Answer :
The question is incomplete, complete question is;
Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction:
[tex]HbO_2(aq) + CO(aq)\rightleftharpoons HbCO(aq) + O_2(aq)[/tex]
Use the reactions and associated equilibrium constants at body temperature to find the equilibrium constant for the above reaction.
[tex]Hb(aq) + O_2(aq)\rightleftharpoons HbO_2(aq) ,K_1=1.8[/tex]
[tex]Hb(aq) + CO(aq)\rightleftharpoons HbCO(aq) ,K_2=306[/tex]
Answer:
The equilibrium constant for the given reaction is 170.
Explanation:
[tex]Hb(aq) + O_2(aq)\rightleftharpoons HbO_2(aq) ,K_1=1.8[/tex]
[tex]K_1=\frac{[HbO_2]}{[Hb][O_2]}[/tex]
[tex][Hb]=\frac{[HbO_2]}{[K_1][O_2]}[/tex]..[1]
[tex]Hb(aq) + CO(aq)\rightleftharpoons HbCO(aq) ,K_2=306[/tex]
[tex]K_2=\frac{[HbCO]}{[Hb][CO]}[/tex]..[2]
[tex]HbO_2(aq) + CO(aq)\rightleftharpoons HbCO(aq) + O_2(aq)[/tex]
[tex]K_c=\frac{[HbCO][O_2]}{[HbO_2][CO]}[/tex]..[3]
Using [1] in [2]:
[tex]K_2=\frac{[HbCO]}{\frac{[HbO_2]}{[K_1][O_2]}\times [CO]}[/tex]
[tex]K_2=K_1\times \frac{[HbCO][O_2]}{[HbO_2][CO]}[/tex]
[tex]K_2=K_1\times K_c[/tex] ( using [3])
[tex]306=1.8\times K_c[/tex]
[tex]K_c=\frac{306}{1.8}=170[/tex]
The equilibrium constant for the given reaction is 170.