Question

Carbon monoxide (CO) is toxic because it binds more strongly to the iron in hemoglobin (Hb) than does oxygen (O2), as indicated by these approximate standard free-energy changes in blood:
reaction A:reaction B:Hb+O2Hb+CO⟶⟶HbO2,HbCO, ΔG∘=−70 kJ/mol ΔG∘=−80 kJ/mol

Estimate the equilibrium constant K at 298 K for the equilibrium


HbO2+CO⇌HbCO+O2

Answer 1

Answer : The value of equilibrium constant K at 298 K is, 56.59

Explanation :

The given chemical reaction are:

(1) $Hb+O_2\rightarrow HbO_2$;     $\Delta G^o_1=-70kJ/mol$

(2) $Hb+CO\rightarrow HbCO$;     $\Delta G^o_2=-80kJ/mol$

First we have to determine the standard free-energy change for the following reaction.

(3) $HbO_2+CO\rightarrow HbCO+O_2$;     $\Delta G^o_3=?$

Now we are reversing the reaction 1 and then adding reaction 1 and 2, we get:

(1) $HbO_2\rightarrow Hb+O_2$;     $\Delta G^o_1=+70kJ/mol$

(2) $Hb+CO\rightarrow HbCO$;     $\Delta G^o_2=-80kJ/mol$

$\Delta G^o_3=\Delta G^o_1+\Delta G^o_2$

$\Delta G^o_3=+70+(-80)=-10kJ/mol$

Now we have to calculate the equilibrium constant K at 298 K.

$\Delta G^o=-RT\times \ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy  = -10kJ/mol = -10000 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 298 K

$K_{eq}$  = equilibrium constant  = ?

Now put all the given values in the above formula, we get:

$-10000J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}$

$K_{eq}=56.59$

Therefore, the value of equilibrium constant K at 298 K is, 56.59