#### Criteria for Spontaneity, Chemistry tutorial

Criteria for Spontaneity:

We have observed that entropy can be employed as a criterion for finding out the spontaneity of a process. We can as well express the criteria for spontaneity in terms of other thermodynamic properties, namely U, H, A and G.

Internal Energy change as a Criterion for Spontaneity:

From the first law of thermodynamics, we are familiar that

dq = dU - dw (true for irreversible or reversible process) and as dqrev = TdS

TdS = dU - dw (for reversible process)

For the irreversible process, dqirrev < dqrev

Therefore, dqirrev < TdS.

Therefore, for an irreversible process, dU - dw < TdS or TdS > dU - dw

Thus for any process:

TdS ≥ dU - dw

Or TdS ≥ dU + (- dw)

If we consider just the pressure-volume work (dw = pdV), then

TdS ≥ dU + pdV

Subtracting TdS from both the sides, 0 > dU + pdV - TdS

Or dU + pdV - TdS < 0

That is, for a process to be feasible, the condition given in the above equation should hold good. At constant volume (dV = 0) and entropy (dS = 0), the equation above therefore reduces to

(∂U)V.S ≤ 0

According to the equation above if the volume and entropy remain constant, the internal energy of a system remains constant in the reversible process [(∂U)V.S = 0] while in an irreversible process the internal energy reduces [(∂U)V.S < 0]

Enthalpy Change as a Criterion for Spontaneity:

We are familiar that H = U + pV

On differentiating the above expression, we obtain

dH = dU + pdV + Vdp

As from the first law, dU = dq + dw

dH = dq + dw + pdV + Vdp

Taking only the pressure-volume work done on the system, we can represent as per equation dw = - pdV; we then have

dH = dq - pdV + pdV + Vdp

Or dH = dq + Vdp

Or dH - Vdp = dq (that is, true for both reversible and irreversible processes).  For a reversible process dqrev = TdS and, for an irreversible process dqirrev < TdS. Therefore for any process, we have:

dH - Vdp ≤ TdS

At constant pressure (dp = 0) and constant entropy (dS = 0)

The above equation reduces to:

(∂H)p.S ≤ 0

According to the above equation, if the pressure and entropy remain constant, the entropy of a system remains constant in the reversible process [(∂H)p.S < 0], while in an irreversible process the enthalpy reduces [(∂H)p.S ≤ 0]

Helmholtz Free Energy change as a Criterion for Spontaneity:

From A = U - TS we have

dA = dU - TdS - SdT

Since dU = dq + dw

dA = dq + dw - TdS - SdT

dA - dw + TdS + SdT = dq (that is, true for both the reversible and irreversible processes)

As, dqrev = TdS and dqirrev < TdS

dA - dw + TdS + SdT ≤ TdS

Or dA - dw + SdT ≤ 0 (cancelling TdS from both the sides)

Taking only pressure-volume work done on the system,

(dw = - pdV), we have

Or dA + pdV + SdT ≤ 0

At constant volume (dV = 0) and constant temperature (dT = 0), the above form reduces to:

(∂A)T,v ≤ 0

According to the above equation, if the temperature and volume remain constant, then the Helmholtz free energy of a system remains constant in the reversible process [(∂A)T,v = 0], while in an irreversible process the Helmholtz free energy reduces [(∂A)T,v < 0]

Gibbs free energy change as a Criterion for Spontaneity:

G = H - TS = U + pV - TS

Or dG = dU + pdV + Vdp - TdS - SdT

As dU = dq + dw

dG = dq + dw + pdV + Vdp - TdS - SdT

Or dG - dw - pdV - Vdp + TdS + SdT = dq (that is, true for the reversible and irreversible processes)

As, dqrev = TdS and dqirrev < TdS

dG - dw - pdV - Vdp + TdS + SdT ≤ TdS

Or dG - dw - pdV - Vdp + SdT ≤ 0 (cancelling TdS terms both sides)

If just pressure-volume work is done (dw = - pdV), we have

dG + pdV - pdV - Vdp + SdT ≤ 0

Or dG - Vdp + SdT ≤ 0

At constant temperature (dT = 0) and constant pressure (dp = 0),

The above equation reduces to:

(∂G)T,P ≤ 0

According to the equation above, if the pressure and temperature remain constant, the Gibbs free energy of a system remains constant in the reversible process [(∂G)p, T = 0], while in an irreversible process, Gibbs free energy reduces [(∂G)p, T < 0].

Therefore the criterion for spontaneity in terms of Gibbs free energy is that the process would be feasible if ΔG is negative.

By taking the equation ΔG = ΔH - TΔS and (∂G)T,P ≤ 0, we can draw some valuable conclusions.

1) If ΔH is negative and ΔS is positive, then ΔG will be negative at all temperatures; therefore, the process would be spontaneous at all temperatures.

2) If ΔH is positive and ΔS is negative, then ΔG will be positive at all temperatures; therefore, the process would not be feasible at any temperature.

3) If ΔH and ΔS are both positive and negative, then ΔG will be positive or negative based on the temperature.

There are mainly four possibilities for ΔG based on the signs of ΔH and ΔS. These are outlined in the table illustrated below:

Table: Criterion for Spontaneous Change

 Case ΔH ΔS ΔG Result 1 - + - Spontaneous at all temperatures 2 - - - + Spontaneous at low temperatures Non-spontaneous at high temperatures 3 + + + - Non-spontaneous at low temperatures Spontaneous at high temperatures 4 + - + Non-spontaneous at all temperatures

The Gibbs free energy of formation:

It is evident from the definition of Gibbs free energy that the absolute value of 'G' is not determinable as the absolute values of 'U' and 'H' are unknown. Though, in most of the cases, this is not a handicap as we need only changes in the Gibbs free energy for a particular process. In this part, we are interested in exhibiting the process of computation of standard Gibbs free energy of a reaction. Prior to that, let us state the standard states of the substances.

The standard states of solids and liquids correspond to their most stable form at 1 bar pressure and the particular temperature. For a gas, the standard state is pure gas at unit fugacity. For ideal gas, fugacity is unity if pressure is 1 bar at a particular temperature. In all such cases, the most stable form consists of the lowest free energy. The standard state convention might be summarized as illustrated below:

a) For a solid: The pure substance at 1 bar external pressure and at a specified temperature.

b) For a liquid: The pure substance at 1 bar external pressure and at a specified temperature.

c) For a gas: An ideal gas at 1 bar partial pressure and at a specified temperature.

d) For a solute: The ideal solution at one molar concentration and at a specified temperature.

The standard Gibbs free energy of formation, ΔfGo, of a substance is stated as the change in the Gibbs free energy which accompanies the formation of one mole of the substance in its standard state from the elements in their standard states. Just similar to ΔfHo, the standard Gibbs free energy of formation of all elements by definition-is zero. Therefore ΔfHo(CO2) is the Gibbs free energy for the reaction,

C (graphite) + O2 (g) → CO2 (g) ArGo = - 394.4 kJ mol-1

It might be illustrated that ΔfGo can be stated at any temperature however usually these values are tabulated at 298.15 K.

The third law of Thermodynamics:

The second law of thermodynamics has been employed to compute entropy changes in have so far remained incalculable. We now try to plan a scale of standard entropies in such a way that we can associate the definite entropies with different states of a given system.

The Nernst Heat Theorem:

We begin with the Gibbs-Helmholtz equation:

ΔG = ΔH + T [∂(ΔG)/∂T]p

This is possible to compute AH by knowing AG at two different temperatures. A question that remained unanswered was whether we can compute ΔG from ΔH data. One thing that is clear from the above equation that as we approach the absolute zero of temperature, ΔG and ΔH should be equivalent unless [∂(ΔG)/∂T]p supposes an infinite value. Richards, though, found, throughout his studies on the electromotive force values of cells, that the quantity [∂(ΔG)/∂T]p in reality decreases as temperature is lowered. This observation prompted the Nernst to conclude that [∂(ΔG)/∂T]p reduces in such a manner that it reaches zero value steadily at absolute zero. This means that ΔG and ΔH not only approach one other near absolute zero, however do so asymptotically (that is, in such a way that their curves overlap at this point as shown in figure below: Fig: Nernst Heat Theorem graph

Mathematically this statement is equal to:

LtT→0 = d(ΔG)dT = LtT→0 = d(ΔH)dT = 0

This is termed as Nernst Heat Theorem and it is strictly applicable only to the pure solids and not to liquids and gases that are not capable of existence at absolute zero.

As we are familiar that [∂(ΔG)/∂T]p = ΔCp

And [∂(ΔG)/∂T]p = - ΔS

Here, ΔS symbolizes the change in entropy accompanying a reaction and ΔCP is the difference between the total heat capacities of the products and those of the reactants at constant pressure. We can place LtT→0 = d(ΔG)dT = LtT→0 = d(ΔH)dT = 0 in the form

LtT→0 = ΔS = LtT→0 ΔCp = 0

In other words, at absolute zero, there is no difference between the heat capacities of reactants and products at constant pressure; this means that the Cp values of all substances at 0 K are similar. Later studies have illustrated that at the absolute zero of temperature, the heat capacity of a solid must be zero.

That is, at 0K, Cp = 0

An identical argument for the entropy change accompanying a reaction leads to the conclusion that all the solids have the similar entropy at absolute zero and according to Max Planck; this value of entropy is zero.

The outcomes so far illustrated can be defined as the Third Law of Thermodynamics: 'The entropy of a pure, perfectly crystalline solid approach to zero, as the temperature approaches to absolute zero.

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