the above reaction.

Answer 1

The question is incomplete, complete question is;

Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction:

$HbO_2(aq) + CO(aq)\rightleftharpoons HbCO(aq) + O_2(aq)$

Use the reactions and associated equilibrium constants at body temperature to find the equilibrium constant for the above reaction.

$Hb(aq) + O_2(aq)\rightleftharpoons HbO_2(aq) ,K_1=1.8$

$Hb(aq) + CO(aq)\rightleftharpoons HbCO(aq) ,K_2=306$

Answer:

The equilibrium constant for the given reaction is 170.

Explanation:

$Hb(aq) + O_2(aq)\rightleftharpoons HbO_2(aq) ,K_1=1.8$

$K_1=\frac{[HbO_2]}{[Hb][O_2]}$

$[Hb]=\frac{[HbO_2]}{[K_1][O_2]}$..[1]

$Hb(aq) + CO(aq)\rightleftharpoons HbCO(aq) ,K_2=306$

$K_2=\frac{[HbCO]}{[Hb][CO]}$..[2]

$HbO_2(aq) + CO(aq)\rightleftharpoons HbCO(aq) + O_2(aq)$

$K_c=\frac{[HbCO][O_2]}{[HbO_2][CO]}$..[3]

Using [1] in [2]:

$K_2=\frac{[HbCO]}{\frac{[HbO_2]}{[K_1][O_2]}\times [CO]}$

$K_2=K_1\times \frac{[HbCO][O_2]}{[HbO_2][CO]}$

$K_2=K_1\times K_c$ ( using [3])

$306=1.8\times K_c$

$K_c=\frac{306}{1.8}=170$

The equilibrium constant for the given reaction is 170.