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Chapter 18: Problem 82

Consider the following reaction carried out at \(1000^{\circ} \mathrm{C}\). $$ \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) $$ Assuming that all gases are at \(1.00 \mathrm{~atm}\), calculate the voltage produced at the given conditions. (Use Appendix 1 and assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with an increase in temperature.)

Short Answer

Expert verified
Question: Calculate the voltage produced by the reaction CO(g) + 1/2 O2(g) -> CO2(g) at 1000°C and a pressure of 1.00 atm, given that ΔH° and ΔS° do not change with an increase in temperature. Answer: To calculate the voltage produced by the reaction at the given temperature and pressure, follow these steps: 1. Write down the given information and note the reaction. 2. Convert the temperature from Celsius to Kelvin: 1000°C + 273.15 = 1273.15 K. 3. Look up the values for the standard heat capacities and entropies for each species in the reaction. Use these values to find ΔH° and ΔS° for the reaction. 4. Calculate ΔG° using the equation: ΔG° = ΔH° - TΔS°. 5. Note that 2 moles of electrons are transferred in the reaction, i.e., n = 2. 6. Use the Nernst equation, E = -ΔG°/(nF), to find the voltage (E) of the reaction. By following these steps and plugging in the necessary values, you can determine the voltage produced by the reaction under the specified conditions.

Step by step solution

01

Write down the given information

The reaction under consideration is: $$ \mathrm{CO}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) $$ The temperature is \(1000^{\circ}C\) and the pressure of all gases is \(1.00 \mathrm{atm}\). We are also given that ΔH° and ΔS° do not change with an increase in temperature.
02

Convert temperature to Kelvin

To ensure we are working in the correct units, we first need to convert the temperature from Celsius to Kelvin. $$ T = 1000^{\circ}C + 273.15 = 1273.15 K $$
03

Find ΔH° and ΔS° from Appendix 1

In this step, we look up the standard heat capacities (\(C_p^{\circ}\)) and entropies (\(S^{\circ}\)) for each species in the reaction. Afterwards, we use these values to calculate the enthalpy change (\(\Delta H^{\circ}\)) and entropy change (\(\Delta S^{\circ}\)) for the reaction.
04

Calculate ΔG° using ΔH° and ΔS°

Once we have found the values for ΔH° and ΔS°, we can now calculate the Gibbs free energy change (ΔG°) for the reaction using the following equation: $$ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} $$ Plug in the values for ΔH°, ΔS°, and T (in Kelvin) to calculate ΔG°.
05

Calculate the number of moles of electrons (n) transferred in the reaction

The balanced equation for the half-reactions is: $$ \mathrm{CO}(g) + \frac{1}{2} \mathrm{O}_{2}(g) + 2e^- \longrightarrow \mathrm{CO}_{2}(g) $$ We can see that 2 moles of electrons are transferred in the reaction. Thus, n = 2.
06

Use the Nernst equation to find the voltage (E) of the reaction

Now, we will use the Nernst equation to calculate the voltage (E) of the reaction, which is as follows: $$ E = -\frac{\Delta G^{\circ}}{nF} $$ Here, \(F\) is the Faraday constant (\(96485 C/mol\)). Plug in the values for ΔG°, n, and F to calculate the voltage produced by the reaction.

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Key Concepts

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

Gibbs Free Energy
Understanding Gibbs free energy is crucial for predicting whether a chemical reaction will occur spontaneously. It represents the balance between enthalpy (heat content) and entropy (degree of disorder) in a system, taking into account the temperature. The formula for Gibbs free energy (GG) is expressed as
\[ \Delta G = \Delta H - T\Delta S \]
where \( \Delta H \) is the enthalpy change, \( T \) is the absolute temperature in Kelvin, and \( \Delta S \) is the entropy change of the reaction. In the exercise, you calculate \( \Delta G \) to determine the driving force of the reaction. A negative \( \Delta G \) indicates a spontaneous process, while a positive \( \Delta G \) suggests that the reaction is non-spontaneous under standard conditions.
Nernst Equation
The Nernst equation is a fundamental equation in electrochemistry that connects the Gibbs free energy change to the electrical potential produced in an electrochemical cell. The equation is given by
\[ E = -\frac{\Delta G}{nF} \]
where \( E \) is the cell potential, \( \Delta G \) is the Gibbs free energy change for the reaction, \( n \) is the number of moles of electrons transferred in the reaction, and \( F \) is the Faraday constant, approximately \( 96485 \text{C/mol} \). The equation is used to calculate the voltage produced, as shown in the exercise. The direction of the charge transfer through the cell, as well as the magnitude of the electrical potential, can be assessed using the Nernst equation.
Enthalpy Change
In thermodynamics, the enthalpy change (\( \Delta H \)) in a reaction reflects the total heat content absorbed or released under constant pressure. If the reaction produces heat, it is exothermic, and \( \Delta H \) is negative. Conversely, if the reaction absorbs heat from the surroundings, it is endothermic, and \( \Delta H \) is positive. The enthalpy change is a component of the Gibbs free energy equation and can be estimated from standard enthalpies of formation, calorimetric data, or tabulated values in reference materials, such as in Appendix 1 referred to in the exercise.

Real-life Implications of Enthalpy


Understanding enthalpy is not only important for calculation purposes but also for practical applications like energy efficiency in industrial reactions. By analyzing enthalpy changes, chemists can predict and control the heating and cooling aspects of various processes.
Entropy Change
Entropy change (\( \Delta S \)) quantifies the disorder or randomness in a chemical system. During a reaction, if the entropy increases, the products are more disordered than the reactants; thus, \( \Delta S \) is positive. If the entropy decreases, order in the system has increased, and \( \Delta S \) is negative. Entropy is a natural tendency of isolated systems to evolve towards thermodynamic equilibrium, which is the state of maximum entropy or disorder.

The Second Law of Thermodynamics


This concept aligns with the Second Law of Thermodynamics, which states that the total entropy of an isolated system can never decrease over time. It is also deeply intertwined with Gibbs free energy, as seen in the equation mentioned earlier. In the context of the exercise, understanding entropy changes allows students to predict the spontaneity of a reaction at any temperature, using the given values in the Appendix.

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Most popular questions from this chapter

In biological systems, acetate ion is converted to ethyl alcohol in a two- step process: \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+3 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{CHO}(a q)+\mathrm{H}_{2} \mathrm{O}\) \(E^{o \prime}=-0.581 \mathrm{~V}\) \(\mathrm{CH}_{3} \mathrm{CHO}(a q)+2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \quad E^{o \prime}=-0.197 \mathrm{~V}\) \(\left(E^{\circ \prime}\right.\) is the standard reduction voltage at \(25^{\circ} \mathrm{C}\) and \(\mathrm{pH}\) of \(7.00 .\) ) (a) Calculate \(\Delta G^{\circ \prime}\) for each step and for the overall conversion. (b) Calculate \(E^{\circ \prime}\) for the overall conversion.

Draw a diagram for a salt bridge cell for each of the following reactions. Label the anode and cathode, and indicate the direction of current flow throughout the circuit. (a) \(\mathrm{Zn}(s)+\mathrm{Cd}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cd}(s)\) (b) \(2 \mathrm{AuCl}_{4}^{-}(a q)+3 \mathrm{Cu}(s) \longrightarrow 2 \mathrm{Au}(s)+8 \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)\) (c) \(\mathrm{Fe}(s)+\mathrm{Cu}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{Cu}(s)+\mathrm{Fe}(\mathrm{OH})_{2}(s)\)

Consider three metals, \(\mathrm{X}, \mathrm{Y}\), and \(\mathrm{Z}\), and their salts, \(\mathrm{XA}, \mathrm{YA}\), and \(\mathrm{ZA}\). Three experiments take place with the following results: \- \(\mathrm{X}+\mathrm{hot} \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2}\) bubbles \(\mathrm{X}+\mathrm{YA} \longrightarrow\) no reaction I \(\mathrm{X}+\mathrm{ZA} \longrightarrow \mathrm{X}\) discolored \(+\mathrm{Z}\) Rank metals \(\mathrm{X}, \mathrm{Y}\), and \(\mathrm{Z}\), in order of decreasing strength as reducing agents.

Calculate \(E^{\circ}\) for the following voltaic cells: (a) \(\mathrm{Pb}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Pb}^{2+}(a q)+2 \mathrm{Ag}(s)\) (b) \(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{Fe}^{2+}(a q)+4 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+4 \mathrm{Fe}^{3+}(a q)\) (c) a Cd-Cd \(^{2+}\) half-cell and a \(\mathrm{Zn}-\mathrm{Zn}^{2+}\) half-cell

Which of the changes below will increase the voltage of the following cell? $$ \text { Co }\left|\mathrm{Co}^{2+}(0.010 M) \| \mathrm{H}^{+}(0.010 \mathrm{M})\right| \mathrm{H}_{2}(0.500 \mathrm{~atm}) \mid \mathrm{Pt} $$ (a) Increase the volume of \(\mathrm{CoCl}_{2}\) solution from \(100 \mathrm{~mL}\) to \(300 \mathrm{~mL}\). (b) Increase \(\left[\mathrm{H}^{+}\right]\) from \(0.010 \mathrm{M}\) to \(0.500 \mathrm{M}\) (c) Increase the pressure of \(\mathrm{H}_{2}\) from \(0.500 \mathrm{~atm}\) to \(1 \mathrm{~atm}\). (d) Increase the mass of the Co electrode from \(15 \mathrm{~g}\) to \(25 \mathrm{~g}\). (e) Increase \(\left[\mathrm{Co}^{2+}\right]\) from \(0.010 \mathrm{M}\) to \(0.500 \mathrm{M}\).
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