Q20.87 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Hemoglobin carries O₂ from the lungs to tissue cells, where the O₂ is released. The protein is represented as Hb in its unoxygenated form and as Hb.O₂ in its oxygenated form. One reason CO is toxic is that it competes with O₂ in binding to Hb: 

$\mathbf{\text{Hb}}\mathbf{\cdot }{\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{\text{CO}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{\text{Hb}}\mathbf{\cdot }\mathbf{\text{CO}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

(a) If $\mathit{\Delta }{\mathit{G}}^{\mathbf{°}}\mathbf{\approx }\mathbf{-}\mathbf{14}\mathbf{ }\mathbf{\text{kJ}}$ at ${\mathbf{37}}^{\mathbf{°}}\mathbf{\text{C}}$ (body temperature), what is the ratio of [Hb.CO] to [Hb.O₂] at ${\mathbf{37}}^{\mathbf{°}}\mathbf{\text{C}}$ with [O₂]=[CO]?

(b) How is Le Châtelier's principle used to treat CO poisoning?

Step-by-Step Solution

Answer

a) The proportion of different haemoglobin types: $\frac{\mid \text{Hb}\cdot \text{CO}]}{[\text{b}\cdot {\text{O}}_2]}=2.286\cdot {10}^2$

b) The higher concentration of oxygen pushes the balance towards creation of the oxygenated Hb form, therefore patients are given oxygen-enriched air.

Step 1: Definition of Delta

It is used in chemistry to indicate a change in enthalpy as well as the addition of heat to the reaction. A triangle symbol is used to represent it.

Step 2: Calculate the ratio of haemoglobin forms

a) 

To find the ratio, we must first express Gibb's energy in terms of the equilibrium constant K:

$\Delta G^{°}=-R\cdot T\cdot$

$InK=\frac{\Delta G^{°}}{-R\cdot T}$

$K=e^{\ln K}$

In this reaction, the equilibrium constant K can be represented as the ratio of products to reactants.

$K=\frac{[Hb\cdot CO]\left[{\text{O}}_2\right]}{[Hb\cdot {\text{O}}_2]\left[\text{CO}\right]}$

Since, [O₂]=[CO], the equation can be simplified as:

$\begin{array}{l}K=\frac{[Hb\cdot CO]\left[CO\right]}{[Hb\cdot O_2]\left[CO\right]}\\ K=\frac{[Hb\cdot CO]}{[Hb\cdot O_2]}\end{array}$

Thus, using the given $\Delta G^{°}$ and temperature, In K can be calculated, where  $KT={37}^{°}\text{C}=310 \text{K}$

$\begin{array}{l}K=\frac{-14.00\cdot {10}^3 \text{J}}{-8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\cdot 310 \text{K}}\\ K=5.432K=e^{5.432}\\ K=228.596=2.286\cdot {10}^2\end{array}$

Thus, the ratio of the haemoglobin forms:

$\frac{[Hb\cdot CO]}{[Hb\cdot {\text{O}}_2]}=2.286\cdot {10}^2$

Step 3: How is Le Châtelier's principle used

b) 

According to the Le Chatelier’s principle, the concentration of the products should be increased to move the equilibrium to the left, toward the oxygenated form of [Hb.O₂] Increasing the oxygen content is a simple approach to do this, as the reaction's equilibrium will be pushed to the left.

Therefore, to reverse the course of CO poisoning, patients are given oxygen-rich air.