Question

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2}\). 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: $$ \mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2} $$

Short answer

For the reaction HgbO₂ + CO ⇌ HgbCO + O₂ at 25°C, the equilibrium constant (K) value is estimated to be approximately 55.6.

Step-by-step solution

Identify the given information

: We are given the following standard free energy changes for two separate reactions: 1. Hgb + O₂ → HgbO₂ with ∆G° = -70 kJ 2. Hgb + CO → HgbCO with ∆G° = -80 kJ And we are asked to determine the equilibrium constant value for the third reaction: 3. HgbO₂ + CO ⇌ HgbCO + O₂

Find the required reaction by combining given reactions

: To find the desired reaction (the third reaction), we need to reverse the first reaction and add it to the second reaction: - Reverse Reaction 1: HgbO₂ → Hgb + O₂ with ∆G° = 70 kJ - Reaction 2: Hgb + CO → HgbCO with ∆G° = -80 kJ - Combine: HgbO₂ + CO ⇌ HgbCO + O₂ with ∆G° = (-80+70) kJ = -10 kJ Now that we have found the ΔG° for the required reaction, we can find the equilibrium constant value.

Determine the equilibrium constant using Gibbs free energy formula

: We can use the following equation to relate the standard free energy change (ΔG°) and the equilibrium constant (K) at a given temperature: ΔG° = -RT ln(K) Where: - ΔG° is the standard free energy change - R is the gas constant (8.314 J/mol*K) - T is the temperature in Kelvin - K is the equilibrium constant First, we need to convert the given temperature from Celsius to Kelvin: T = 25°C + 273.15 = 298.15 K Now, let's calculate the equilibrium constant by rearranging the formula and plugging in the values: ln(K) = -ΔG° / (RT) ln(K) = -(-10 × 10³) / (8.314 × 298.15) ln(K) ≈ 4.02 To find the K value, we need to take the exponential of both sides: K ≈ e⁴.⁰² ≈ 55.6

State the equilibrium constant value for the reaction

: For the reaction HgbO₂ + CO ⇌ HgbCO + O₂ at 25°C, the equilibrium constant (K) value is estimated to be approximately 55.6.

Key concepts

Gibbs free energy

Gibbs free energy is a fundamental concept in thermodynamics, often symbolized by \( G \). It combines enthalpy, entropy, and temperature in a way that helps predict the direction of chemical reactions. The key equation for Gibbs free energy is:\[ G = H - TS \]where:

• \( G \) is the Gibbs free energy
• \( H \) is the enthalpy
• \( T \) is the temperature (in Kelvin)
• \( S \) is the entropy

Gibbs free energy provides insight into whether a reaction is spontaneous. If \( \Delta G < 0 \), the reaction proceeds spontaneously in the forward direction. If \( \Delta G > 0 \), the reaction is non-spontaneous as written. A \( \Delta G = 0 \) indicates a system at equilibrium."
In the case of the hemoglobin reaction, the standard free energy change helps us evaluate the tendency of carbon monoxide to bind more strongly to hemoglobin than oxygen, highlighting its toxicity.

standard free energy change

The standard free energy change, denoted as \( \Delta G^{\circ} \), is a measure of a reaction's tendency under standard conditions, usually 1 atm pressure and 298 K temperature.
It is calculated using the equation:\[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \]where \( \Delta H^{\circ} \) is the standard enthalpy change and \( \Delta S^{\circ} \) is the standard entropy change.
In the provided exercise, the standard free energy changes for the reactions are:

• \( \Delta G^{\circ} = -70 \) kJ for the reaction forming hemoglobin with oxygen.
• \( \Delta G^{\circ} = -80 \) kJ for the reaction forming hemoglobin with carbon monoxide.

The negative values indicate that these reactions have a strong tendency to proceed as written.
By manipulating given conditions and calculations, we were able to assess the tendencies of compounds to form in the presence of others under standard conditions.

temperature conversion

Converting temperature from Celsius to Kelvin is a straightforward but crucial step in thermodynamic calculations. The formula for this conversion is:
\[ T(\text{K}) = T(\text{°C}) + 273.15\]In the exercise, we converted 25°C to Kelvin, resulting in a temperature of 298.15 K. Using Kelvin in calculations is essential because the gas constant \( R \), as used in the Gibbs free energy equation, is in units of J/mol*K.
This ensures consistency in units when solving for equilibrium constants or any thermodynamic property.

chemical equilibrium

At chemical equilibrium, the forward and reverse reactions occur at the same rate, so the concentrations of reactants and products remain constant over time.
The equilibrium constant, \( K \), quantifies this balance and is given by the expression:\[K = \frac{[\text{products}]}{[\text{reactants}]}\]based on the stoichiometry of the reaction.
In the challenge provided, we calculated \( K \) using the relationship between Gibbs free energy change and the equilibrium constant. This was achieved by:

• Using \( \Delta G^{\circ} = -RT \ln(K) \)
• Where \( \Delta G^{\circ} = -10 \) kJ, calculated by combining reactions appropriately
• Determining \( K \) as approximately 55.6, indicating a strong tendency towards product formation at equilibrium.

Understanding these concepts helps predict and manipulate reactions in various fields such as chemistry and biochemistry.