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* \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)* (298K)* \ln K_(eq)`

`K_(eq)=56.59`

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