(a) Calculate the molar frec energy change when the partial pressure of \(\mathrm{NO}(\mathrm{g})\) in a mixture of gases is increased from \(5.00\) bar to \(15.00\) bar at constant temperature and volume. (b) Calculate the molar frec energy change when the partial pressure of HCN(g) in a mixture of gases is increased from \(5.00\) bar to 15.00 bar. (c) Calculare the molar free energy change when the partial pressure of \(\mathrm{O}_{2}(\mathrm{~g})\) in a mixture of gases is decreased from \(3.00\) bar to \(1.50\) bar. (d) Calculate the molar free energy change when the partial pressure of \(\mathrm{O}_{2}(\mathrm{~g})\) in a mixture of gases is decreased from \(9.00\) bar to \(4.50\) bar.

Short Answer

Expert verified
The molar Gibbs free energy change for (a) \(NO(g)\) and (b) \(HCN(g)\) is \(RT \ln(3)\), for (c) \(O_{2}(g)\) it's \(RT \ln(0.5)\), and for (d) \(O_{2}(g)\) it's also \(RT \ln(0.5)\). Exact numerical values require the temperature \(T\) in Kelvin.

Step by step solution

01

- Understanding the Concept

The molar Gibbs free energy change (\Delta G\textsubscript{m}\textsuperscript{\textdegree}\textsuperscript{\text{\textprime}}) for a gas can be calculated at constant temperature using the formula: \[\Delta G_m' = nRT \ln\left(\frac{P_2}{P_1}\right)\] where \(n\) is the number of moles (since we're calculating the molar free energy change, \(n = 1\)), \(R\) is the ideal gas constant (\(8.314 \text{ J/(mol K)}\)), \(T\) is the temperature in Kelvin, \(P_1\) is the initial pressure, and \(P_2\) is the final pressure. The fact that the problem specifies 'molar' free energy change allows us to consider a single mole throughout the calculations.
02

- Calculating Change for NO(g)

Apply the formula for \(NO(g)\) increasing from \(5.00\) bar to \(15.00\) bar: \[\Delta G_m' = RT \ln\left(\frac{15.00}{5.00}\right)\] Convert pressures to the same units if needed (in this case the units are the same, so we can use them as is).
03

- Calculating Change for HCN(g)

The calculation for \(HCN(g)\) is identical to that for \(NO(g)\) as the change in partial pressure is the same: \[\Delta G_m' = RT \ln\left(\frac{15.00}{5.00}\right)\] The temperature \(T\) is constant and doesn't change the outcome, so it's not given explicitly.
04

- Calculating Change for O2(g) - Case 1

For the decrease from \(3.00\) bar to \(1.50\) bar for \(O_2(g)\): \[\Delta G_m' = RT \ln\left(\frac{1.50}{3.00}\right)\] Again, no need to convert pressure as the units are consistent.
05

- Calculating Change for O2(g) - Case 2

For the decrease from \(9.00\) bar to \(4.50\) bar for \(O_2(g)\): \[\Delta G_m' = RT \ln\left(\frac{4.50}{9.00}\right)\] As in the previous cases, we still don't need to consider units of pressure as they cancel out, and we're using the consistent unit bar.
06

- Calculating R and T values

To carry out the calculations, we need the value of \(R\) and \(T\). Since the temperature \(T\) isn't given in the problem, we cannot compute the exact numerical value for \(\Delta G_m'\) in Joules. For actual computation, the value of \(T\) must be known. However, we can express our answer in terms of \(RT\), which will give a general equation that can be used to find \(\Delta G_m'\) once \(T\) is known.

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Key Concepts

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

Partial Pressure
When discussing gases, especially in a mixture, the concept of partial pressure is crucial. Partial pressure refers to the pressure that a single gas in a mixture would exert if it occupied the entire volume alone at the same temperature. Mathematically, it can be expressed as a fraction of the total pressure, which is proportional to the mole fraction of the gas in the mixture. This concept is based on Dalton's Law of Partial Pressures, and is integral to solving problems related to gas mixtures in thermodynamics.

Understanding partial pressure is essential because it influences how gases behave and react with each other. In chemical reactions, changes in partial pressures of reactants or products can drive the reaction forward or backward, affecting the overall chemical equilibrium and the resulting Gibbs free energy change.
Ideal Gas Constant
The ideal gas constant, represented by the symbol R, is a fundamental physical constant that appears in the equation of state for ideal gases. Its value, which is approximately 8.314 J/(mol K), relates the energy scale to the temperature scale, and when multiplied by temperature, gives an energy per mole of gas. It's instrumental in many thermodynamic equations, including the calculation of Gibbs free energy change.

R assumes that the gas being described behaves ideally, which means the gas particles do not interact with each other except through elastic collisions and occupy no volume themselves. While real gases do not perfectly adhere to these assumptions, the ideal gas law and constants like R provide a good approximation for many gases under a range of conditions.
Gibbs Free Energy Formula
The Gibbs free energy formula is vital for understanding the spontaneity of processes. It is given by the equation \[G = H - TS\] where G is Gibbs free energy, H is enthalpy, T is absolute temperature, and S is entropy. The equation reflects the trade-off between enthalpy and entropy that dictates whether a process can happen spontaneously.

For changes in the state of a system, the change in Gibbs free energy is given by: \[\Delta G = \Delta H - T\Delta S\] Here, a negative value of \(\Delta G\) implies a spontaneous process at constant pressure and temperature. When dealing with gases, and particularly changes in partial pressure, the formula simplifies to the one used in the textbook exercise: \[\Delta G_m' = RT \ln\left(\frac{P_2}{P_1}\right)\] which is specifically tailored for the molar free energy changes resulting from pressure changes in gases.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy, particularly the rules governing the conversion of heat into mechanical work and vice versa. It involves studying systems at macroscopic scales and determining the directional flow of energy and how it affects the system's properties.

Four fundamental laws of thermodynamics establish the principles regarding energy, heat, work, and entropy, laying the groundwork for countless scientific and engineering processes, including engines, refrigerators, and even living cells. The principles of thermodynamics are integral to the understanding of Gibbs free energy and the chemical potential, providing the framework for predicting the spontaneity and equilibrium position of chemical reactions.
Chemical Potential
The term chemical potential refers to the energy that can be released (or the increase in thermodynamic potential) when the number of particles increases by one at constant pressure and temperature. It is essentially the 'molar Gibbs free energy' of a substance and critically influences the behavior of substances in different phases or within a mixture of gases.

Chemical potential is key to understanding how particles move from areas of high concentration to low concentration (diffusion), how phases change (melting, vaporization), and how substances react in chemical reactions. When a system reaches equilibrium, the chemical potential for a substance must be the same everywhere within the system, ensuring no net movement of particles and no net change in the state of the system.

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Most popular questions from this chapter

A mixture consisting of \(1.000 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) and \(1.000 \mathrm{~mol} \mathrm{CO}(\mathrm{g})\) is placed in a \(10.00-\mathrm{L}\) reaction vessel at \(800 \mathrm{~K}\). At equilibrium, \(0.665 \mathrm{~mol} \mathrm{CO}_{2}(\mathrm{~g})\) is present as a result of the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})=\) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\). What are (a) the equilibrium concentrations for all substances and (b) the value of \(K_{e}\) ?

A 0.500-L reaction vessel at \(700 \mathrm{~K}\) contains \(1.20 \times 10^{-3} \mathrm{~mol} \mathrm{SO}_{2}(\mathrm{~g}), 5.0 \times 10^{-4} \mathrm{~mol} \mathrm{O}_{2}(\mathrm{~g})\), and \(1.0 \times 10^{-4} \mathrm{~mol} \mathrm{SO}_{3}(\mathrm{~g}) .\) At \(700 \mathrm{~K}, K_{c}=1.7 \times 10^{6}\) for the cquilibrium \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}) .\) (a) Calculate the reaction quotient \(Q_{e}\) (b) Will more \(\mathrm{SO}_{3}(\mathrm{~g})\) tend to form?

Predict whether each of the following equilibria will shift toward products or reactants with a temperature increase. (a) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g})\), \(\Delta H^{\circ}=+206 \mathrm{~kJ}\) (b) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})=\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\), \(\Delta H^{2}=-41 \mathrm{~kJ}\) (c) \(2 \mathrm{SO}_{2}\) (g) \(+\mathrm{O}_{2}\) (g) \(=2 \mathrm{SO}_{3}\) (g), \(\Delta H^{\circ}=-198 \mathrm{~kJ}\)

Explain why the following cquilibria are heterogeneous and write the reaction quotient \(Q\) for each one. (a) \(2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{FcCl}_{3}(\mathrm{~s})\) (b) \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)(\mathrm{s}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})\) (c) \(2 \mathrm{KNO}_{3}\) (s) \(\rightleftharpoons 2 \mathrm{KNO}_{2}\) (s) \(+\mathrm{O}_{2}\) (g)

For the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})=2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(400 \mathrm{~K}, K=41\). Find the value of \(\mathrm{K}\) for each of the following reactions at the same temperature. (a) \(2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\) (b) \(\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})\) (c) \(2 \mathrm{~N}_{2}(\mathrm{~g})+6 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NH}_{3}(\mathrm{~g})\)

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