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 dq_{rev} = TdS
TdS = dU  dw (for reversible process)
For the irreversible process, dq_{irrev} < dq_{rev }
Therefore, dq_{irrev} < 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 pressurevolume 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 pressurevolume 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 dq_{rev} = TdS and, for an irreversible process dq_{irrev }< 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, dq_{rev} = TdS and dq_{irrev} < TdS
dA  dw + TdS + SdT ≤ TdS
Or dA  dw + SdT ≤ 0 (cancelling TdS from both the sides)
Taking only pressurevolume 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, dq_{rev }= TdS and dq_{irrev} < TdS
dG  dw  pdV  Vdp + TdS + SdT ≤ TdS
Or dG  dw  pdV  Vdp + SdT ≤ 0 (cancelling TdS terms both sides)
If just pressurevolume 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),
(∂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
Nonspontaneous at high temperatures
3
Nonspontaneous at low temperatures
Spontaneous at high temperatures
4
Nonspontaneous 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, Δ_{f}G^{o}, 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 Δ_{f}H^{o}, the standard Gibbs free energy of formation of all elements by definitionis zero. Therefore Δ_{f}H^{o}(CO_{2}) is the Gibbs free energy for the reaction,
C (graphite) + O_{2} (g) → CO_{2} (g) A_{r}G^{o} =  394.4 kJ mol^{1}
It might be illustrated that Δ_{f}G^{o }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 GibbsHelmholtz 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:
Lt_{T→0 }= d(ΔG)dT = Lt_{T→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} = ΔC_{p}
And [∂(ΔG)/∂T]_{p }=  ΔS
Here, ΔS symbolizes the change in entropy accompanying a reaction and ΔC_{P} is the difference between the total heat capacities of the products and those of the reactants at constant pressure. We can place Lt_{T→0 }= d(ΔG)dT = Lt_{T→0 }= d(ΔH)dT = 0 in the form
Lt_{T→0} = ΔS = Lt_{T→0 }ΔC_{p} = 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 C_{p} 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, C_{p }= 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.
Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)
Expand your confidence, grow study skills and improve your grades.
Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, oneonone online tutoring.
Using an advanced developed tutoring system providing little or no wait time, the students are connected ondemand with a tutor at www.tutorsglobe.com. Students work oneonone, in realtime with a tutor, communicating and studying using a virtual whiteboard technology. Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.
Free to know our price and packages for online Chemistry tutoring. Chat with us or submit request at info@tutorsglobe.com
18,76,764
Questions Asked
21,311
Experts
9,67,568
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!