**Spontaneous Process:**

- The spontaneous process is defined as a process that takes place on its own without the external influence.
- A spontaneous process does not mean fast process. For example, take the case of a combination of hydrogen and oxygen. These gases may be mixed at room temperature and left for many years without observing any perceptible change. Although the reaction is taking place between them, it is at an extremely slow rate. It is still called spontaneous reaction. So spontaneity means having the potential to proceed without the assistance of an external agency.
- spontaneity does not tell about the rate of the reaction or process. Another aspect of spontaneous reaction or process, as we see is that these cannot reverse their direction on their own. We may summarize it as follows: A spontaneous process is an irreversible process and may only be reversed by some external agency.
- All spontaneous process takes place in the direction to decrease in energy and to attain equilibrium.

#### Examples of Spontaneous Processes:

- Water always flows from higher level to lower level on its own but cannot flow from lower level to higher level on its own.
- Heat always flows from body at higher temperature to body at lower temperature on its own it cannot flow from body at lower temperature to body at higher temperature on its own

**Characteristics of Spontaneous Process:**

- Once the spontaneous process starts, it proceeds without the continuous external help.
- For spontaneity of reaction, the products must be more stable than the reactants.
- For spontaneity of reaction, the product should have less energy than the reactants.
- All spontaneous process takes place in the direction to decrease in energy and to attain equilibrium.
- Spontaneous process increase disorder in the system thereby increasing the entropy of the system.
- the spontaneous process cannot reverse their direction on their own. Thus a spontaneous process is an irreversible process and may only be reversed by some external agency.
- The spontaneous process proceeds till an equilibrium is reached.
- Generally, spontaneous processes are exothermic but there are some processes which are spontaneous but endothermic. e.g. melting of ice.
- The rate of a spontaneous process may be fast or slow.

**Non-spontaneous Process:**

- A non-spontaneous process is defined as a process which does not take place on its own but can take place under the external influence.

**Characteristics of Non-Spontaneous Process:**

- Non-spontaneous processes are not natural.
- They can not takes place on their own.
- An external agency is required to carry out such reactions.
- These processes are generally accompanied by absorption of energy. i.e. generally they are endothermic
- In the non-spontaneous process, there is an increase in enthalpy, decrease in entropy and increase in Gibb’s energy.

**Energy and Spontaneity:**

- The substances having more energy are less stable than the substances having low energy.
- In spontaneous reaction, the substances with more energy are converted into substances of low energy.
- In general, all spontaneous process takes place in the direction to decrease in energy and to attain equilibrium.
- Thus reaction to be spontaneous, it should be exothermic.
**Example:**Acid-base neutralization is spontaneous because it is exothermic. Exothermic nature assists spontaneity but is not sure criteria of spontaneity. Melting of ice and the formation of a solution of NaCl in water are endothermic processes but are spontaneous.

**Concept of Entropy:**

- Entropy is a thermodynamic property which determines spontaneity of the process. Consider melting of ice. In this process, the ordered solid state gets converted into a disordered liquid state. Similarly, during vaporization of water, the ordered liquid state gets converted into the disordered gaseous state. Thus in both the cases discussed above, there is an increase in molecular disorder. This randomness of molecules is measured by a thermodynamic property called entropy.

- When there is an increase in disorderedness then change in ΔS is positive. ΔS > 0. When there is a decrease in disorderedness then change in ΔS is negative. ΔS < 0.
- The absolute value of entropy cannot be calculated. The entropy change of a system in a process is equal to the amount of heat transferred to it in a reversible manner divided by the temperature at which the transfer takes place. When heat is supplied to a system the disorder increases. Thus the change in entropy ΔS is directly proportional to heat supplied. It is also observed that the change in entropy ΔS is inversely proportional to the temperature at which the heat addition takes place.

- Entropy is a state function. The unit of ΔS is J K
^{-1}mol^{-1}.

**Entropy and Spontaneity:**

- In most of the cases, the entropy of a system increases in a spontaneous process. But there are some spontaneous processes in which it decreases. This discrepancy is explained in the second law of thermodynamics which states that “the total entropy of the system and its surroundings (universe) increase in spontaneous process
- Thus,S
_{Universe}= S_{Total}= ΔS_{Systm}+ ΔS_{Surroundings} - For spontaneous process ΔS
_{Total}> 0. - The spontaneity of reaction can be determined by following relations.
- If ΔS
_{Total}> 0, the process is spontaneous - If ΔS
_{Total}< 0, the process is non-spontaneous - If ΔS
_{Total}= 0, the process is in equilibriums

- If ΔS

**Entropies of Phase Transformation:**

**Entropy of Fusion:**

- It is defined as the entropy change taking place when one mole of a substance changes from a solid state into a liquid state at its melting point.

Solid ⇌ Liquid

Since ΔS_{Fusion }is positive. Hence entropy of liquid state is greater than that of solid state.

**Entropy of Vapourization:**

- It is defined as the entropy change taking place when one mole of a substance changes from a liquid state into a gaseous state at its boiling point.

Liquid ⇌ Gas

Since ΔS_{vap} is positive. Hence entropy of gaseous state is greater than that of the liquid state.

**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

We have 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

We have 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

We have 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

We have 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 non spontaneous.

#### 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 non spontaneous.

#### Example – 7:

- Determine ΔS
_{Total }for the reaction and discuss its spontaneity at 298 K.

Fe_{2}O_{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 log_{10}K

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

**Relation Between Δ****G 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 (S_{0} = 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} – S_{0}

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} – S_{0}

Where, S_{T} = Absolute entropy of substance at temperature T

S_{0} = Absolute entropy of substance at temperature 0 K = 0

ΔS = S_{T} – S_{0} = 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.