/*! 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 98 Using the data in Appendix \(C\)... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 98

Using the data in Appendix \(C\) and given the pressures listed, calculate \(\Delta G^{\circ}\) for each of the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \\ \quad P_{\mathrm{N}_{2}}=2.6 \mathrm{~atm}, P_{\mathrm{H}_{2}}=5.9 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=1.2 \mathrm{~atm} \\ \text { (b) } 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \\ \quad P_{\mathrm{N}_{2} \mathrm{H}_{4}}=P_{\mathrm{NO}_{2}}=5.0 \times 10^{-2} \mathrm{~atm} \\ \quad P_{\mathrm{N}_{2}}=0.5 \mathrm{~atm}, P_{\mathrm{H}_{2} \mathrm{O}}=0.3 \mathrm{~atm} \\ \text { (c) } \mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \\ \quad P_{\mathrm{N}_{2} \mathrm{H}_{4}}=0.5 \mathrm{~atm}, P_{\mathrm{N}_{2}}=1.5 \mathrm{~atm}, P_{\mathrm{H}_{2}}=2.5 \mathrm{~atm} \end{array} $$

Short Answer

Expert verified
For each reaction, we can find the standard Gibbs free energy change (\(\Delta G^{\circ}\)) by first calculating the reaction quotients: Reaction (a): \(Q_p = \frac{ (1.2)^2 }{ (2.6) \cdot (5.9)^3 }\) Reaction (b): \(Q_p = \frac{ (0.5)^3 \cdot (0.3)^4 }{ (5.0 \times 10^{-2})^2 \cdot (5.0 \times 10^{-2})^2 }\) Reaction (c): \(Q_p = \frac{ (1.5) \cdot (2.5)^2 }{ (0.5) }\) Then, we can use the equation \(\Delta G = \Delta G^{\circ}+RT\ln Q_p\) to solve for \(\Delta G^{\circ}\) using the values provided in Appendix C and assuming T = 298 K, R = 8.314 J/mol 路 K.

Step by step solution

01

To calculate the reaction quotient Q, we will use the equation \(Q_p = \frac{[C]^c[D]^d}{[A]^a[B]^b}\), where A, B, C and D represent the reactants and products and their corresponding coefficients, and their pressures are given by [A], [B], [C], and [D]. For each reaction, let's calculate the Q value: #For reaction (a)# \(Q_p = \frac{ P_{NH_3}^2 }{ P_{N_2} \cdot P_{H_2}^3 }\) \(Q_p = \frac{ (1.2)^2 }{ (2.6) \cdot (5.9)^3 }\) #For reaction (b)# \(Q_p = \frac{ P_{N_2}^3 \cdot P_{H_2O}^4 }{ P_{N_2H_4}^2 \cdot P_{NO_2}^2 }\) \(Q_p = \frac{ (0.5)^3 \cdot (0.3)^4 }{ (5.0 \times 10^{-2})^2 \cdot (5.0 \times 10^{-2})^2 }\) #For reaction (c)# \(Q_p = \frac{ P_{N_2} \cdot P_{H_2}^2 }{ P_{N_2H_4} }\) \(Q_p = \frac{ (1.5) \cdot (2.5)^2 }{ (0.5) }\) #Step 2: Solve for 螖G掳 using the equation 螖G = 螖G掳 + RTlnQ#

To find the standard Gibbs free energy change (\(\Delta G^{\circ}\)), we'll first need to use the given 螖G values from Appendix C for each reaction involved in this exercise. Then, we will use the equation \(\Delta G = \Delta G^{\circ}+RT\ln Q\) to solve for \(\Delta G^{\circ}\). For each reaction, let's plug the reaction quotient values and other known values into the equation: (Note: Assume T = 298 K, R = 8.314 J/mol 路 K) #For reaction (a)# \(\Delta G^{\circ} = \Delta G - RT\ln Q_p\) \(\Delta G^{\circ} = \) (Value of \(\Delta G\) for reaction (a) from Appendix C) - \((8.314)(298) \ln Q_p\) #For reaction (b)# \(\Delta G^{\circ} = \Delta G - RT\ln Q_p\) \(\Delta G^{\circ} = \) (Sum of values of \(\Delta G\) for reaction (b) from Appendix C) - \((8.314)(298) \ln Q_p\) #For reaction (c)# \(\Delta G^{\circ} = \Delta G - RT\ln Q_p\) \(\Delta G^{\circ} = \) (Value of \(\Delta G\) for reaction (c) from Appendix C) - \((8.314)(298) \ln Q_p\) Calculating the standard Gibbs free energy change (\(\Delta G^{\circ}\)) for each reaction will give you the desired results.

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Key Concepts

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

Reaction Quotient
The reaction quotient (\textbf{Q}) plays a crucial role in understanding the progress of a reaction at a given moment in time. It is a snapshot of a reaction's ratio of products to reactants, each raised to the power of their stoichiometric coefficients. The equation for a reaction in the gas phase, at pressures P, is:

\[\[\begin{align*}Q_p = \frac{{P_{\text{{products}}}^\text{{stoichiometric coefficient}} }}{{ P_{\text{{reactants}}}^\text{{stoichiometric coefficient}} }}\end{align*}\]\]
Where \textbf{P} represents the partial pressures of the reactants and products. The values of Q can help predict whether a reaction will proceed forward or reverse to achieve chemical equilibrium. If Q is less than the equilibrium constant (\textbf{K}), the reaction tends to produce more products. Conversely, if Q is greater than K, the reaction favors the formation of reactants.
Standard State
The standard state of a substance is a reference point used to define its thermodynamic properties under specified conditions of pressure and temperature. For gases, the standard state is typically 1 atmosphere (atm) of pressure and usually a temperature of 298 K (25掳C), unless otherwise specified. The standard Gibbs free energy change (\textbf{\[\[\begin{align*}circle\end{align*}\]\]}), denoted by the superscript 掳, indicates the amount of energy change when reactants transform into products at standard state conditions. Understanding standard state is vital because it sets a baseline from which we can understand how variations in conditions like pressure and temperature affect the free energy and directionality of a chemical reaction.
Chemical Equilibrium
Chemical equilibrium is a state of balance in which the forward and reverse reactions occur at equal rates, so the concentrations of the reactants and products remain constant over time. At equilibrium, the reaction quotient (\textbf{Q}) equals the equilibrium constant (\textbf{K}), and there is no net change in the Gibbs free energy of the system.\[\[\begin{align*}circle\end{align*}\]\]Understanding this concept is vital because it helps to predict the behavior of reactions under different conditions and is the basis for optimizing reaction conditions in industrial processes. Moreover, the concept of equilibrium does not imply that reactants and products are in equal concentrations, but rather that their rates of formation are equivalent.
Partial Pressure
Partial pressure is the pressure exerted by a single gas in a mixture of gases. Each gas in the mixture behaves as if it alone occupies the entire volume of the mixture, and its partial pressure is a measure of its concentration in the gas phase. It's a direct way to express the amount of each gas involved in a reaction. In chemical equilibrium and reaction quotient calculations, the partial pressures are used to represent the concentration of gaseous reactants and products. It's important for predicting how changes in pressure can shift equilibrium and alter the Gibbs free energy change of a reaction.
Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat, work, temperature, and energy. In chemistry, it's essential for understanding how energy changes during chemical reactions. The Gibbs free energy, a thermodynamic potential, predicts whether a reaction will happen spontaneously under constant pressure and temperature. The sign of the Gibbs free energy change (\textbf{\[\[\begin{align*}circle\end{align*}\]\]}) determines the spontaneity of a reaction: if negative, the reaction is spontaneous; if positive, it requires external energy to proceed. This allows chemists to assess the feasibility of reactions and to manipulate conditions to favor the production of desired products.

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Most popular questions from this chapter

For a certain chemical reaction, \(\Delta H^{\circ}=-35.4 \mathrm{~kJ}\) and \(\Delta S^{\circ}=-85.5 \mathrm{~J} / \mathrm{K} .\) (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K} .(\mathbf{d})\) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

Consider what happens when a sample of the explosive TNT (Section 8.8: "Chemistry Put to Work: Explosives and Alfred Nobel") is detonated under atmospheric pressure. (a) Is the detonation a spontaneous process? (b) What is the sign of \(q\) for this process? (c) Can you determine whether \(w\) is positive, negative, or zero for the process? Explain. (d) Can you determine the sign of \(\Delta E\) for the process? Explain.

Suppose we vaporize a mole of liquid water at \(25^{\circ} \mathrm{C}\) and another mole of water at \(100{ }^{\circ} \mathrm{C}\). (a) Assuming that the enthalpy of vaporization of water does not change much between \(25^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C},\) which process involves the larger change in entropy? (b) Does the entropy change in either process depend on whether we carry out the process reversibly or not? Explain.

(a) Give two examples of endothermic processes that are spontaneous. (b) Give an example of a process that is spontaneous at one temperature but nonspontaneous at a different temperature.

The following data compare the standard enthalpies and free energies of formation of some crystalline ionic substances and aqueous solutions of the substances: $$ \begin{array}{lrr} \text { Substance } & \Delta \boldsymbol{H}_{f}^{\circ}(\mathbf{k} \mathbf{J} / \mathbf{m o l}) & \Delta \mathbf{G}_{f}^{\circ}(\mathbf{k J} / \mathbf{m o l}) \\ \hline \mathrm{AgNO}_{3}(s) & -124.4 & -33.4 \\ \mathrm{AgNO}_{3}(a q) & -101.7 & -34.2 \\ \mathrm{MgSO}_{4}(s) & -1283.7 & -1169.6 \\ \mathrm{MgSO}_{4}(a q) & -1374.8 & -1198.4 \end{array} $$ (a) Write the formation reaction for \(\mathrm{AgNO}_{3}(s) .\) Based on this reaction, do you expect the entropy of the system to increase or decrease upon the formation of \(\mathrm{AgNO}_{3}(s) ?\) (b) Use \(\Delta H_{f}^{\circ}\) and \(\Delta G_{f}^{\circ}\) of \(\mathrm{AgNO}_{3}(s)\) to determine the entropy change upon formation of the substance. Is your answer consistent with your reasoning in part (a)? (c) Is dissolving \(\mathrm{AgNO}_{3}\) in water an exothermic or endothermic process? What about dissolving \(\mathrm{MgSO}_{4}\) in water? (d) For both \(\mathrm{AgNO}_{3}\) and \(\mathrm{MgSO}_{4},\) use the data to calculate the entropy change when the solid is dissolved in water. (e) Discuss the results from part (d) with reference to material presented in this chapter and in the "A Closer Look" box on page 540 .
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