The free energy change \(\Delta G=0\), when (a) the system is at equilibrium (b) catalyst is added (c) reactants are initially mixed thoroughly (d) the reactants are completely consumed

Short Answer

Expert verified
\(\Delta G = 0\) when the system is at equilibrium (a), while adding a catalyst (b) does not change free energy. When reactants are initially mixed (c), \(\Delta G\) is not necessarily zero, and if reactants are completely consumed (d), it suggests completion rather than equilibrium, unless no further change is possible at that state.

Step by step solution

01

Understanding Free Energy Change at Equilibrium

The free energy change, denoted as \(\Delta G\), is a thermodynamic quantity that indicates the spontaneity of a process. At equilibrium, the process is at a state where it can no longer proceed spontaneously in either the forward or reverse direction, which means \(\Delta G = 0\) because there is no net change in free energy.
02

Analyzing the Effect of a Catalyst

A catalyst is a substance that increases the rate of a reaction without being consumed by it. It works by lowering the activation energy, but it does not affect the thermodynamics of the reaction. Since it only affects the kinetics (rate) of the reaction and not the state functions like free energy, the presence of a catalyst does not affect the value of \(\Delta G\), which remains the same before and after adding a catalyst.
03

Consideration of Initial Mixing of Reactants

When reactants are initially mixed thoroughly, the reaction begins and progresses towards equilibrium. \(\Delta G\) is not necessarily zero because the initial mixing doesn’t define equilibrium; it merely initiates the reaction process. The value of \(\Delta G\) will be determined by whether the reaction is spontaneous or non-spontaneous and by the distance from equilibrium.
04

Assessing the Scenario of Complete Consumption of Reactants

When the reactants are completely consumed, it means that the reaction has gone to completion. At this point, the system may not be at equilibrium; instead, the products would dominate. If the reaction is perfectly complete, no reverse reaction can occur, indicating a state of no change. However, this does not comply with the definition of equilibrium regarding \(\Delta G = 0\), unless at the end point, no further net change is possible without the addition of more reactants.

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Key Concepts

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

Thermodynamics in Chemistry
Understanding thermodynamics in chemistry is crucial for grasping how reactions occur and predicting whether they will proceed. Thermodynamics revolves around several key concepts, including energy, heat, work, and the laws of thermodynamics. One of the central components of this field is free energy, symbolized as \( \Delta G \).

Free energy change reflects the energy available to do work during a chemical reaction and is a measure of the inherent potential of a reaction to occur spontaneously. The sign of \( \Delta G \) helps us determine a process's spontaneity: if it's negative, the reaction tends to occur on its own, while a positive value indicates non-spontaneity. A zero value, as in equilibrium conditions, indicates that the system is stable and no further work can be extracted without additional changes.

These principles allow chemists to predict the direction and extent of chemical reactions and are foundational for everything from biochemistry to materials science.
Chemical Equilibrium
Chemical equilibrium represents a state in which the rates of the forward and reverse reactions are equal, leading to no net change in the concentrations of reactants and products over time. It's crucial to note that this doesn't mean the reactants and products are in equal concentrations, but rather that they are no longer changing.

At equilibrium, \( \Delta G = 0 \), signifying that the system has reached a state of balance. In the context of a chemical reaction, achieving equilibrium doesn't necessarily mean that the reaction stops entirely; it continues but with reactants converting to products at the same rate as products revert to reactants.

Understanding equilibrium is vital for various applications, such as synthesis reactions in the pharmaceutical industry, where maximizing yield is key, as well as in understanding environmental systems where reactions often occur at equilibrium.
Reaction Spontaneity
Reaction spontaneity is reflected in whether a chemical reaction will proceed without the need for external energy. Not to be confused with the speed of the reaction, spontaneity is about the inherent potential for a process to occur. Thermodynamic quantities, primarily free energy change \( \Delta G \), entropy change \( \Delta S \), and enthalpy change \( \Delta H \), determine this spontaneity.

Considering the Gibbs free energy equation, \( \Delta G = \Delta H - T\Delta S \), where \( T \) is temperature, predictions about reaction spontaneity are possible. Negative values of \( \Delta G \) indicate spontaneous reactions under certain conditions, typically favored at lower temperatures for exothermic reactions (\( \Delta H < 0 \) and \( \Delta S > 0 \)) and at higher temperatures for endothermic reactions (\( \Delta H > 0 \) and \( \Delta S > 0 \)).

Understanding these relationships helps chemists control reaction conditions to obtain desirable outcomes—a fundamental aspect of chemical engineering and design.
Effect of Catalyst on Reaction
The role of a catalyst in a chemical reaction is to lower the activation energy—the energy barrier that must be overcome for a reaction to proceed. Consequently, this accelerates the reaction rate, enabling it to reach equilibrium more quickly. However, it is crucial to emphasize that adding a catalyst does not affect the equilibrium position or the free energy change \( \Delta G \).

The presence of a catalyst neither shifts the equilibrium concentrations of reactants and products nor alters the thermodynamic properties of the reaction system. What changes is how rapidly the equilibrium is approached. This understanding is beneficial in industrial chemistry, where catalysts are used to increase the efficiency of chemical processes, enabling faster production rates without necessitating alterations to the chemical's free energy landscape.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}, 1\) mole of \(\mathrm{MgSO}_{4}\) was dissolved in water, the heat evolved was found to be \(91.2 \mathrm{~kJ}\). One mole of \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) on dissolution gives a solution of the same composition accompanied by an absorption of \(13.8 \mathrm{~kJ}\). The enthalpy of hydration, i.e., \(\Delta H\) for the reaction $$ \mathrm{MgSO}_{4}(s)+7 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}(s) \text { is : } $$ (a) \(-105 \mathrm{~kJ} / \mathrm{mol}\) (b) \(-77.4 \mathrm{~kJ} / \mathrm{mol}\) (c) \(105 \mathrm{~kJ} / \mathrm{mol}\) (d) None of these

Out of molar entropy (I), specific volume (II), heat capacity (III), volume (IV), extensive properties are : (a) I, II (b) I, II, IV (c) II, III (d) III, IV

Calculate \(\Delta S\) for following process: $$ \underset{\text { at } 100 \mathrm{~K}}{X(s)} \longrightarrow \underset{\text { at } 200 \mathrm{~K}}{X(l)} $$ Given : Melting point of \(X_{(s)}=100 \mathrm{~K} ; \Delta H_{\text {Fusion }}=20 \mathrm{~kJ} / \mathrm{mol} ; C_{p, m}(X, l)=10 \mathrm{~J} / \mathrm{mol} \mathrm{K}\) (a) \(26.93 \mathrm{~J} / \mathrm{K}\) (b) \(206.93 \mathrm{~J} / \mathrm{K}\) (c) \(203 \mathrm{~J} / \mathrm{K}\) (d) \(206.93 \mathrm{~kJ} / \mathrm{K}\)

The entropies of \(\mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{H}(\mathrm{g})\) are \(130.6\) and \(114.6 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) respectively at \(298 \mathrm{~K}\). Using the data given below calculate the bond energy of \(\mathrm{H}_{2}\) (in \(\mathrm{kJ} / \mathrm{mol}\) ): \(\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g) ; \Delta G^{\circ}=406.6 \mathrm{~kJ}\) (a) \(377.2\) (b) \(436.0\) (c) \(425.5\) (d) \(430.5\)

5 mole of an ideal gas expand isothermally and irreversibly from a pressure of 10 atm to 1 atm against a constant external pressure of 1 atm. \(w_{\mathrm{irr}}\) at \(300 \mathrm{~K}\) is : (a) \(-15.921 \mathrm{~kJ}\) (b) \(-11.224 \mathrm{~kJ}\) (c) \(-110.83 \mathrm{~kJ}\) (d) None of these

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