Problem 89 - Chemistry An Atoms First Approach

Question

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does O_2. Consider the following reactions and approximate standard free energy changes: $$\begin{array}{cl} \mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} & \Delta G^{\circ}=-70 \mathrm{kJ} \\ \mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} & \Delta G^{\circ}=-80 \mathrm{kJ} \end{array}$$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$

Step-by-Step Solution

Answer

The estimated equilibrium constant value at \(25^{\circ}\mathrm{C}\) for the given reaction \(\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}\) is \(1.10 \times 10^3\).

Step 1: Calculate the standard free energy change for the given reaction

We are given the standard free energy changes for reactions (1) and (2) as follows: (1) \(\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2}, \Delta G^{\circ}_1=-70 \mathrm{kJ}\). (2) \(\mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO}, \Delta G^{\circ}_2=-80 \mathrm{kJ}\). We want to find the standard free energy change for the following reaction: \[ \mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}. \] To get the desired reaction, we have to reverse reaction (1) and add it to reaction (2): \(-1 \times \mathrm{(1)}\) : \[\mathrm{HgbO}_{2} \longrightarrow \mathrm{Hgb}+\mathrm{O}_{2}, \Delta G^{\circ}_1'=70 \mathrm{kJ}\]. \(+1 \times \mathrm{(2)}\) : \[\mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO}, \Delta G^{\circ}_2=-80 \mathrm{kJ}\]. Adding the two reactions: \[\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}, \Delta G^{\circ}=\Delta G^{\circ}_1'+\Delta G^{\circ}_2=70 \mathrm{kJ}- 80 \mathrm{kJ} = -10 \mathrm{kJ}\]. The standard free energy change for the given reaction is -10 kJ.

Step 2: Calculate the equilibrium constant

The relation between standard free energy change and the equilibrium constant is given by: \[\Delta G^{\circ}=-RT \ln K\] Here, \(R = 8.3145 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}\) is the gas constant, \(T = 25^{\circ}\mathrm{C} = 298\mathrm{K}\) is the temperature, and \(K\) is the equilibrium constant. We need to find the value of \(K\). First, convert the standard free energy change from kJ to J: \[\Delta G^{\circ}=-10\ \mathrm{kJ} = -10,000\ \mathrm{J}\]. Now, rearrange the equation and solve for \(K\): \[K = e^{-\frac{\Delta G^{\circ}}{RT}} = e^{\frac{10,000\ \mathrm{J}}{(8.3145\ \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}})(298\ \mathrm{K})}} \approx 1.10 \times 10^3\]. The estimated equilibrium constant value at \($25^{\circ}\mathrm{C}\) for the given reaction is \(1.10 \times 10^3\).

Key Concepts

Standard Free Energy Change

The concept of Standard Free Energy Change, denoted as \( \Delta G^{\circ} \), is crucial in determining the spontaneity of a chemical reaction. It represents the change in free energy when a reaction occurs under standard state conditions (1 bar pressure and concentrations of 1 mol/L for all substances involved, typically at 25°C).

When \( \Delta G^{\circ} \) is negative, the reaction is spontaneous, meaning it can occur without external energy input. Conversely, a positive \( \Delta G^{\circ} \) indicates a non-spontaneous reaction, requiring energy to proceed. In the case of hemoglobin discussed, the standard free energy changes for the reactions favor the formation of \( \text{HgbCO} \) over \( \text{HgbO}_2 \).

• \( \Delta G^{\circ} = -70 \text{ kJ} \) for the formation of \( \text{HgbO}_2 \)
• \( \Delta G^{\circ} = -80 \text{ kJ} \) for the formation of \( \text{HgbCO} \)

In thermodynamics, the most stable state is the one with the lowest free energy, explaining why hemoglobin tends to bind more to CO than O₂ under these conditions.

Equilibrium Constant

The equilibrium constant, \( K \), is a number that expresses the ratio of the concentrations of products to reactants at equilibrium, with each concentration raised to the power of its stoichiometric coefficient. It provides insight into the position of equilibrium and the extent of a reaction under given conditions.

The relationship between \( \Delta G^{\circ} \) and \( K \) is described by the equation \( \Delta G^{\circ} = -RT \ln K \), where \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. A reaction with a large \( K \) value has products favored at equilibrium, while a small \( K \) means reactants are favored.

In this context, using \( \Delta G^{\circ} = -10,000 \text{ J} \), the equilibrium constant is calculated as approximately \( 1.10 \times 10^3 \). This value indicates the reaction strongly favors the formation of \( \text{HgbCO} \) over \( \text{HgbO}_2 \) under standard conditions.

Thermodynamics

Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy. It aims to describe how energy conversion affects matter. In the context of chemical reactions, it helps to understand and predict the behavior of reactions in terms of energy changes.

The principles of thermodynamics apply universally, including biological systems like hemoglobin's reaction with oxygen and carbon monoxide. Key thermodynamic terms include:

• **System**: The part of the universe being studied, often isolated from the surroundings. In this example, the system could be the hemoglobin molecule reacting with gases.
• **Surroundings**: Everything outside the system. This often includes the medium in which a reaction is occurring.
• **Open, Closed, and Isolated Systems**: These classifications define how energy and matter can exchange with the surroundings.

Understanding thermodynamic concepts allows us to analyze energy transfer, helping to determine reaction feasibility by calculating the standard free energy changes and equilibria.