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Chapter 6: Problem 174

The incorrect expression among the following is (a) \(\ln \mathrm{K}=\frac{\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}}{\mathrm{RT}}\) (b) In isothermal process \(\mathrm{W}_{\text {reversible }}=-\mathrm{nRT} \operatorname{In} \frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{1}}\) (c) \(\frac{\Delta \mathrm{G}_{\text {System }}}{\Delta \mathrm{S}_{\text {total }}}=-\mathrm{T}\) (d) \(\mathrm{K}=\mathrm{e}^{\Delta \mathrm{G}^{\circ} / \mathrm{RT}}\)

Short Answer

Expert verified
Expressions (a) and (c) are incorrect.

Step by step solution

01

Analyze each expression for correctness according to thermodynamic equations and principles

**(a) \( \ln \mathrm{K}=\frac{\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}}{\mathrm{RT}} \)**This expression relates the equilibrium constant \( \mathrm{K} \) with the change in enthalpy \( \Delta \mathrm{H}^{\circ} \) and entropy \( \Delta \mathrm{S}^{\circ} \) using temperature \( \mathrm{T} \) and the gas constant \( \mathrm{R} \). The Gibbs free energy equation at standard conditions is used: \[ \Delta \mathrm{G}^{\circ} = \Delta \mathrm{H}^{\circ} - \mathrm{T} \Delta \mathrm{S}^{\circ} \]The correct form for \( \ln \mathrm{K} \) is:\[ \ln \mathrm{K}=\frac{-\Delta \mathrm{G}^{\circ}}{\mathrm{RT}} \]Since the expression in (a) does not match, it is incorrect.
02

Confirm correctness of expressions (b), (c), and (d)

**(b) In an isothermal process \( \mathrm{W}_{\text {reversible }}=-\mathrm{nRT} \operatorname{In} \frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{1}} \):**This is the equation for the work done in a reversible isothermal process for an ideal gas. It is consistent with thermodynamic principles.**(c) \( \frac{\Delta \mathrm{G}_{\text {System }}}{\Delta \mathrm{S}_{\text{total }}}=-\mathrm{T} \):**This expression is incorrect according to the Second Law of Thermodynamics. For spontaneity, the correct equation is:\[ \Delta \mathrm{G}_{\text{System}} = -T \Delta S_{\text{total}} \]Thus, it simplifies to \( -\Delta S_{\text{total}} = \frac{\Delta G_{\text{System}}}{T} \) and must be correlated with the sign of \( \Delta G_{\text{System}} \), not the division expression provided.**(d) \( \mathrm{K}=\mathrm{e}^{\Delta \mathrm{G}^{\circ} / \mathrm{RT}} \):**This represents the relation between equilibrium constant \( \mathrm{K} \) and standard Gibbs free energy change \( \Delta \mathrm{G}^{\circ} \), which is correctly described by:\[ \mathrm{K}=\mathrm{e}^{-\Delta \mathrm{G}^{\circ} / \mathrm{RT}} \]Indicating the exponent should be negative.

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Key Concepts

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

Gibbs Free Energy
Gibbs free energy is a vital concept in thermodynamics, signifying the maximum reversible work a thermodynamic system can perform at constant temperature and pressure. The equation connecting Gibbs free energy to other thermodynamic quantities is:\[ \Delta G = \Delta H - T \Delta S \]where:
  • \( \Delta G \) is the change in Gibbs free energy,
  • \( \Delta H \) is the change in enthalpy,
  • \( T \) is the temperature in Kelvin, and
  • \( \Delta S \) is the change in entropy.
A negative \( \Delta G \) indicates a spontaneous process. The relationship with the equilibrium constant \( K \) is crucial because it demonstrates the system's tendency to reach a particular equilibrium. The expression \( \ln K = -\Delta G^{\circ} / RT \) shows that Gibbs free energy connects to both the thermodynamic stability and reaction feasibility: a more negative \( \Delta G \) equates to a higher \( K \), implying a more favorable reaction. Understanding these relationships helps in predicting how systems react under varying conditions of pressure and temperature.
Equilibrium Constant
The equilibrium constant \( K \) depicts the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction. It serves as an indicator of the favorability of a reaction's completion.
In thermodynamic terms, \( K \) is directly related to Gibbs free energy using the formula:\[ K = e^{-\Delta G^{\circ} / RT} \]This connection is pivotal because it links the concepts of thermodynamics with chemical equilibrium. When \( \Delta G^{\circ} \) is negative, it results in a large positive \( K \), signifying that the products are favored.
The temperature dependency of the equilibrium constant can be clearly differentiated using the Van 't Hoff equation:\[ \ln K = \frac{-\Delta H^{\circ} + T \Delta S^{\circ}}{RT} \]This indicates how changes in temperature affect \( K \), essentially amending the favorability of either the products or reactants at equilibrium. A thorough understanding of \( K \) solidifies one's grip over the dynamic nature of chemical systems.
Entropy
Entropy \( S \) is a measure of the disorder or randomness in a thermodynamic system. The concept becomes essential when discussing spontaneity and the direction of processes. Entropy change \( \Delta S \) can be understood as the measure of energy dispersion in a system. The Second Law of Thermodynamics states that the total entropy of an isolated system can never decrease over time.
Thermodynamically:\[ \Delta G = \Delta H - T \Delta S \]The balance of these components, enthalpy \( \Delta H \) and entropy \( \Delta S \), dictates the spontaneity of a reaction. Processes with positive \( \Delta S \) are often spontaneous because they conform to natural tendencies of systems to disperse energy.
  • High entropy configurations signify likely states under equilibrium.
  • Entropy changes also determine how temperature affects reaction spontaneity.
Hence, understanding entropy's role assists in explaining why and how chemical processes occur in nature, making it essential knowledge for chemistry enthusiasts.
Enthalpy
Enthalpy \( H \) is a measure of the total heat content of a system, functionally defined as \( H = U + PV \), where \( U \) is internal energy, and \( P \) and \( V \) are pressure and volume, respectively.
It reflects the heat absorbed or released under constant pressure when a process occurs. A positive \( \Delta H \) indicates an endothermic process, while a negative \( \Delta H \) reflects an exothermic process.
In terms of Gibbs free energy, enthalpy plays a direct role in calculating spontaneity:\[ \Delta G = \Delta H - T \Delta S \]A lower enthalpy generally drives systems toward equilibrium by reducing \( \Delta G \) when not counteracted by entropy changes. The connection between enthalpy changes \( \Delta H \) and the equilibrium constant \( K \) is also described by the Van 't Hoff equation, which examines how temperature variations impact system balance.
  • \( \Delta H \) helps predict heat flow tendencies.
  • Detailed knowledge of \( \Delta H \) guides the studying of reaction energetics and pathways.
Understanding enthalpy is crucial for predicting how different systems evolve under constant pressure conditions, especially in chemical reaction contexts.

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Most popular questions from this chapter

Which are the intensive properties? (a) Volume (b) Enthalpy (c) Temperature (d) Refractive index

Which of the following is /are true about the isothermal expansion of an ideal gas? (a) \(\Delta \mathrm{U}=0\) (b) \(\Delta \mathrm{T}=0\) (c) \(\mathrm{q}=2.303 \mathrm{nRT} \log _{10}\left(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}\right)\) (d) \(\mathrm{q}=0\)

Two moles of an ideal gas are compressed at \(300 \mathrm{~K}\) from a pressure of 1 atm to a pressure of 2 atm. The change in free energy is (a) \(5.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(2.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(3.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(8.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The dissociation energies of \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) to convert them into gaseous atoms are 360 and \(620 \mathrm{kcal}\) mol respectively. The bond energy of \(\mathrm{C}-\mathrm{C}\) bond is (a) \(280 \mathrm{kcal} \mathrm{mol}^{-1}\) (b) \(240 \mathrm{kcal} \mathrm{mol}^{-1}\) (c) \(160 \mathrm{kcal} \mathrm{mol}^{-1}\) (d) \(80 \mathrm{kcal} \mathrm{mol}^{-1}\)

At \(300 \mathrm{~K}\) and \(1 \mathrm{~atm}, 15 \mathrm{~mL}\) a gaseous hydrocarbon requires \(375 \mathrm{~mL}\) air containing \(20 \% \mathrm{O}_{2}\) by volume for complete combustion. After comustion the gases occupy \(330 \mathrm{~mL}\). Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: (a) \(\mathrm{C}_{3} \mathrm{H}_{8}\) (b) \(\mathrm{C}_{4} \mathrm{H}_{8}\) (c) \(\mathrm{C}_{4} \mathrm{H}_{10}\) (d) \(\mathrm{C}_{3} \mathrm{H}_{6}\)
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