/*! 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 122 The overall reaction and equilib... [FREE SOLUTION] | 91影视

91影视

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at 298 \(\mathrm{K}\) is $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad K=1.28 \times 10^{83}$$ a. Calculate \(8^{\circ}\) and \(\Delta G^{\circ}\) at 298 \(\mathrm{K}\) for the fuel cell reaction. b. Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the fuel cell reaction. c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or remain the same? Explain.

Short Answer

Expert verified
To summarize, at 298 K, 螖G掳 for the fuel cell reaction is -245943 J/mol. Both 螖H掳 and 螖S掳 are negative, indicating an exothermic reaction with a decrease in entropy. As temperature increases, the maximum amount of work obtained from the fuel cell reaction decreases, since the negative value of 螖G掳 becomes less negative due to the increasing positive magnitude of T螖S掳.

Step by step solution

01

Part a: Calculating 螖G掳

We use the relation between 螖G掳 and the equilibrium constant, K: 螖G掳 = -RT ln(K) where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin (298 K), and K is the equilibrium constant (1.28 脳 10^83). 螖G掳 = - (8.314 J/mol K) 脳 (298 K) 脳 ln(1.28 脳 10^83) 螖G掳 = - (245943 J/mol) So, the value of 螖G掳 is -245943 J/mol.
02

Part b: Predicting the signs of 螖H掳 and 螖S掳

We know that, at a constant temperature, 螖G掳 is the combination of 螖H掳 and 螖S掳. 螖G掳 = 螖H掳 - T螖S掳 We have already calculated the value of 螖G掳. Now, we need to analyze the reaction. This is an exothermic reaction because hydrogen and oxygen combine to produce water, which releases energy. Therefore, 螖H掳 should be negative. Since the reaction involves the conversion of gases to a liquid, the system becomes more ordered. This increase in order leads to a decrease in entropy, and thus, 螖S掳 should be negative. Therefore, the signs of 螖H掳 and 螖S掳 are both negative for the fuel cell reaction.
03

Part c: The effect of temperature on the maximum amount of work obtained

To understand how temperature affects the maximum amount of work obtained from the fuel cell reaction, we must analyze the effect of temperature on the equilibrium constant K and on 螖G掳. Since the reaction is exothermic (螖H掳 is negative), an increase in temperature will shift the equilibrium towards the reactants. This implies that the equilibrium constant K will decrease with increasing temperature. Recall that 螖G掳 is the maximum amount of work that can be extracted from the system. From the equation: 螖G掳 = 螖H掳 - T螖S掳 As the temperature increases, the term T螖S掳 will also increase. Since both 螖H掳 and 螖S掳 are negative, T螖S掳 will be a positive quantity with a larger magnitude as temperature increases. This causes 螖G掳 to become less negative as the temperature increases. Therefore, as the temperature increases, the maximum amount of work obtained from the fuel cell reaction will decrease.

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.

Gibbs Free Energy
Gibbs Free Energy, denoted as \(\Delta G^{\circ}\), plays a crucial role in understanding how fuel cells work. It's a thermodynamic function that helps us determine the spontaneity of a chemical reaction. When \(\Delta G^{\circ}\) is negative, it implies the reaction can proceed spontaneously, releasing free energy that can potentially be harnessed to do work, such as powering a device. In fuel cells like the hydrogen-oxygen type, the calculation of \(\Delta G^{\circ}\) gives insight into the energy available to produce electricity. For the reaction presented, where hydrogen reacts with oxygen to form water: - The equilibrium constant \(K\) is extremely large \((1.28 \times 10^{83})\), indicating a strong tendency to form products. - Using the relation \(\Delta G^{\circ} = -RT \ln(K)\), where \(R\) is the gas constant and \(T\) is the temperature, we can calculate \(\Delta G^{\circ} = -245943 \ \,\text{J/mol}\) for this reaction, illustrating its high spontaneity at 298 K.
Equilibrium Constant
In thermodynamics, the equilibrium constant \(K\) expresses the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients, at equilibrium. For the hydrogen-oxygen reaction in a fuel cell, the equilibrium constant is astronomically large: \(1.28 \times 10^{83}\). This value signifies that, at 298 K, the reaction strongly favors the formation of water from hydrogen and oxygen. Key aspects include: - A large equilibrium constant generally indicates a product-favored reaction. This means nearly all hydrogen and oxygen are converted into water. - The equilibrium constant also assists in calculating \(\Delta G^{\circ}\), establishing the work potential of the reaction. - Changes in \(K\) with temperature can alter the direction of reaction shifts, impacting the reaction dynamics.
Exothermic Reaction
An exothermic reaction is characterized by the release of energy, typically in the form of heat. In the context of a hydrogen-oxygen fuel cell, where the reaction \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \) occurs, energy is released as water is formed. Significant points include: - The negative \(\Delta H^{\circ}\) denotes that the reaction is exothermic, implying an output of energy as thermal energy. - The released energy is harnessed to perform electrical work, functioning as an efficient energy conversion in the fuel cell. - Exothermic reactions like this contribute to the decreasing energy requirement of the system to sustain the reaction, enhancing fuel cell efficiency.
Entropy Change
Entropy change, denoted as \(\Delta S^{\circ}\), indicates the disorder or randomness change of a system during a reaction. In chemical reactions, it signifies how the internal arrangement of particles changes.For the hydrogen-oxygen fuel cell reaction, gases (\(2 \mathrm{H}_{2}(g)\) and \(\mathrm{O}_{2}(g)\)) convert into a liquid (water), leading to: - A decrease in entropy \((\Delta S^{\circ} < 0)\), as a more ordered liquid is formed from gaseous reactants. - This coincides with the typical trend where reactions resulting in fewer gaseous products than reactants see decreased entropy. Other considerations include: - The interplay between \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) influences \(\Delta G^{\circ}\), our measure of reaction spontaneity.- The negative entropy change is typical for exothermic reactions that shift from less orderly gas states to more orderly liquid states, which can affect the overall energy output efficiency of the fuel cell.

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

Hydrogen peroxide can function either as an oxidizing agent or as a reducing agent. At standard conditions, is \(\mathrm{H}_{2} \mathrm{O}_{2}\) a better oxidizing agent or reducing agent? Explain.

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C} :\) $$3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.957 \mathrm{V}$$ $$\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.775 \mathrm{V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and NO \(_{2}\) mixture with only 0.20\(\% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

Combine the equations $$\Delta G^{\circ}=-n F \mathscr{E}^{\circ} \text { and } \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$$ to derive an expression for \(\mathscr{E}^{\circ}\) as a function of temperature. Describe how one can graphically determine \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) from measurements of \(\mathscr{E}^{\circ}\) at different temperatures, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. What property would you look for in designing a reference half-cell that would produce a potential relatively stable with respect to temperature?

You want to 鈥減late out鈥 nickel metal from a nickel nitrate solution onto a piece of metal inserted into the solution. Should you use copper or zinc? Explain.

When aluminum foil is placed in hydrochloric acid, nothing happens for the first 30 seconds or so. This is followed by vigorous bubbling and the eventual disappearance of the foil. Explain these observations.

See all solutions

Recommended explanations on Chemistry Textbooks

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.