/*! 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 37 \(\mathrm{N}_{2} \mathrm{O}_{3}\... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 37

\(\mathrm{N}_{2} \mathrm{O}_{3}\) dissociates according to the equilibrium \(\mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) .\) At \(298 \mathrm{K}\) and one bar pressure, the degree of dissociation defined as the ratio of moles \(\operatorname{of} \mathrm{NO}_{2}(g)\) or \(\mathrm{NO}(g)\) to the moles of the reactant assuming no dissociation occurs is \(3.5 \times 10^{-3}\). Calculate \(\Delta G_{R}^{\circ}\) for this reaction.

Short Answer

Expert verified
The Gibbs free energy change for the dissociation of nitrogen trioxide at 298 K and one bar pressure is approximately \(-5168.92~J/mol\).

Step by step solution

01

Write the balanced chemical equation

First, the balanced chemical equation for the dissociation of nitrogen trioxide is given by: \[N_2O_3(g) \rightleftharpoons NO_2(g) + NO(g)\]
02

Relate the degree of dissociation to the equilibrium constant

Let initially 1 mole of N鈧侽鈧� be present and no dissociation has occurred. Since the degree of dissociation is 3.5 脳 10鈦宦�, at equilibrium: - Moles of N鈧侽鈧�: 1 - 3.5 脳 10鈦宦� - Moles of NO鈧�: 3.5 脳 10鈦宦� - Moles of NO: 3.5 脳 10鈦宦� For the dissociation reaction, the equilibrium constant, K, can be written as: \[K = \frac{[NO_2][NO]}{[N_2O_3]}\] Substituting the moles for the concentrations, we have: \[K = \frac{(3.5 \times 10^{-3})(3.5 \times 10^{-3})}{(1 - 3.5 \times 10^{-3})}\]
03

Determine the relationship between 螖G岬Ｂ� and the equilibrium constant

The relationship between the Gibbs free energy change at standard conditions (螖G岬Ｂ�) and the equilibrium constant K is given by the equation: \[\Delta G_{R}^{\circ} = -RT \ln K\] Where R is the universal gas constant (8.314 J/mol路K) and T is the temperature in Kelvin (298 K).
04

Calculate 螖G岬Ｂ� for the reaction

Substitute the equilibrium constant and the given temperature into the 螖G岬Ｂ� equation: \[\Delta G_{R}^{\circ} = -8.314 \times 298 \times \ln \left( \frac{(3.5 \times 10^{-3})(3.5 \times 10^{-3})}{(1 - 3.5 \times 10^{-3})} \right)\] Now, calculate 螖G岬Ｂ�: \[\Delta G_{R}^{\circ} = -8.314 \times 298 \times \ln \left( \frac{(3.5 \times 10^{-3})(3.5 \times 10^{-3})}{(1 - 3.5 \times 10^{-3})} \right) = -5168.92~J/mol\] Therefore, the Gibbs free energy change for the dissociation of nitrogen trioxide at 298 K and one bar pressure is approximately -5168.92 J/mol.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentrations of reactants and products remain constant over time. This condition doesn't imply that the reactants and products are in equal concentrations, but rather that their ratios do not change.

In the given exercise, N鈧侽鈧� is in equilibrium with NO鈧� and NO. At equilibrium, there is no net change in the concentrations of these species, even though the reactions continue to occur in both directions. This concept is essential when predicting the outcome of chemical reactions and understanding how various factors can shift the equilibrium.
Degree of Dissociation
The degree of dissociation is a measure of the extent to which a compound separates into its constituents in a particular process, such as a chemical reaction. It is defined as the ratio of the quantity dissociated to the total quantity present. In the context of the provided exercise, the degree of dissociation of N鈧侽鈧� into NO鈧� and NO is given as 3.5 脳 10鈦宦�.

Understanding the degree of dissociation helps in determining the concentrations of all species at equilibrium. It's crucial because it provides insight into how completely a reactant is converted into products under specific conditions.
Equilibrium Constant
The equilibrium constant, represented by the symbol K, is a number that expresses the ratio of the concentration of products to reactants at chemical equilibrium, each raised to the power of their respective stoichiometric coefficients in the balanced equation. The equilibrium constant is a reflection of how far a reaction will proceed before reaching equilibrium.

In the solved example, we calculate the equilibrium constant (K) using the degree of dissociation. A larger K value typically indicates that more products are formed, representing a reaction that tends to favor the products at equilibrium. Knowing K is fundamental for predicting the position of equilibrium and for calculating other important thermodynamic quantities, such as Gibbs free energy.
Thermodynamics
Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy. In chemical reactions, thermodynamics helps to predict whether a process will occur spontaneously under a set of given conditions. The concept that connects thermodynamics to chemical equilibrium is the Gibbs free energy change (螖G).

Gibbs free energy change is used to determine the spontaneity of a process. If 螖G is negative, the process or reaction occurs spontaneously. This value can be calculated from the equilibrium constant as showcased in the step-by-step example, providing a quantitative measure of the system's tendency to reach equilibrium and how external conditions like temperature affect this balance.

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Most popular questions from this chapter

Calculate \(\Delta G\) for the isothermal expansion of \(2.25 \mathrm{mol}\) of an ideal gas at \(325 \mathrm{K}\) from an initial pressure of 12.0 bar to a final pressure of 2.5 bar.

Consider the equilibrium \(\mathrm{C}_{2} \mathrm{H}_{6}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\) \(\mathrm{H}_{2}(g)\). At \(1000 .\) K and a constant total pressure of \(1.00 \mathrm{bar}\) \(\mathrm{C}_{2} \mathrm{H}_{6}(g)\) is introduced into a reaction vessel. The total pressure is held constant at 1 bar and at equilibrium the composition of the mixture in mole percent is \(\mathrm{H}_{2}(g): 26.0 \%\) \(\mathrm{C}_{2} \mathrm{H}_{4}(g): 26.0 \%\) and \(\mathrm{C}_{2} \mathrm{H}_{6}(g): 48.0 \%\) a. Calculate \(K_{P}\) at \(1000 .\) K. b. If \(\Delta H_{R}^{\circ}=137.0 \mathrm{kJ} \mathrm{mol}^{-1}\), calculate the value of \(K_{P}\) at \(298.15 \mathrm{K}\) c. Calculate \(\Delta G_{R}^{\circ}\) for this reaction at \(298.15 \mathrm{K}\)

In Example Problem \(6.9, K_{P}\) for the reaction \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) was calculated to be \(3.32 \times 10^{3}\) at \(298.15 \mathrm{K}\). At what temperature is \(K_{P}=5.50 \times 10^{3} ?\) What is the highest value that \(K_{P}\) can have by changing the temperature? Assume that \(\Delta H_{R}^{\circ}\) is independent of temperature.

Calculate \(\mu_{O_{2}}^{\text {mixture}}(298.15 \mathrm{K}, 1\) bar) for oxygen in air, assuming that the mole fraction of \(\mathrm{O}_{2}\) in air is \(0.210 .\) Use the conventional molar Gibbs energy defined in Section 6.17.

Show that $$\left[\frac{\partial(A / T)}{\partial(1 / T)}\right]_{V}=U$$ Write an expression analogous to Equation (6.36) that would allow you to relate \(\Delta A\) at two temperatures.
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