Chapter 13: Problem 24
For the reaction \(\mathrm{CO}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}, \Delta G^{\circ}=-257 \mathrm{~kJ}\) per mole of \(\mathrm{CO}\) at \(298 \mathrm{~K}\). Find \(E^{\circ}\) and the equilibrium constant for the reaction.
Short Answer
Expert verified
\( K \approx 2.03 \times 10^{45} \), \( E^{\circ} \approx 1.33 \text{ V} \).
Step by step solution
01
Understand Gibbs Free Energy Change
Gibbs free energy change, \( \Delta G^{\circ} \), helps determine the spontaneity of a chemical reaction. Its standard value at \(298 \text{ K}\) is given as \(-257 \text{ kJ} \) per mole for this reaction.
02
Use Gibbs Free Energy Equation
The relationship between Gibbs free energy change and the equilibrium constant \( K \) is given by the equation \( \Delta G^{\circ} = -RT \ln K \) where \( R = 8.314 \text{ J/mol K} \) and \( T = 298 \text{ K} \). First, convert \( \Delta G^{\circ} \) to J/mol: \(-257,000 \text{ J/mol}\).
03
Calculate Equilibrium Constant
Rearrange the formula to solve for \( K \): \[\ln K = -\frac{\Delta G^{\circ}}{RT} \Rightarrow K = e^{-\frac{-257,000}{8.314 \times 298}}.\] Evaluate \( K \) to find its value.
04
Evaluate Natural Logarithm and Exponential
Compute the exponent: \[ \frac{257,000}{8.314 \times 298} \approx 104.01.\]Thus, \[ K = e^{104.01}.\]Calculate \( K \) to find \( e^{104.01} \approx 2.03 \times 10^{45}.\)
05
Understand the Standard Cell Potential
The standard cell potential \( E^{\circ} \) for a galvanic cell can be related using \( \Delta G^{\circ} = -nFE^{\circ} \), where \( n \) is the number of moles of electrons and \( F \) is Faraday's constant (96485 C/mol). Here \( n = 2 \) because CO to COâ‚‚ involves two electrons.
06
Calculate Standard Cell Potential
Rearrange to find \( E^{\circ} \): \[ E^{\circ} = -\frac{\Delta G^{\circ}}{nF} = -\frac{-257000}{2 \times 96485} \text{ V}. \]Calculate \( E^{\circ} \approx 1.33 \text{ V} \).
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gibbs Free Energy
Gibbs Free Energy, denoted as \( \Delta G \), is an incredibly useful concept in chemistry to predict the likelihood of a reaction occurring spontaneously. It combines enthalpy, entropy, and temperature into one single value. If \( \Delta G \) is negative, it typically indicates that the process can occur on its own, without additional energy.
For standard conditions, which include 1 atm pressure and 298 K temperature, this value is symbolized as \( \Delta G^{\circ} \). In the reaction of carbon monoxide with oxygen to form carbon dioxide, \( \Delta G^{\circ} \) is given as \(-257 \text{ kJ/mol}\). This shows that under these conditions, the reaction is favorable, meaning it can happen spontaneously.
For standard conditions, which include 1 atm pressure and 298 K temperature, this value is symbolized as \( \Delta G^{\circ} \). In the reaction of carbon monoxide with oxygen to form carbon dioxide, \( \Delta G^{\circ} \) is given as \(-257 \text{ kJ/mol}\). This shows that under these conditions, the reaction is favorable, meaning it can happen spontaneously.
- Negative \( \Delta G \, = \) spontaneous reaction
- Positive \( \Delta G \, = \) non-spontaneous reaction
Standard Cell Potential
The Standard Cell Potential, represented as \( E^{\circ} \), is a measure of the voltage or electric potential difference of a cell under standard conditions. It shows how strong a particular redox reaction is in converting chemical energy into electrical energy.
In the case of our carbon monoxide and oxygen reaction, the relevant equation is \( \Delta G^{\circ} = -nFE^{\circ} \), linking Gibbs Free Energy to the cell potential. Here, \( n \) is the number of moles of electrons exchanged (which is 2), and \( F \) is Faraday's Constant \( 96485 \text{ C/mol} \). After rearranging and inserting values, we calculate \( E^{\circ} \approx 1.33 \text{ V}\), indicating a strong ability to drive the reaction under standard conditions.
In the case of our carbon monoxide and oxygen reaction, the relevant equation is \( \Delta G^{\circ} = -nFE^{\circ} \), linking Gibbs Free Energy to the cell potential. Here, \( n \) is the number of moles of electrons exchanged (which is 2), and \( F \) is Faraday's Constant \( 96485 \text{ C/mol} \). After rearranging and inserting values, we calculate \( E^{\circ} \approx 1.33 \text{ V}\), indicating a strong ability to drive the reaction under standard conditions.
Chemical Reaction Spontaneity
A reaction's spontaneity tells us whether it happens by itself or if it needs external energy. Spontaneity is mainly determined by \( \Delta G \), which, as mentioned, when negative indicates that a reaction is spontaneous.
Let’s connect this with equilibrium. The spontaneous conversion of CO and \( \text{O}_2 \) into \( \text{CO}_2 \) has a highly favorable equilibrium constant, \( K \), calculated from the equation \( \Delta G^{\circ} = -RT\ln K\). The very large \( K \approx 2.03 \times 10^{45} \) in this case directly points to a reaction that heavily favors product formation, demonstrating its high spontaneity under standard conditions.
Let’s connect this with equilibrium. The spontaneous conversion of CO and \( \text{O}_2 \) into \( \text{CO}_2 \) has a highly favorable equilibrium constant, \( K \), calculated from the equation \( \Delta G^{\circ} = -RT\ln K\). The very large \( K \approx 2.03 \times 10^{45} \) in this case directly points to a reaction that heavily favors product formation, demonstrating its high spontaneity under standard conditions.
- \( \Delta G < 0 \) and large \( K \) \( = \) spontaneous
- \( \Delta G > 0 \) and small \( K \) \( = \) non-spontaneous
Faraday's Constant
Faraday's Constant, denoted as \( F \), is a central figure in electrochemistry representing the charge of one mole of electrons. Its value is approximately \( 96485 \text{ C/mol} \).
In the context of the standard cell potential, Faraday's Constant helps convert Gibbs Free Energy change into potential energy of a cell. Here, it's used in the equation \( \Delta G^{\circ} = -nFE^{\circ} \). Given \( n \) equals 2 for the CO and \( \text{O}_2 \) reaction, Faraday’s Constant allows us to compute \( E^{\circ} \) efficiently by providing a link between electric and chemical energies.
In the context of the standard cell potential, Faraday's Constant helps convert Gibbs Free Energy change into potential energy of a cell. Here, it's used in the equation \( \Delta G^{\circ} = -nFE^{\circ} \). Given \( n \) equals 2 for the CO and \( \text{O}_2 \) reaction, Faraday’s Constant allows us to compute \( E^{\circ} \) efficiently by providing a link between electric and chemical energies.
- \( F = 96485 \text{ C/mol}\)
- Connects electron transfer processes to energy calculations