/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 99 Carbon monoxide is toxic because... [FREE SOLUTION] | 91影视

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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 鈬 HgbCO + O2 at 25掳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 鈬 HgbCO + O2. We can do this by manipulating the given reactions as follows: 1. Reverse the first reaction: HgbO2 鈫 Hgb + O2; 鈭咷掳 = +70 kJ 2. Add the reversed first reaction to the second reaction: (Hgb + O2) + (Hgb + CO) 鈫 (HgbO2) + (HgbCO) Now let's calculate the new standard free energy change for the reaction: 鈭咷掳(new) = 鈭咷掳(first reaction, reversed) + 鈭咷掳(second reaction) 鈭咷掳(new) = (+70 kJ) + (-80 kJ) 鈭咷掳(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路mol) and T is the temperature in Kelvin (25掳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 鈬 HgbCO + O2 at 25掳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鈥檚 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鈥檚 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

Ethanethiol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH} ; \text { also called ethyl mercaptan) is }\right.\) commonly added to natural gas to provide the 鈥渞otten egg鈥 smell of a gas leak. The boiling point of ethanethiol is \(35^{\circ} \mathrm{C}\) and its heat of vaporization is 27.5 \(\mathrm{kJ} / \mathrm{mol}\) . What is the entropy of vaporization for this substance?

You have a 1.00 -L sample of hot water \(\left(90.0^{\circ} \mathrm{C}\right)\) sitting open in a \(25.0^{\circ} \mathrm{C}\) room. Eventually the water cools to \(25.0^{\circ} \mathrm{C}\) while the temperature of the room remains unchanged. Calculate \(\Delta S_{\mathrm{sur}}\) for this process. Assume the density of water is 1.00 \(\mathrm{g} / \mathrm{cm}^{3}\) over this temperature range, and the heat capacity of water is constant over this temperature range and equal to 75.4 \(\mathrm{J} / \mathrm{K} \cdot\) mol.

At \(100 .^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm}, \Delta H^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol}\) for the vaporization of water. Estimate \(\Delta G^{\circ}\) for the vaporization of water at \(90 .^{\circ} \mathrm{C}\) and \(110 .^{\circ} \mathrm{C}\) . Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) at \(100 .^{\circ} \mathrm{C}\) and 1.00 \(\mathrm{atm}\) do not depend on temperature.

Choose the substance with the larger positional probability in each case. a. 1 mole of \(\mathrm{H}_{2}\) (at \(\mathrm{STP} )\) or 1 mole of \(\mathrm{H}_{2}\left(\text { at } 100^{\circ} \mathrm{C}, 0.5 \mathrm{atm}\right)\) b. 1 mole of \(\mathrm{N}_{2}(\text { at } \mathrm{STP})\) or 1 mole of \(\mathrm{N}_{2}(\text { at } 100 \mathrm{K}, 2.0 \mathrm{atm})\) c. 1 mole of \(\mathrm{H}_{2} \mathrm{O}(s)\) (at \(0^{\circ} \mathrm{C} )\) or 1 \(\mathrm{mole}\) of \(\mathrm{H}_{2} \mathrm{O}(l)\left(\mathrm{at} 20^{\circ} \mathrm{C}\right)\)

For mercury, the enthalpy of vaporization is 58.51 kJ/mol and the entropy of vaporization is 92.92 J/K ? mol. What is the normal boiling point of mercury?

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