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Chapter 3: Problem 34

Consider the reaction $$ 2 \mathrm{NH}_{3}=\mathrm{N}_{2}+3 \mathrm{H}_{2} $$ which occurs in the gas phase. Start initially with \(2 \mathrm{~mol}\) of \(\mathrm{NH}_{3}\) and 0 mol each of \(\mathrm{H}_{2}\) and \(\mathrm{N}_{2}\). Assume that the reaction occurs at temperature \(T\) and pressure \(P\). Use ideal gas equations for the chemical potential. (a) Compute and plot the Gibbs free energy, \(G(T, P\), (5), as a function of the degree of reaction, \(\xi\), for (i) \(P=1\) atm and \(T=298 \mathrm{~K}\) and (ii) \(P=1 \mathrm{~atm}\) and \(T\). \(=894 \mathrm{~K}\). (b) Compute and plot the affinity, \(A(T, P, \xi)\), as a function of the degree of reaction, \(\xi\), for (i) \(P=1 \mathrm{~atm}\) and \(T=298 \mathrm{~K}\) and (ii) \(P=1\) atm and \(T=894 \mathrm{~K}\). (c) What is the degree of reaction, \(\xi\), at chemical equilibrium for \(P=1 a t m\) and temperature \(T=894\) K? How many moles of \(\mathrm{NH}_{3}, \mathrm{H}_{2}\), and \(\mathrm{N}_{2}\) are present at equilibrium? (d) If initially the volume is \(V_{0}\), what is the volume at equilibrium for \(P=\) \(1 \mathrm{~atm}\) and \(T=894 \mathrm{~K}\) ? (e) What is the heat of reaction for \(P=1 \mathrm{~atm}\) and \(T=894 \mathrm{~K}\) ?

Short Answer

Expert verified
Plot \( G(T, P, \xi) \) and \( A(T, P, \xi) \) as functions of \( \xi \). Solve \( \Delta G = 0 \) to find \( \xi \) at equilibrium. Calculate moles and volume at equilibrium using \( \xi_{eq} \) and ideal gas law.

Step by step solution

01

- Write the reaction and define the Gibbs free energy

The reaction is given as: $$2 \mathrm{NH}_{3}=\mathrm{N}_{2}+3 \mathrm{H}_{2}$$ The change in Gibbs free energy (\( \Delta G \)) for the reaction can be written in terms of the standard Gibbs free energy change (\( \Delta G^0 \)) and the reaction quotient (Q) as: \( \Delta G = \Delta G^0 + RT\ln Q \) Where \( Q = \frac{{[\mathrm{N}_2][\mathrm{H}_2]^3}}{{[\mathrm{NH}_3]^2}} \).
02

- Express the concentration in terms of the degree of reaction

Let the initial moles of NH3 be 2 mol. The change in the number of moles due to the degree of reaction (\( \xi \)) is: - NH3: \( 2 - 2\xi \) - N2: \( \xi \) - H2: \( 3\xi \) With these, the reaction quotient can be rewritten as: \( Q = \frac{{\xi (3\xi)^3}}{{(2 - 2\xi)^2}} \)
03

- Calculate the standard Gibbs free energy change

The standard Gibbs free energy change \( \Delta G^0 \) can be found using: \( \Delta G^0 = \Delta H^0 - T \Delta S^0 \) Where \( \Delta H^0 \) and \( \Delta S^0 \) are the standard enthalpy and entropy changes, respectively.
04

- Compute Gibbs free energy for given temperatures and pressures

Substitute the values for \( T = 298 \mathrm{~K} \) and \( T = 894 \mathrm{~K} \) and \( P = 1 \mathrm{~atm} \) into the equation \( \Delta G = \Delta G^0 + RT\ln Q \). Plot \( G(T, P, \xi) \) as a function of \( \xi \) for each temperature.
05

- Calculate the affinity

The affinity A(T,P,\xi) is given by: \( A = -\Delta G \). Compute and plot \( A(T, P, \xi) \) as a function of \( \xi \) for both temperatures \( T = 298 \mathrm{~K} \) and \( T = 894 \mathrm{~K} \).
06

- Determine the degree of reaction at equilibrium

At equilibrium, the Gibbs free energy change \( \Delta G \) is zero. Set \( \Delta G = 0 \) and solve for \( \xi \) using the equation: \( \Delta G^0 + RT \ln Q = 0 \).
07

- Calculate moles at equilibrium

Using the value of \( \xi \) at equilibrium obtained from Step 6: - Moles of NH3 = \( 2 - 2\xi_{eq} \) - Moles of N2 = \( \xi_{eq} \) - Moles of H2 = \( 3\xi_{eq} \)
08

- Compute the volume at equilibrium

Apply the ideal gas law: \( PV = nRT \) with \( P = 1 \mathrm{~atm} \), \( T = 894 \mathrm{~K} \), and total moles \( n = 2 - 2\xi_{eq} + \xi_{eq} + 3\xi_{eq} = 2 + 2\xi_{eq} \). Solve for the volume \( V \) at equilibrium.
09

- Determine the heat of reaction

The heat of reaction (\( \Delta H \)) at \( P = 1 \mathrm{~atm} \) and \( T = 894 \mathrm{~K} \) can be found using the standard enthalpy change and/or using calorimetry data.

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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 the state in which both reactants and products are present in concentrations that have no further tendency to change with time. In a dynamic equilibrium, the forward and reverse reactions occur at the same rate. For our reaction: \[2\mathrm{NH}_3 = \mathrm{N}_2 + 3\mathrm{H}_2\], at equilibrium, the concentrations of \( \mathrm{NH}_3 \), \( \mathrm{N}_2 \), and \( \mathrm{H}_2 \) will remain constant. However, they are not necessarily equal. The position of equilibrium is determined by the Gibbs free energy, which becomes zero at equilibrium (\(\Delta G=0\)). This allows us to solve for the degree of reaction \(\xi\) at equilibrium.
reaction quotient
The reaction quotient (\(Q\)) is a measure of the relative amounts of reactants and products present during a reaction at a given point in time. For our reaction, it is defined as: \( Q = \frac{[\mathrm{N}_2][\mathrm{H}_2]^3}{[\mathrm{NH}_3]^2} \). This quotient helps us compare the initial response of the system to its condition at equilibrium. When \( Q = K_{eq} \) (the equilibrium constant), the system is at equilibrium. In our case, we can express reactant and product concentrations in terms of the degree of reaction \( \xi \), transforming the equation to: \( Q = \frac{\xi (3\xi)^3}{(2 - 2\xi)^2} \). This makes it easier to calculate and plot \(Q\) for specific values of \( \xi \).
ideal gas law
The ideal gas law is a fundamental equation in chemistry, expressed as \( PV = nRT \), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. We use this law to calculate the volume of gases at equilibrium. For our reaction occurring at 1 atm pressure and 894 K, the total number of moles at equilibrium is \(n = 2 + 2\xi_{eq}\), where \(\xi_{eq}\) is the degree of reaction at equilibrium. Plugging in our values into \( PV = nRT \) will give us the volume at equilibrium.
standard Gibbs free energy change
The standard Gibbs free energy change (\(\Delta G^0\)) represents the free energy change for a reaction under standard conditions (298 K, 1 atm pressure, and 1 M concentration). It is calculated using \( \Delta G^0 = \Delta H^0 - T \Delta S^0 \), where \( \Delta H^0 \) and \( \Delta S^0 \) are the standard enthalpy and entropy changes, respectively. For our reaction, we substitute these values into the equation \(\Delta G = \Delta G^0 + RT\ln Q\), to calculate the Gibbs free energy at different temperatures and pressures, helping us understand the spontaneity of the reaction. When \(\Delta G = 0\), the system is at equilibrium, and we can solve for \(\xi\).

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Most popular questions from this chapter

Experimentally one finds that for a rubber band $$ \begin{aligned} &\left(\frac{\partial J}{\partial L}\right)_{\mathrm{T}, M}=\frac{a T}{L_{0}}\left[1+2\left(\frac{L_{0}}{L}\right)^{3}\right] \quad \text { and } \\\ &\left(\frac{\partial J}{\partial T}\right)_{L, M}=\frac{a L}{L_{0}}\left[1-\left(\frac{L_{0}}{L}\right)^{3}\right] \end{aligned} $$ where \(J\) is the tension, \(a=1.0 \times 10^{3} \mathrm{dyn} / \mathrm{K}\), and \(L_{0}=0.5 \mathrm{~m}\) is the length of the band when no tension is applied. The mass \(M\) of the rubber band is held fixed. (a) Compute \((\partial L / \partial T)_{J, M}\) and discuss its physical meaning. (b) Find the equation of state and show that \(\mathrm{d} J\) is an exact differential. (c) Assume that the heat capacity at constant length is \(C_{L}=1.0 \mathrm{~J} / \mathrm{K}\). Find the work necessary to stretch the band reversibly and adiabatically to a length of \(1 \mathrm{~m}\). Assume that when no tension is applied, the temperature of the band is \(T=290 \mathrm{~K}\). What is the change in temperature?

A material is found to have a thermal expansivity \(\alpha_{P}=R / P v+a / R T^{2} v\) and an isothermal compressibility \(K_{T}=1 / v(T f(P)+(b / P))\) where \(v=V / n\) is the molar volume. (a) Find \(f(P)\). (b) Find the equation of state. (c) Under what conditions is this material mechanically stable?

Two containers, each of volume \(V\), contain ideal gas held at temperature \(T\) and pressure \(P\). The gas in chamber 1 consists of \(N_{1, a}\) molecules of type \(a\) and \(N_{1, b}\) molecules of type \(b\). The gas in chamber 2 consists of \(N_{2, a}\) molecules of type \(a\) and \(N_{2, b}\) molecules of type \(b\). Assume that \(N_{1, a}+N_{1, b}=\) \(N_{2, a}+N_{2, b}\). The gases are allowed to mix so the final temperature is \(T\) and the final pressure is \(P\). (a) Compute the entropy of mixing. (b) What is the entropy of mixing if \(N_{1, \mathrm{a}}=N_{2, \mathrm{a}}\) and \(N_{1, \mathrm{~b}}=N_{2, \mathrm{~b}}\). (c) What is the entropy of mixing if \(N_{1, \mathrm{a}}=N_{2, \mathrm{~b}}\) and \(N_{1, \mathrm{~b}}=N_{2, \mathrm{a}}=0\). Discuss your results for (b) and (c).

A heat engine uses blackbody radiation as its operating substance. The equation of state for blackbody radiation is \(P=1 / 3 a T^{4}\) and the internal energy is \(U=a V T^{4}\), where \(a=7.566 \times 10^{-16}\) \(\mathrm{J} /\left(\mathrm{m}^{3} \mathrm{~K}^{4}\right)\) is Stefan's constant, \(P\) is pressure, \(T\) is temperature, and \(V\) is volume. The engine cycle consists of three steps. Process \(1 \rightarrow 2\) is an expansion at constant pressure \(P_{1}=P_{2} .\) Process \(2 \rightarrow 3\) is a decrease in pressure from \(P_{2}\) to \(P_{3}\) at constant volume \(V_{2}=V_{3}\). Process \(3 \rightarrow 1\) is an adiabatic contraction from volume \(V_{3}\) to \(V_{1}\). Assume that \(P_{1}=3.375 P_{3}, T_{1}=2000 \mathrm{~K}\), and \(V_{1}=10^{-3} \mathrm{~m}^{3}\). (a) Express \(V_{2}\) in terms of \(V_{1}\) and \(T_{1}=T_{2}\) in terms of \(T_{3}\) (b) Compute the work done during each part of the cycle. (c) Compute the heat absorbed during each part of the cycle. (d) What is the efficiency of this heat engine (get a number)? (e) What is the efficiency of a Carnot engine operating between the highest and lowest temperatures.

An insulated box with fixed total volume \(V\) is partitioned into \(m\) insulated compartments, each containing an ideal gas of a different molecular species. Assume that each compartment has the same pressure but a different number of moles, a different temperature, and ( \(\left.\left.P, \mathrm{n}_{i}, T_{\mathrm{i}}, V_{\mathrm{i}}\right)_{-}\right)\)a different volume. (The thermodynamic variables for the ith compartment are If all partitions are suddenly removed and the system is allowed to reach equilibrium: (a) Find the final temperature and pressure, and the entropy of mixing. (Assume that the particles are monatomic.) (b) For the special case of \(m=2\) and parameter \(\mathrm{n}_{1}=1 \mathrm{~mol}, T_{1}=300 \mathrm{~K}, V_{1}=1 \mathrm{l}, \mathrm{n}_{2}=3 \mathrm{~mol}_{1}\) and \(V_{2}=2\) 1, obtain numerical values for all parameters in part (a).
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