/*! 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 89 Carbon monoxide is toxic because... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 89

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: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$

Short Answer

Expert verified
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 by step solution

01

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.
02

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 \(1.10 \times 10^3\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

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掳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 O鈧� 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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the following reaction at \(800 . \mathrm{K}:\) $$\mathrm{N}_{2}(g)+3 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g)$$ An equilibrium mixture contains the following partial pressures: \(P_{\mathrm{N}_{2}}=0.021 \mathrm{atm}, P_{\mathrm{F}_{2}}=0.063 \mathrm{atm}, P_{\mathrm{NF}_{3}}=0.48\) atm. Calculate \(\Delta G^{\circ}\) for the reaction at \(800 .\) K.

Consider the reaction $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ where \(\Delta H^{\circ}=-103.8 \mathrm{kJ} / \mathrm{mol} .\) In a particular experiment, equal moles of \(\mathrm{H}_{2}(g)\) at 1.00 atm and \(\mathrm{Br}_{2}(g)\) at 1.00 atm were mixed in a 1.00-L flask at \(25^{\circ} \mathrm{C}\) and allowed to reach equilibrium. Then the molecules of \(\mathrm{H}_{2}\) at equilibrium were counted using a very sensitive technique, and \(1.10 \times 10^{13}\) molecules were found. For this reaction, calculate the values of \(K, \Delta G^{\circ},\) and \(\Delta S^{\circ} .\)

What information can be determined from \(\Delta G\) for a reaction? Does one get the same information from \(\Delta G^{\circ},\) the standard free energy change? \(\Delta G^{\circ}\) allows determination of the equilibrium constant \(K\) for a reaction. How? How can one estimate the value of \(K\) at temperatures other than \(25^{\circ} \mathrm{C}\) for a reaction? How can one estimate the temperature where \(K=1\) for a reaction? Do all reactions have a specific temperature where \(K=1 ?\)

a. Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{\mathrm{f}} / k_{\mathrm{r}},\) where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{\mathrm{a}} / R T}\right).\) b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction.鈥�

Given the following data: $$2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\Delta G^{\circ}=-6399 \mathrm{kJ}$$ $$\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta G^{\circ}=-394 \mathrm{kJ}$$ $$\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta G^{\circ}=-237 \mathrm{kJ}$$ calculate \(\Delta G^{\circ}\) for the reaction $$6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$$
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.