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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: $$\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} \quad \Delta G^{\circ}=-70 \mathrm{kJ}$$ $$\mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} \quad \Delta G^{\circ}=-80 \mathrm{kJ} $$ 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

Expert verified
The equilibrium constant (K) for the reaction HgbO2 + CO ⇌ HgbCO + O2 at 25°C is approximately 56.64.

Step by step solution

01

Calculate the standard free energy change for the desired reaction

We start by calculating the standard free energy change for the reaction of interest: HgbO2 + CO ⇌ HgbCO + O2. We can do this by manipulating the given reactions as follows: 1. Reverse the first reaction: HgbO2 → Hgb + O2; ∆G° = +70 kJ 2. Add the reversed first reaction to the second reaction: (Hgb + O2) + (Hgb + CO) → (HgbO2) + (HgbCO) Now let's calculate the new standard free energy change for the reaction: ∆G°(new) = ∆G°(first reaction, reversed) + ∆G°(second reaction) ∆G°(new) = (+70 kJ) + (-80 kJ) ∆G°(new) = -10 kJ
02

Calculate the equilibrium constant using the standard free energy change

Now that we have the standard free energy change for the desired reaction, we can calculate the equilibrium constant (K) using the following formula: \[ K = e^{(-\Delta G^{\circ} / RT)} \] where R is the gas constant (8.314 J/K·mol) and T is the temperature in Kelvin (25°C = 298K). Plugging the values into the formula, we get: \[ K = e^{(-(-10,000) / (8.314 \times 298))} \]
03

Calculate the equilibrium constant

By simplifying the expression, we obtain the value of the equilibrium constant: \[ K = e^{(10,000 / 2477.612)} \] \[ K = e^{(4.036)} \] \[ K \approx 56.64 \] Therefore, the equilibrium constant (K) for the reaction HgbO2 + CO ⇌ HgbCO + O2 at 25°C is approximately 56.64.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Free Energy Change
When we talk about the standard free energy change, we're discussing how energy varies during a chemical reaction under standard conditions. It’s represented by the symbol \( \Delta G^{\circ} \) and usually expressed in kilojoules per mole (kJ/mol). It measures the maximum reversible work a system can perform at constant temperature and pressure. If \( \Delta G^{\circ} \) is negative, it means the reaction is spontaneous and will likely proceed without input energy. A positive \( \Delta G^{\circ} \) suggests that energy needs to be added for the reaction to go forward. In our exercise, the reactions involving hemoglobin have different \( \Delta G^{\circ} \) values:
  • From \( \text{Hgb} + \text{O}_2 \rightarrow \text{HgbO}_2 \), \( \Delta G^{\circ} = -70\,\text{kJ} \)
  • From \( \text{Hgb} + \text{CO} \rightarrow \text{HgbCO} \), \( \Delta G^{\circ} = -80\,\text{kJ} \)
These negative values indicate that both formations of hemoglobin compounds are spontaneous under standard conditions.
Hemoglobin
Hemoglobin is a protein in your blood responsible for transporting oxygen from the lungs to the rest of your body and carrying carbon dioxide back to the lungs for expulsion. It’s composed of heme groups that contain iron, which allows it to bind to oxygen molecules effectively. The interaction with other gases can affect its normal function. When carbon monoxide (CO) binds to hemoglobin, forming carboxyhemoglobin (HgbCO), it hinders the ability of hemoglobin to carry oxygen, leading to potential health risks.Carbon monoxide binds to hemoglobin about 240 times more strongly than oxygen (\( \text{O}_2 \)), causing severe implications when inhaled in significant amounts. This strong binding affinity results in fewer hemoglobin molecules being available for oxygen transport.
Toxic Gas Interaction
Carbon monoxide is considered a toxic gas due to its detrimental effects on oxygen transport in the bloodstream. As mentioned, it binds strongly to hemoglobin, leading to the formation of HgbCO, and this formation is much stronger than the binding of oxygen, with a longer-lasting interaction. The high affinity and binding with hemoglobin reduce the overall capacity of red blood cells to carry oxygen. This means that even small amounts of CO can have significant impacts on the body, resulting in symptoms of CO poisoning: headache, dizziness, and even death in extreme cases. Cleaning agents, vehicle exhaust systems, and smoking are common sources of carbon monoxide. Precautionary measures, such as adequate ventilation and regular checking of potential CO emission sources, are essential for safety.
Reaction Manipulation
Reaction manipulation involves changing the path a chemical process takes, or its endpoint, by altering the conditions or components involved in a reaction. This can involve reversing reactions, combining them, or even altering conditions such as temperature or concentration.In the exercise provided, we manipulated reactions to find the desired equilibrium constant.
  • The first reaction, \( \text{HgbO}_2 \rightarrow \text{Hgb} + \text{O}_2 \), was reversed to change its \( \Delta G^{\circ} \) to positive.
  • This reversed reaction was then added to the second reaction \( \text{Hgb} + \text{CO} \rightarrow \text{HgbCO} \) to compute the overall \( \Delta G^{\circ} \) of the new, desired reaction.
Such manipulations help in calculating the values like the equilibrium constant, \( K \), guiding us whether the reaction favors products or reactants under standard conditions, and potential ways to shift the equilibrium if needed.

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Most popular questions from this chapter

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

True or false: High temperatures are favorable to a reaction both kinetically and thermodynamically. Explain.

Which of the following reactions (or processes) are expected to have a negative value for \(\Delta S^{\circ} ?\) a. \(\mathrm{SiF}_{6}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)+\mathrm{SiF}_{4}(g)\) b. \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\) c. \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)\) d. \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) e. \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\)

In the text, the equation $$\Delta G=\Delta G^{\circ}+R T \ln (Q)$$ was derived for gaseous reactions where the quantities in \(Q\) were expressed in units of pressure. We also can use units of mol/L for the quantities in \(Q\)specifically for aqueous reactions. With this in mind, consider the reaction $$\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)$$ for which \(K_{\mathrm{a}}=7.2 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) . Calculate \(\Delta G\) for the reaction under the following conditions at \(25^{\circ} \mathrm{C} .\) a. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \mathrm{M}\) b. \([\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M\) c. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} \mathrm{M}\) d. \([\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M\) e. \([\mathrm{HF}]=0.52 M,\left[\mathrm{F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} \mathrm{M}\) Based on the calculated DG values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?

Human DNA contains almost twice as much information as is needed to code for all the substances produced in the body. Likewise, the digital data sent from Voyager II contained one redundant bit out of every two bits of information. The Hubble space telescope transmits three redundant bits for every bit of information. How is entropy related to the transmission of information? What do you think is accomplished by having so many redundant bits of information in both DNA and the space probes?

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