Gibb’s Energy or Free Energy:

Gibb’s energy or free energy is a thermodynamic function which helps us in the development of a criterion of spontaneity or feasibility of a process. It refers to the capacity of the system to do useful work.

Gibb’s energy is defined as the amount of energy available from a system at a given set of conditions that can be put into useful work. It is also called free energy.

Mathematically,   G = U + PV – TS

Where      G = Gibb’s energy

U = Internal energy of the system

P = Pressure of the system

V = Volume of the system

T = Absolute temperature of the system

S = Entropy of the system

The absolute value of Gibb’s energy cannot be calculated. But the change in it can be calculated as

ΔG = ΔU + Δ(PV) – Δ(TS)

For constant pressure and constant temperature process

ΔG = ΔU + PΔV – TΔS

But ΔU + PΔV = ΔH

∴ ΔG = ΔH – TΔS

This relation is known as Gibb’s Helmholtz equation.

Where,      ΔG = Change in Gibb’s energy

ΔH = Change in enthalpy of the system

T = Absolute temperature of the system

ΔS = Change in entropy of the system

The spontaneity of a Reaction w.r.t. Gibb’s Energy and Total Entropy:

Let ΔS_Total be the total enthalpy of the system and ΔG_T,P is Gibb’s energy at constant temperature and pressure.

• If ΔS_Total > 0 (positive) or ΔG_T,P < 0 (negative), the process will be spontaneous and proceeds in a backward direction. It nears to completion. (K > 1)
• If ΔS_Total < 0 (negative) or ΔG_T,P > 0 (positive), the process will be non-spontaneous. and favoured in backward direction. (K < 1)
• If ΔS_Total = 0 or ΔG_T,P = 0, the process will be in equilibrium state. (K = 1)

Relation Between Gibb’s Energy and Chemical Equilibrium:

For spontaneous process Gibb’s energy is negative. For a reversible reaction, there is a decrease in Gibb’s energy during the course of reaction whether we start from reactants or products. Let us consider a hypothetical reaction

A ⇌ B

A graph is drawn by taking Gibb’s energy on the y-axis and the change in the composition of the reacting mixture with time on the x-axis. The graph is as follows.

The minima in the curve correspond to the composition of the reaction mixture at the equilibrium state at which Gibb’s energy is minimum.

From graph following points should be noted.

• In reaching the equilibrium state whether from A or from B, the ΔG is negative.
• At equilibrium state, there is no change in Gibb’s energy. i.e. ΔG at equilibrium = 0.
• If minima of the curve lie very close to the products, then it means that the equilibrium composition strongly favours the products and hence K >> 1. i.e. reaction will proceed to completion.
• If minima of the curve lie very close to the reactants, then it means that the equilibrium composition strongly favours the reactants and hence K < 1. i.e. the reaction hardly proceeds.

Relation Between Standard Gibb’s Energy change and Equilibrium Constant:

Let us consider a hypothetical reaction

A + B  ⇌  C + D

The relation between Gibb’s energy change (ΔG)and standard Gibb’s energy change ΔG^o is given by

ΔG = ΔG^o  + RT ln Q

Where      R = Universal gas constant

T = Absolute temperature of the system

Q = Concentration coefficient

At equilibrium ΔG = 0 and Q = K_c

∴   0 = ΔG^o  + RT ln K_c

∴    ΔG^o  = – RT ln K_c

∴    ΔG^o  = – 2.303 RT log ₁₀ K_c

This is the relation between the standard Gibb’s energy change and equilibrium constant. In exponential form, the expression can be written as

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The Concept of Gibb’s Free Energy (G):

By the second law of thermodynamics, S_Total = ΔS_Systm + ΔS_Surroundings

Thus to decide spontaneity of the process we have to determine the change in entropy of the system and the change in entropy of the surroundings. In chemistry, we are not concerned about the change in entropy of the surroundings. Hence to overcome this problem J. W. Gibbs introduces another thermodynamic property called Gibbs energy without considering the change in entropy of the surroundings.

Gibbs energy is defined as G = H – TS

Where H = Enthalpy of a system, T = Temperature of the system, S = Entropy of a system

H, T and S are state functions hence G is also a state function. Unit of G is J mol-1.

At constant temperature ΔG =  ΔH – T ΔS

Derivation of Gibb’s Helmholtz Equation:

Gibbs energy is defined as G = H – TS  …………..(1)

But H = U = PV   ……………(2)

Where,   H = Enthalpy of the system, T = Temperature of the system, S = Entropy of the system

U = Internal energy of the system, P = Pressure of the system,    V = Volume of the system

From equations (1) and (2) we get

G = U + PV – TS

The change in Gibb’s energy in the process is given by

ΔG = ΔU + Δ(PV) – Δ(TS)

At constant pressure and temperature

ΔG = ΔU + PΔV – TΔS ….(3)

But at constant pressure

ΔU + PΔV = ΔH

Substituting in equation (3) we get

ΔG = ΔH – TΔS

This equation is known as Gibb’s Helmholtz Equation.

Gibb’s Energy and Spontaneity:

The total entropy changed in the process is given by

S_Total = ΔS_Systm + ΔS_Surroundings

S_Total = ΔS + ΔS_Surroundings

ΔS_Total = ΔS + ΔS_Surroundings ………………. (1)

Here, ΔS = ΔS_Systm

If ΔH is enthalpy change which accompanies the process, then the enthalpy change in surroundings is – ΔH.

By definition of entropy,

Substituting in equation (1)

As T is always positive we can say that ΔG and ΔS have opposite signs. Thus in the non-spontaneous process, Gibbs’s energy increases while in spontaneous process Gibbs’s energy decreases. The spontaneity of reaction can be determined by following relations.

• If ΔG < 0, the process is spontaneous
• If ΔG > 0, the process is non-spontaneous
• If ΔG = 0, the process is in equilibriums

Example – 1:

Determine whether the reaction with ΔH = – 40 kJ and ΔS = +135 J K^-1 at 300 K is spontaneous or not. Also, predict the nature exothermic or endothermic.

Solution:

Given: ΔH = – 40 kJ and ΔS = +135 J K^-1 = + 0.135 kJ K^-1, T = 300 K.

To Find: Nature of reaction

Gibb’s energy is given by ΔG = ΔH – TΔS

∴   ΔG = – 40 kJ – 300 K × 0.135 kJ K^-1 = – 40 kJ –  40.5 kJ = -80.5 kJ

ΔG is negative i.e. ΔG < 0, the process is spontaneous

ΔH is negative i.e. ΔH < 0, the process is exothermic.

Example – 2:

Determine whether the reaction with ΔH = – 60 kJ and ΔS = – 160 J K^-1 at 400 K is spontaneous or not. Also, predict the nature exothermic or endothermic.

Solution:

Given: ΔH = – 60 kJ and ΔS = – 160 J K^-1 = – 0.160 kJ K^-1, T = 400 K.

To Find: Nature of reaction

Gibb’s energy is given by ΔG = ΔH – TΔS

∴   ΔG = – 60 kJ – 400 K × (-0.160 kJ K^-1) =  – 60 kJ +  64 kJ = + 4 kJ

ΔG is positive i.e. ΔG > 0, the process is non-spontaneous

ΔH is negative i.e. ΔH < 0, the process is exothermic.

Example – 3:

Determine whether the reaction with ΔH = – 110 kJ and ΔS = + 40 J K^-1 at 400 K is spontaneous or not. Also, predict the nature exothermic or endothermic.

Solution:

Given: ΔH = – 110 kJ and ΔS = +40 J K^-1 = + 0.040 kJ K^-1, T = 400 K.

To Find: Nature of reaction

Gibb’s energy is given by ΔG = ΔH – TΔS

∴   ΔG = – 110 kJ – 400 K × 0.040 kJ K^-1 = – 110 kJ –  16 kJ = – 126 kJ

ΔG is negative i.e. ΔG < 0, the process is spontaneous

ΔH is negative i.e. ΔH < 0, the process is exothermic.

Example – 4:

Determine whether the reaction with ΔH = + 50 kJ and ΔS =  -130 J K^-1 at 250 K is spontaneous or not. Also, predict the nature exothermic or endothermic.

Solution:

Given: ΔH =+ 50 kJ and ΔS = – 130 J K^-1 = – 0.130 kJ K^-1, T = 250 K.

To Find: Nature of reaction

Gibb’s energy is given by ΔG = ΔH – TΔS

∴   ΔG = + 50 kJ – 250 K × ( -0.130 kJ K^-1) =  + 50 kJ +  32.5 kJ = + 82.5 kJ

ΔG is positive i.e. ΔG > 0, the process is non-spontaneous

ΔH is positive i.e. ΔH > 0, the process is endothermic.

Temperature of Equilibrium:

At equilibrium, the process is neither spontaneous nor non-spontaneous and ΔG = 0.

This is the temperature at which change over between spontaneous and non-spontaneous behaviour occurs.

Example – 5:

For a certain reaction, ΔH = – 25 kJ and ΔS =  -40 J K^-1 at what temperature will it change from spontaneous to non-spontaneous.

Given: ΔH = – 25 kJ and ΔS = – 40 J K^-1 = – 0.040 kJ K^-1,

To Find: T =?

We have T = ΔH / ΔS

T = – 25 / – 0.040 = 625 K

Ans: At a temperature of 625 K the reaction change from spontaneous to nonspontaneous.

Example – 6:

For a certain reaction, ΔH = – 224 kJ and ΔS = – 153 J K^-1 at what temperature will it change from spontaneous to non-spontaneous.

Given: ΔH = – 224 kJ and ΔS = – 153 J K^-1 = – 0.153 kJ K^-1,

To Find: T =?

We have T = ΔH / ΔS

T = – 224 / – 0.153 = 1464 K

Ans: At a temperature of 1464 K the reaction change from spontaneous to nonspontaneous.

Example – 7:

Determine ΔS_Total  for the reaction and discuss its spontaneity at 298 K.

Fe₂O_3(s) + 3CO_(g) →  2Fe_(s) + 3 CO_2(g), ΔH° = -24.8 kJ, ΔS° = 15 J K^-1.

Given: ΔH° = – 24.8 kJ, ΔS° = 15 J K^-1, T = 298 K

To Find:  ΔS_Total = ?

We have ΔS_Surr = – ΔH° / T

ΔS_Surr = – ΔH° / T = – (-24.8 kJ)/ 298 K = + 0.0832 kJ K^-1  = + 83.2 J K^-1

Now ΔS_Sys  = ΔS° = 15 J K^-1.

Now, ΔS_Total  = ΔS_Surr + ΔS_Sys =  + 83.2 J K^-1 +  15 J K^-1  = 98.2 15 J K^-1 .

ΔS_Total  is positive i.e.  ΔS_Total  > 0, hence reaction is spontaneous at 298 K

Ans: ΔS_Total  = 98.2 15 J K^-1, The reaction is spontaneous at 298 K

Example – 8:

Determine  ΔS_Total  for the reaction and discuss its spontaneity at 298 K.

_HgS(s) + O_2(g) →  Hg_(l) + SO_2(g), ΔH° = -238.6 kJ, ΔS° = + 36.7 J K^-1.

Given: ΔH° = -238.6 kJ, ΔS° = + 36.7 J K^–,  T = 298 K

To Find:  ΔS_Total  = ?

We have ΔS_Surr = – ΔH° / T

ΔS_Surr = – ΔH° / T = – (- 238.6 kJ)/ 298 K = + 0.8006 kJ K^-1  = + 800.6 J K^-1

Now ΔS_Sys  = ΔS° = + 36.7 J K^-1.

Now, ΔS_Total  = ΔS_Surr + ΔS_Sys =  + 800.6 J K^-1 +  36.7 J K^-1  = + 837.3 J K^-1 .

ΔS_Total  is positive i.e.  ΔS_Total  > 0, hence reaction is spontaneous at 298 K

Ans: ΔS_Total  = + 837.3 J K^-1, The reaction is spontaneous at 298 K.

ΔG and Equilibrium Constant:

The change in Gibb’s energy of reaction, related to standard Gibb’s energy is given by

ΔG = ΔG° + RT ln Q

At equilibrium ΔG = 0 and Q = K, thus equation becomes

ΔG° = – RT ln K

∴   ΔG° = – 2.303 RT log₁₀K

This equation gives the relation between standard Gibb’s energy of the reaction and its equilibrium constant.

Relation Between ΔG and Non Mechanical Work:

According to First law of Thermodynamics

DU = q + W

Here W includes two types of work pressure-volume work (Mechanical) and non-expansive work (non mechanical).

ΔU = q – PΔV + W_Non-Exp

∴ q = ΔU + PΔV – W_Non-Exp

∴ q = ΔH – W_Non-Exp     ………. (1)

For isothermal and reversible condition we have

Thus Gibb’s energy gives us the measure of non-expansion work done by the system.

Third Law of Thermodynamics:

It states that the entropy of a perfectly ordered crystalline substance is zero at absolute zero of temperature. Thus S = 0 at T = 0 for perfectly crystalline substance.

If crystal contains some impurity or some disorder in its structure its entropy is always greater than zero at T = 0. Such entropy of the substance is called residual entropy of the system. The third law helps in determination of absolute entropy of any substance either solid, liquid or in the gaseous state.

The significance of Third Law of Thermodynamics:

• It gives absolute datum from which entropy can be measured.
• Using this law absolute entropy of any substance either solid, liquid or in the gaseous state at a temperature above 0 K can be obtained.
• Standard entropy change ΔS° for a reaction can be calculated and hence the spontaneity of reaction can be determined.
• The standard enthalpy S of a pure substance can be measured at 25 °C and 1 atm pressure.

Note:

For perfectly crystalline substance absolute entropy is zero (S₀ = 0) at absolute zero (0 K)

As the temperature increases say to T K its absolute entropy changes to S_T.

Change in entropy is given by ΔS =  S_T –  S₀

i.e. ΔS =  S_T –  0 = S_T.

This shows that it impossible for any substance to have an absolute entropy zero at temperature greater than 0 K

Entropy Change Due to Increase in Temperature:

The increase in entropy is given by

ΔS = S_T – S₀

Where, S_T = Absolute entropy of substance at temperature T

S₀ = Absolute entropy of substance at temperature 0 K = 0

ΔS = S_T – S₀ = S_T –  0=  S_T

S_T can be determined by the relation

Where C_P = Molar heat capacity at constant pressure.

Standard Molar Entropy:

We know that entropy is the measure of disorderedness. Disorder of any substance depends on the mass of the substance, its molecular and its condition of temperature and pressure.

The absolute entropy(S) of 1 mole of pure substance at 1 atm pressure and 25 °C is called the standard molar entropy of the substance.By knowing the values of standard molar entropy of all reactants and products in chemical reaction we can calculate ΔS° of the reaction using the relation

ΔS° = ∑ ΔS°_Products  –    ∑  ΔS°_Reactants

The standard molar entropy is useful in comparison of entropies of different substances under the same conditions of temperature and pressure.

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