/*! 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 64 Calculate the pressure of \(\mat... [FREE SOLUTION] | 91影视

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Chapter 18: Problem 64

Calculate the pressure of \(\mathrm{O}_{2}\) (in atm) over a sample of \(\mathrm{NiO}\) at \(25^{\circ} \mathrm{C}\) if \(\Delta G^{\circ}=212 \mathrm{~kJ} / \mathrm{mol}\) for the reaction $$ \mathrm{NiO}(s) \rightleftharpoons \mathrm{Ni}(s)+\frac{1}{2} \mathrm{O}_{2}(g) $$

Short Answer

Expert verified
The pressure of \( \mathrm{O}_2 \) is approximately \( 3.06 \times 10^{-74} \text{ atm} \).

Step by step solution

01

Understand the Reaction

The given reaction is \( \mathrm{NiO}(s) \rightleftharpoons \mathrm{Ni}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \). The task is to find the pressure of \( \mathrm{O}_2 \) using the provided \( \Delta G^{\circ} \). The reaction involves solids (\(\mathrm{NiO}\) and \(\mathrm{Ni}\)) and a gas (\(\mathrm{O}_2\)).
02

Use Gibbs Free Energy Equation

The relationship between Gibbs free energy change \( \Delta G^\circ \), reaction quotient \( Q \), and the equilibrium constant \( K \) at standard conditions is given by \( \Delta G^\circ = -RT \ln K \). Since the reaction is at equilibrium, \( Q = K \).
03

Solve for Equilibrium Constant K

Rearrange the equation \( \Delta G^\circ = -RT \ln K \) to find \( K \):\[K = e^{-\Delta G^\circ / (RT)}\]where \( R = 8.314 \text{ J/mol K} \) and temperature \( T = 298 \text{ K} \) (since \( 25^\circ \text{C} = 298 \text{ K} \)).
04

Substitute Values and Calculate K

Substitute \( \Delta G^\circ = 212,000 \text{ J/mol} \), \( R = 8.314 \text{ J/mol K} \), and \( T = 298 \text{ K} \) into the equation:\[K = e^{-\frac{212000}{8.314 \times 298}} = e^{-85.5}\]This results in a very small value for \( K \), which approximates to \( K = 1.75 \times 10^{-37} \).
05

Relate K to Pressure of O2

Since only \( \mathrm{O}_2 \) is in the gas phase, the equilibrium expression for the reaction is \( K = (P_{\mathrm{O}_2})^{1/2} \). Therefore, \( K = (P_{\mathrm{O}_2})^{1/2} \). Solve for \( P_{\mathrm{O}_2} \):\[P_{\mathrm{O}_2} = K^2 = (1.75 \times 10^{-37})^2\]which simplifies to approximately \( 3.06 \times 10^{-74} \text{ atm} \).
06

Conclusion

The extremely small value of \( P_{\mathrm{O}_2} \) indicates that the reaction greatly favors the formation of \( \mathrm{NiO}(s) \) at equilibrium, resulting in almost no loss of \( \mathrm{O}_2 \) into the atmosphere at the given conditions.

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, often symbolized as \( K \), is crucial in predicting the direction and extent of a chemical reaction. It tells us how far a reaction has proceeded towards equilibrium. Let's consider its calculation in the context of our reaction.
  • The equilibrium constant expression is based on the balanced chemical equation.
  • For the given reaction \( \mathrm{NiO}(s) \rightleftharpoons \mathrm{Ni}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \), only the gas, \( \mathrm{O}_2 \), contributes to \( K \).
Indeed, the solids do not affect the equilibrium constant, as their concentrations are constant. Thus, the expression simplifies, leading to \( K = (P_{\mathrm{O}_2})^{1/2} \). By calculating \( K \) from \( \Delta G^\circ \), we understand the pressure of \( \mathrm{O}_2 \), which in this case turns out to be very small, showing the favorability of reactants at equilibrium.
Reaction Quotient
The reaction quotient \( Q \) is similar to the equilibrium constant \( K \), but it applies to a system that is not at equilibrium. It shows the ratio of product and reactant concentrations at any given point in time.
  • If \( Q < K \), the reaction will shift toward the products to reach equilibrium.
  • If \( Q > K \), the reverse is true: the reaction will move towards the reactants.
  • For \( Q = K \), the system is at equilibrium.
In equilibrium calculations, as shown in our problem, \( Q \) becomes irrelevant because the system is already at equilibrium, and so \( Q = K \). This equality simplifies our calculations as we directly equate the calculated \( K \) with the expression involving the gas-phase component.
Pressure Calculation
Calculating the pressure of a gas in equilibrium reactions involves using the ideal gas law or related expressions in physical chemistry.
  • We used the relation \( K = (P_{\mathrm{O}_2})^{1/2} \) to find \( P_{\mathrm{O}_2} \).
  • Given a very small \( K \) (\( 1.75 \times 10^{-37} \)), resolving \( P_{\mathrm{O}_2} \) involves squaring \( K \).
The outcome, \( 3.06 \times 10^{-74} \text{ atm} \), showcases that at equilibrium, the pressure of \( \mathrm{O}_2 \) is negligibly small. This suggests minimal formation of \( \mathrm{Ni} \) and \( \mathrm{O}_2 \) under the conditions, and practically, \( \mathrm{NiO} \) remains largely unreacted.
Thermodynamics
Thermodynamics provides a framework for understanding energy changes, such as Gibbs Free Energy \( \Delta G \), underlying chemical reactions.
  • \( \Delta G^\circ \) quantifies the spontaneity of a reaction: negative values indicate spontaneous reactions under standard conditions.
  • For our reaction, \( \Delta G^\circ = 212 \mathrm{~kJ} / \, \) indicates non-spontaneity since it is positive.
  • This ties into our small \( K \) value, showing equilibrium lying towards reactants.
Thermodynamics links directly with kinetics and equilibrium to predict whether and to what extent a reaction will proceed. In essence, getting familiar with Gibbs Free Energy helps demystify the reaction's fate and relate it to observable quantities like concentration or pressure.

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Most popular questions from this chapter

The equilibrium constant \(K_{P}\) for the reaction $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) $$ is \(5.62 \times 10^{35}\) at \(25^{\circ} \mathrm{C} .\) Calculate \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{COCl}_{2}\) at \(25^{\circ} \mathrm{C} .\)

Give a detailed example of each of the following, with an explanation: (a) a thermodynamically spontaneous process; (b) a process that would violate the first law of thermodynamics; (c) a process that would violate the second law of thermodynamics; (d) an irreversible process; (e) an equilibrium process.

Why is it more convenient to predict the direction of a reaction in terms of \(\Delta G_{\text {sys }}\) instead of \(\Delta S_{\text {univ }}\) ? Under what conditions can \(\Delta G_{\text {sys }}\) be used to predict the spontaneity of a reaction?

Consider two carboxylic acids (acids that contain the \(-\mathrm{COOH}\) group \(): \mathrm{CH}_{3} \mathrm{COOH}\) (acetic acid, \(K_{\mathrm{a}}=1.8 \times 10^{-5}\) ) and \(\mathrm{CH}_{2} \mathrm{ClCOOH}\) (chloroacetic acid, \(K_{\mathrm{a}}=1.4 \times 10^{-3}\) ). (a) Calculate \(\Delta G^{\circ}\) for the ionization of these acids at \(25^{\circ} \mathrm{C}\) (b) From the equation \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ},\) we see that the contributions to the \(\Delta G^{\circ}\) term are an enthalpy term \(\left(\Delta H^{\circ}\right)\) and a temperature times entropy term \(\left(T \Delta S^{\circ}\right) .\) These contributions are listed below for the two acids: $$ \begin{array}{lcc} & \Delta \boldsymbol{H}^{\circ}(\mathbf{k} \mathbf{J} / \mathbf{m o l}) & \boldsymbol{T} \Delta \boldsymbol{S}^{\circ}(\mathbf{k} \mathbf{J} / \mathbf{m o l}) \\ \hline \mathrm{CH}_{3} \mathrm{COOH} & -0.57 & -27.6 \\ \mathrm{CH}_{2} \mathrm{ClCOOH} & -4.7 & -21.1 \\ \hline \end{array} $$ Which is the dominant term in determining the value of \(\Delta G^{\circ}\) (and hence \(K_{\mathrm{a}}\) of the acid)? (c) What processes contribute to \(\Delta H^{\circ} ?\) (Consider the ionization of the acids as a Br贸nsted acid-base reaction.) (d) Explain why the \(T \Delta S^{\circ}\) term is more negative for \(\mathrm{CH}_{3} \mathrm{COOH}\).

Consider the reaction $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Given that \(\Delta G^{\circ}\) for the reaction at \(25^{\circ} \mathrm{C}\) is 173.4 \(\mathrm{kJ} / \mathrm{mol},\) (a) calculate the standard free energy of formation of \(\mathrm{NO},\) and \((\mathrm{b})\) calculate \(K_{P}\) of the reaction. (c) One of the starting substances in smog formation is NO. Assuming that the temperature in a running automobile engine is \(1100^{\circ} \mathrm{C}\), estimate \(K_{P}\) for the above reaction. (d) As farmers know, lightning helps to produce a better crop. Why?
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