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Explain the difference between \(\Delta G\) and \(\Delta G^{\circ}\).

Short Answer

Expert verified
\(\Delta G\) and \(\Delta G^{\circ}\) both measure the maximum reversible work a system can do at constant temperature and pressure. \(\Delta G\) describes this energy under any condition and is influenced by concentrations or pressures of the reactants and products, while \(\Delta G^{\circ}\) refers to these quantities under standard conditions.

Step by step solution

01

Definition of \(\Delta G\)

\(\Delta G\), or Gibbs free energy change, is a thermodynamic quantity that measures the maximum reversible work that a system can perform at constant temperature and pressure. It is given as \(\Delta G = \Delta H - T\Delta S\), where \(\Delta H\) is the change in enthalpy (or heat content), \(T\) is the absolute temperature and \(\Delta S\) is the change in entropy (or disorder). The sign of \(\Delta G\) dictates the spontaneity of a process. If \(\Delta G < 0\), the process is spontaneous; if \(\Delta G > 0\), the process is non-spontaneous.
02

Definition of \(\Delta G^{\circ}\)

\(\Delta G^{\circ}\) is the standard Gibbs free energy change, which is the energy change when reactants and products are in their standard states (usually 1 bar or 1 atm for gases, 1 M for solutions). It is given as \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), where \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are the standard enthalpy change and standard entropy change respectively.
03

Differences between \(\Delta G\) and \(\Delta G^{\circ}\)

The difference between \(\Delta G\) and \(\Delta G^{\circ}\) lies mainly in their use and interpretation. While \(\Delta G\) indicates the spontaneity of a process under any condition, \(\Delta G^{\circ}\) reveals the spontaneity under standard conditions. Furthermore, \(\Delta G\) is influenced by the concentrations (or pressures) of the reactants and products, while \(\Delta G^{\circ}\) refers to situations where these are in their standard states.

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat, work, temperature, and energy. In terms of chemistry and the study of reactions, thermodynamics focuses on how energy is transferred and transformed during chemical processes. The laws of thermodynamics govern the principles that dictate whether a reaction can occur spontaneously, which leads directly to understanding Gibbs free energy.

The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. The second law introduces the concept of entropy, a measure of the disorder or randomness in a system. It states that the entropy of the universe tends to increase over time, meaning that systems naturally tend towards disorder unless energy is input to maintain or create order.

In the realm of chemical reactions, thermodynamics helps us calculate whether a reaction is energetically feasible, and if so, whether it will occur spontaneously. This is where Gibbs free energy comes into play, providing a critical thermodynamic function that predicts the direction and extent of chemical reactions.
Chemical Spontaneity
Chemical spontaneity refers to whether a chemical reaction can occur without needing to be driven by an outside force or energy. The concept of spontaneity in chemistry is closely linked to Gibbs free energy, denoted by \( \Delta G \). Gibbs free energy is a thermodynamic property that combines enthalpy (the heat content of a system) and entropy (the system's disorder) to determine the feasibility and spontaneity of a reaction.

A spontaneous reaction is one that can proceed on its own, without any continuous input of energy. According to Gibbs free energy, a reaction is spontaneous if \( \Delta G < 0 \) – meaning that the free energy of the system decreases. On the contrary, if \( \Delta G > 0 \) the reaction is not spontaneous and requires energy to proceed. Equally important is the condition of \( \Delta G = 0 \) which indicates that the system is at equilibrium and the reaction can proceed in either direction with no net change over time.

Understanding chemical spontaneity helps scientists and engineers predict how reactions will behave under different conditions, enabling the design of processes and products that align with these natural tendencies.
Standard State Conditions
Standard state conditions are a set of agreed-upon reference conditions used to report thermodynamic properties, such as enthalpy (\( \Delta H^\circ \)), entropy (\( \Delta S^\circ \)), and Gibbs free energy (\( \Delta G^\circ \)). These conditions include a pressure of 1 bar (or 1 atm in some contexts), a concentration of 1 molarity for substances in solution, and a specified temperature, oftentimes 298.15 K (25 °C). The purpose of these conditions is to provide a consistent and comparable basis for measuring and communicating the properties of substances.

When chemists refer to any of the 'standard' thermodynamic functions, they are referencing the system's behavior when all components are in their standard states. For example, \( \Delta G^\circ \) represents the Gibbs free energy change when reactants and products are at standard state conditions. This allows comparing different reactions under the same baseline conditions.

Using standard states is crucial for establishing a common reference that aids in calculating the spontaneity of a reaction under actual conditions. This also assists in constructing tables of standard thermodynamic data which are immensely useful for predicting reaction behavior and for calculating equilibrium constants.

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Most popular questions from this chapter

(a) Calculate \(\Delta G^{\circ}\) and \(K_{P}\) for the following equilibrium reaction at \(25^{\circ} \mathrm{C}\). The \(\Delta G_{f}^{\circ}\) values are 0 for \(\mathrm{Cl}_{2}(g),-286 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{PCl}_{3}(g),\) and \(-325 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{PCl}_{5}(g)\) $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$. (b) Calculate \(\Delta G\) for the reaction if the partial pressures of the initial mixture are \(P_{\mathrm{PCl}_{5}}=0.0029 \mathrm{~atm}\) \(P_{\mathrm{PCl}_{3}}=0.27 \mathrm{~atm},\) and \(P_{\mathrm{Cl}_{2}}=0.40 \mathrm{~atm}\).

Under what conditions does a substance have a standard entropy of zero? Can a substance ever have a negative standard entropy?

Predict the signs of \(\Delta H, \Delta S,\) and \(\Delta G\) of the system for the following processes at \(1 \mathrm{~atm}:\) (a) ammonia melts at \(-60^{\circ} \mathrm{C},\) (b) ammonia melts at \(-77.7^{\circ} \mathrm{C},\) (c) ammonia melts at \(-100^{\circ} \mathrm{C}\). (The normal melting point of ammonia is \(\left.-77.7^{\circ} \mathrm{C} .\right)\)

For reactions carried out under standard-state conditions, Equation (18.10) takes the form \(\Delta G^{\circ}=\Delta H^{\circ}-\) \(T \Delta S^{\circ} .\) (a) Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are independent of temperature, derive the equation $$\ln \frac{K_{2}}{K_{1}}=\frac{\Delta H^{\circ}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$$ where \(K_{1}\) and \(K_{2}\) are the equilibrium constants at \(T_{1}\) and \(T_{2}\), respectively. (b) Given that at \(25^{\circ} \mathrm{C} K_{\mathrm{c}}\) is \(4.63 \times 10^{-3}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=58.0 \mathrm{~kJ} / \mathrm{mol} $$ calculate the equilibrium constant at \(65^{\circ} \mathrm{C}\)

The equilibrium constant \(\left(K_{P}\right)\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ is 4.40 at \(2000 \mathrm{~K}\). (a) Calculate \(\Delta G^{\circ}\) for the reaction. (b) Calculate \(\Delta G\) for the reaction when the partial pressures are \(P_{\mathrm{H}_{2}}=0.25 \mathrm{~atm}, P_{\mathrm{CO}_{2}}=0.78 \mathrm{~atm}\) \(P_{\mathrm{H}_{2} \mathrm{O}}=0.66 \mathrm{~atm},\) and \(P_{\mathrm{CO}}=1.20 \mathrm{~atm}\)

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