Problem 174 - A Complete Resource Book in Chemistry for JEE Main

Question

The incorrect expression among the following is \([2012]\) (a) \(\ln \mathrm{K}=\frac{\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}}{\mathrm{RT}}\) (b) In isothermal process \(\mathrm{W}_{\text {revesible }}=-\mathrm{nRT} \operatorname{In} \frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{1}}\) (c) \(\frac{\Delta \mathrm{G}_{\text {sysem }}}{\Delta \mathrm{S}_{\text {loel }}}=-\mathrm{T}\) (d) \(\mathrm{K}=\mathrm{e}^{\Delta \mathrm{G}^{*} / \mathrm{RT}}\)

Step-by-Step Solution

Answer

Options (c) and (d) are incorrect expressions.

Step 1: Analyze Option (a)

The expression given is \( \ln K = \frac{\Delta H^\circ}{RT} - \frac{T \Delta S^\circ}{RT} \). This can be rewritten as \( \ln K = \frac{\Delta H^\circ - T \Delta S^\circ}{RT} \), which is the correct Gibbs free energy relationship for equilibrium constants, so it is correct.

Step 2: Analyze Option (b)

The provided expression is \( W_{\text{reversible}} = -nRT \ln \frac{V_f}{V_1} \). This formula corresponds to the work done during an isothermal and reversible expansion or compression of an ideal gas, so it is correct.

Step 3: Analyze Option (c)

The expression \( \frac{\Delta G_{\text{sysem}}}{\Delta S_{\text{loel}}} = -T \) seems incorrect as \( \Delta G \) should relate to \( \Delta S_{\text{total}} \) not \( \Delta S_{\text{local}} \), and the equation does not match known thermodynamic identities. This option is incorrect.

Step 4: Analyze Option (d)

The expression given is \( K = e^{\frac{\Delta G^*}{RT}} \). The correct relationship between Gibbs free energy change and equilibrium constant \( K \) is \( K = e^{-\frac{\Delta G^\circ}{RT}} \), factoring in the negative sign. Therefore, this expression is incorrect.

Key Concepts

Gibbs Free Energy

Gibbs Free Energy, denoted by \( \Delta G \), is a valuable concept in thermodynamics that measures the "useful" energy in a system that can do work at constant temperature and pressure. It determines the spontaneity of a process.
To predict whether a reaction occurs spontaneously,

• If \( \Delta G < 0 \), the process is spontaneous.
• If \( \Delta G = 0 \), the system is at equilibrium, meaning no net change occurs.
• If \( \Delta G > 0 \), the process is non-spontaneous and will not occur without external energy.

This formula ties into the equilibrium constant \( K \) where \( \Delta G = \Delta H - T \Delta S \). Here, \( \Delta H \) is the change in enthalpy and \( \Delta S \) is the change in entropy. A significant equation showing this tie is \( K = e^{-\frac{\Delta G^\circ}{RT}} \), which connects Gibbs free energy to equilibrium constant \( K \). The negative sign indicates the inversely proportional relationship between \( \Delta G^\circ \) and \( K \).

Equilibrium Constant

The Equilibrium Constant, \( K \), is a fascinating aspect of chemical reactions. It's a dimensionless value that indicates the ratio of product concentrations to the reactant concentrations at equilibrium, where each concentration is raised to the power of its stoichiometric coefficient.
This constant can be expressed mathematically as

• \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

In this expression, \([A]\) and \([B]\) are the concentrations of the reactants, while \([C]\) and \([D]\) represent the concentrations of the products.
Equilibrium constants also play a critical role in determining reaction spontaneity and direction. They are linked with Gibbs Free Energy through the equation:

• \( \ln K = -\frac{\Delta G^\circ}{RT} \)

This ties \( K \) directly to how favorable a reaction is under standard state conditions.

Isothermal Process

An isothermal process is very intriguing, highlighting a scenario where a system's temperature remains constant throughout the process. This characteristic makes it different from adiabatic and other thermodynamic processes.
In an isothermal process involving an ideal gas, any heat absorbed by the system matches the work done by the system on the surroundings or vice versa. This means there's a continuous exchange of heat energy to maintain the temperature.
For an isothermal process involving ideal gases, the work done by or on the system is derived using the equation:

• \( W = -nRT \ln \frac{V_f}{V_i} \)

Here, \( n \) represents the number of moles, \( R \) is the gas constant, \( T \) denotes the temperature, while \( V_f \) and \( V_i \) are the final and initial volumes, respectively. The logarithmic term arises from the integration of the combined gas law under constant temperature conditions.

Reversible Expansion

Reversible Expansion is a captivating form of thermodynamic transformation that operates under conditions allowing the system to reverse the process at any point by an infinitesimally small change. This ensures the system is always in equilibrium with its surroundings, maximizing work.In a reversible expansion, an ideal gas is gradually expanded against a constraining pressure, such that it can remain quasi-static, i.e., at equilibrium.For reversible isothermal expansion of an ideal gas, the maximum useful work is given by:

• \( W_{rev} = -nRT \ln \frac{V_f}{V_i} \)

This equation mirrors that of the work calculation for isothermal processes, showcasing how tightly these concepts are connected. Such expansions are theoretical ideals, essential for understanding bounds of real-world processes and efficiencies. They set the benchmark for comparing actual processes, providing insight into possible improvements.