Derivation of the conditional equation of chemical equilibrium

Derivation of the conditional equation of chemical equilibrium

The condition under which a chemical reaction is in equilibrium is when the entropy of the reacting system, including the reacting system and the outside world, does not change. In other words, the following conditions hold

From this condition,

Conditional equation for chemical equilibrium：

⊿∑G_f⁰ = －RT・Ln{(Concentration product of product form) / (Concentration product of the original form)}

We want to derive this. We will therefore deform the thermodynamic function.

Review the energy breakdown diagram from the previous course (reaction between hydrogen and oxygen) to see that the following relationship holds

The amount of this minute change is noted.

The first law of thermodynamics is written in terms of minute changes.

The definition of entropy change is written in terms of minute changes：

→    TdS = dQ = dU + PdV             Substitute this into (Equation 1-1).

If the temperature is kept constant before and after the chemical reaction, dT = 0,

From the equation of state of gas: PV = RT (at 1 mole),

If we integrate this from "partial pressure 1 in standard state" to "partial pressure P in any state",

This G_standard is the standard generated Gibbs energy G_f⁰ that has emerged so far.In other words,

As mentioned earlier, we assumed per mole in the equation of state of the gas, so what this equation means is the Gibbs energy per mole.

In chemistry, the "Gibbs energy per mole" is called the chemical potential.

Now consider a simple reaction system (substance A ⇆ substance B). The standard Gibbs energy of formation is noted below the reaction equation below.

                                                                      Origin form                       Generative form

                                                                      Substance A          ⇆        SubstanceB

The standard Gibbs energy of formation      G_f⁰_{(A)generation}                          G_f⁰_(B)origin

For substance A, the energy required to change from the standard state to an arbitrary concentration (partial pressure) and temperature is  RTLnP_A. 

The Gibbs energy for an arbitrary state is expressed as

Conditions of equilibrium between the generated and original forms：⊿G = G_{(B)generation} － G_(A)origin = 0

A variant of this equation is,

This should be obtained,

　Here we have only considered one substance each in its original and product form, but it is the same if there is more than one. Let's also write down the chemical potential in μ and develop the equation in more detail.

Do the same derivation of the equation again, using the chemical potential (µ).

Return to the earlier starting point.

Integrate this to calculate the change in Gibbs energy when n moles of a substance are changed from state (0) to state (1).

Let state (0) be the standard state and P₀ = 1 atm, assuming that the system contains only the substance concerned.

The change in Gibbs energy (⊿G) when changing from the standard state (0) to state (1) (partial pressure P₁) is expressed as follows

The Gibbs energy per mole is called the chemical potential (μ).

(In chemistry, the Gibbs energy per mole is called the chemical potential; in physics, the Gibbs energy per particle is called the chemical potential.)

By convention, the subscript "0" represents the standard state. The subscript "1" is omitted when representing an arbitrary state (arbitrary partial pressure P) at the destination of the change.

In other words, the change in Gibbs energy when one mole of a substance goes from the standard state to pressure P can be expressed using the chemical potential as follows.

This formula means

(This μ₀ is the standard generated Gibbs energy.)

Consider a chemical reaction (original form before the reaction, production form after the reaction).

Gives energy of each substance in its original form

              Substance Q ： G_Q = q・(μ_Q⁰   +   RT・lnP_Q )

              Substance R ： G_R = r・(μ_R⁰   +   RT・lnP_R )

              Substance S ： G_S = s・(μ_S⁰   +   RT・lnP_S )

              ...

Gives energy for each substance in its production form

              Substance X ： G_X = x・(μ_X⁰   +   RT・lnP_x )

              Substance Y ： G_Y = y・(μ_Y⁰   +   RT・lnP_Y )

              Substance Z ： G_Z = z・(μ_Z⁰   +   RT・lnP_Z )

              ...

　The "standard state chemical potential" of each substance, summarised on the left-hand side, is equal to the "standard production Gibbs energy" of each substance. We write this as G_f⁰, which is an acronym for formation. (Note that the abbreviation varies from book to book.)

The above equation for the equilibrium condition can be summarised as,

【Supplement】

　What's that?　It is the 'partial pressure' of each substance that goes into the equilibrium constant K, is it not?　Why can I also use it as a proxy for molar concentration in solution chemistry?　You may wonder.

　In solution chemistry, we consider the standard Gibbs energy of formation (G_f⁰) of an ion dissolved in water; since we cannot make a solution containing only one type of ion, the G_f⁰ of an ion is always relative. The G_f⁰G_f⁰ of the hydrogen ion H+ in aqueous solution is defined as 0, while the G_f⁰ of the other ions is expressed as relative values. The reason there is no problem with relative values is that K is expressed as a ratio of concentration product/concentration product.

　Therefore, '⊿G_f⁰ ＝ －R・T・lnK' can be applied directly to solution chemistry.

　To be precise, the equilibrium constant K is not expressed in terms of the concentration or partial pressure of each substance, but K is expressed as a product of the 'active mass' of each substance. The active mass is the proportion of molecules that are active in a chemical reaction. To describe a "ratio", something must be set to 1. A more detailed explanation of active quantities will be given in the section "Ionic Strength and Debye Radius" next to Chemical Equilibrium. For now, remember that

・　The molar concentration a (mol/L) of the ideal solution is the active volume a.

・　The partial pressure b (atm) of an ideal gas is the active volume b.

・　The electron activity is set to 1.

・　When water molecules are involved in the reaction in aqueous solution, the water activity is set to 1.

・　When solids are present in a gas- or liquid-phase reaction, the activity of the solids is 1.

・　When a liquid is produced in a gas-phase reaction, the active volume of the liquid is 1.

　In non-ideal (real) solutions and non-ideal (real) gases, not all substances of a certain concentration are involved in chemical reactions. In concentrated solutions and high-pressure gases, the proportion of substances that are not involved in chemical reactions and is larger.The coefficient to compensate for this ratio is called the activity coefficient. For the moment, it is only necessary to understand it in ideal conditions.