Joule first time demonstrated that work can be converted into heat and vice versa. After the detailed research on Joule’s demonstration, it is concluded that “all work can be converted into heat, but all heat can not be converted into work.”

According to the law of conservation of energy, energy can neither be created nor be destroyed but can be transferred from one form to another. However, during this interconversion, a lot of energy is converted into useless form. i.e. it is changed into that form of energy, which is not used to form the useful work. Thus, we can say that, during transformation of energy, it is converted into two forms i.e. i) converted into useful energy and ii) into useless energy. The useful energy is called enthalpy and useless energy is called entropy. Or simply, enthalpy is that form of energy which is converted into work and entropy is that form of energy which is not converted into work.

DEFINITION: The measure of a system’s thermal energy per unit temperature that is unavailable for doing useful work is called entropy. Or measures of the disorderness in a system or measures of the energy unavailable to do work is called entropy.

EXPLANATION: Entropy is first time introduced by Rudolf Clausius in the mid-19th century, to explain the irreversibility of natural processes and the direction of spontaneity that is the second law of thermodynamics. The entropy of highly organised system is very low, as the system change into disordered form its entropy increases. If we want to create an order in one place it will result to produce disorder in another place. Thus, entropy will increase in every case. Or in other words, this world is structured in that way that, every natural process proceeds with increase in entropy. Even, in every day, the entropy increases, and this increment related to the revolution of the universe. This increase in entropy can understand by a simple observation that, it is very easy to break everything (increase entropy) but it is impossible to generate the original thing back (decrease entropy). Thus, this universe is structured by nature with increasing entropy. If entropy does not exist, world would not evaluate and as a result, life and time is not be possible thus making it a fundamental reason why time seems to move only in one direction i.e. forward. Similarly, all processes in this universe can occur only, if the entropy increases, otherwise, no life-supporting processes take place in this universe.

            Thus, it is concluded that the entropy of the universe is continuously increases, and there was a day when entropy of the universe will be the maximum. That day would be the last day of the universe.

MATHEMATICAL EXPRESSION: The symbol used for entropy is S, and change in entropy is written as ΔS. Mathematically, it is written as ΔS = ΔQ/T, where Q is the heat and T is the temperature. when we add heat to any system, the entropy will increase at a given temperature and when heat is released from a system, entropy of system decreases (surrounding increases).

By changing the physical state of a material from solids to liquids or vapours entropy will increase because molecules gain freedom: Thus, S_solid <S_liquid < S_gas.

The formulas used for phase changes are:

• The entropy changes for fusion or melting = ΔH_fus/ΔT.
• Similarly, the entropy changes for vaporization = ΔH_vap/ΔT

The entropy change of a reaction is represented as, ΔS = ∑S_products−∑S_{reactants .} For ideal gases or ideal solutions, mixing different species increases entropy while mixing identical species does not increase entropy.   

Second Law: In any spontaneous process, the total entropy of the universe (system + surroundings) increases i.e. ΔS_universe = ΔS_system + ΔS_surroundings ≥ 0

Third Law: At absolute zero (0 K), the entropy of a perfect crystalline substance is zero, defining a baseline for absolute entropy.

 ENTROPY CHANGE AT VARIOUS THERMODYNAMICAL PROCESSES

For any reversible process: ΔS = Qrev/T

For different processes, dq can be expressed in terms of specific heat capacities (Cp or Cv) and temperature changes.

General Change in Entropy (For an ideal gas)

For an ideal gas, the general formula for entropy change of an ideal gas can be obtained by first law of thermodynamics with the ideal gas and integrating the resulting expression for dS.

Derivation: The general formula for entropy change is dS = dQ/T. for an ideal gas, the value of Q is obtained from the first law of thermodynamics, i.e. dE = dq-pdV or dq = dE+pdV.

We also know that dE = nCvdT and from the ideal gas law pV =nRT. Thus, on putting, p = nRT/V and pdV = nRTdV/V into dq.

Thus, dq = dE+pdV → dq = nCvdT+nRTdV/V. Now putting the value of dq into the formula of entropy, i.e. dS = dQ/T  →   dS =  (nCvdT+nRTdV/V)/T. On simplifying the value of dS. dS=nCvdT/T+nRdV/V. Integrating the value of dS. So, ΔS = ∫(nCvdT/T+∫nRdV/V) → ΔS=nC_VlnT₂/T₁+nRlnV₂/V₁. Alternatively, using pressure instead of volume ΔS=nC_PlnT₂/T₁-nRlnP₂/P₁.

Isobaric Process (P₁=P₂) in anisobaric process, as pressure remains constant, so: dq = nCpdT, substituting this value intothegeneral formula, ΔS= ∫(nC_PdT/T) = nC_P ∫(dT/T) = nC_PlnT₂/T₁. Thus, ΔS=nC_PlnT₂/T₁. 

Isothermal Process (T₁=T₂) In anisothermal process, temperature remains constant, so,

ΔS=nRlnV₂/V₁ or ΔS = -nRlnP₂/P₁   

Isochoric Process (V₁=V₂) As volume remains constant, so: ΔS=nCvlnT₂/T₁         

Adiabatic Process (Reversible) in adiabatic processes, dq=0, so ΔS = 0. A reversible adiabatic process is isentropic (an isentropic process is a process in which the entropy of a system remains constant). So, ΔS=0​

Entropy Change in Mixing (Ideal Gases) When mixing gases, A and B at the same or constant temperature and pressure, the entropy is increased which is ΔS_mix = −R(nAlnxA – nBlnxB).

Derivation: let’sconsider the mixing of two ideal gases, A and B, at constant temperature and pressure. For gas A, ΔSA = nARlnV_f/V_A,

For gas B, ΔSB = nBRlnV_f/V_B,

 The total entropy = ΔS_total = ΔSA + ΔSB → ΔS_total = nARln (Vf/VA) + nBRln(Vf/VB). Here, Vf = VA+VB.

On simplify the expression, ΔS_total = nARln (VA+VB/VA) + nBRln(VA+VB/VB). On expressing in terms of mole fractions. xA = nA/(nA+nB) and xB = nB/(nA+nB).  ΔS_total = -nARln(xA)-nBRln(xB). For a mixture of multiple components: ΔS_total = -RƩniln(xi).

  IMPORTANT NUMERICALS RELATED TO ENTROPY

Q: 2.00 mol of an ideal gas expands isothermally and reversibly from V₁​ to V₂​=2V₁​. Find ΔS of the gas.

Soln: Formula: ΔS=nRlnV₂/V₁   

Calculation: ΔS=2.00×8.3144626×ln(2) → ln2=0.693147 → 2.00×8.3144626×0.693147 = 11.526JK⁻¹.

Q:  1.00 mol of ice melts at 0.00 °C (273.15 K). Molar enthalpy of fusion of water ΔHfus = 6.01 kJ mol⁻¹ .  Find ΔS?.

Soln: Formula: ΔS=ΔHfus/T.

Calculation: ΔS=6.010 J mol⁻¹/273.15​=22.003 JK⁻¹mol⁻¹.

Q: Mix 1.00 mol ideal gas A and 1.00 mol ideal gas B (different gases) at constant T, P. Find entropy change on mixing.

Soln: xA = nA / (nA + nB) = 1/2, xB = nB / (nA + nB) = 1/2

Entropy of Mixing, ΔSmix = – R [nA ln(xA) + nB ln(xB)]

= – 8.314 J/mol·K [1 ln(1/2) + 1 ln(1/2)] = – 8.314 J/mol·K [2 ln(1/2)]

= – 8.314 J/mol·K [-1.3863] = 11.53 J/K

The entropy change on mixing is 11.53 J/K.

Formula (ideal gases): ΔSmix = -ntotR∑xilnxi, Here ntot​=2 mol, xA=xB=0.5

Q: 1.00 mol of an ideal gas (assume constant molar Cp=29.1 J mol⁻¹K⁻¹ is heated from 300 K to 600 K. Find ΔS.   

Formula: ΔS=CplnT₂/T₁

Soln:  ΔS= 29.1×ln (600/300) = 29.1×ln2 = 29.1×0.693147 = 29.1×0.693147   29.1×0.693147=20.171 JK⁻¹. ΔS=20.17 J K⁻¹

Q: 1.00 mol of an ideal gas expands freely into a vacuum from V₁​ to V₂=3V₁ Find ΔS

Soln: For an ideal gas, entropy depends only on the volume ratio for same T: ΔS=nRlnV₂/V₁

Calculation: ΔS=1.00×8.3144626×ln3 → ln3=1.0986123 
So ΔS=8.3144626×1×1.0986123=9.1344 J K⁻¹ 

Q: Mix 2.00 mol A and 1.00 mol B (ideal gases) isothermally. Find Δs_mix 

Soln: nA = 2.00 mol, nB = 1.00 mol

Mole Fractions: xA = nA / (nA + nB) = 2/3, xB = nB / (nA + nB) = 1/3

Entropy of Mixing ΔSmix = – R [nA ln(xA) + nB ln(xB)]

= – 8.314 J/mol·K [2 ln(2/3) + 1 ln(1/3)] = – 8.314 J/mol·K [-0.8109 + -1.0986]

= – 8.314 J/mol·K [-1.9095] = 15.88 J/K

Q: 50.0 g of ice (at 0.00 °C) melts to water at 0.00 °C. ΔHfus=6.01 kJ mol⁻¹ Find ΔS.

Soln: Moles of water: n=50.0 g/18.015 g mol⁻¹=2.7759 mol  ΔS=nΔHfus/T (for the determination of entropy from the phase transition of whole material we use the formula, ΔS=ΔHfus/T. however for the determination of entropy from the specific amount of material than the following formula will be used, ΔS=nΔHfus/T). So, in this particular case, the value of entropy will be =2.7759×6.010 J mol⁻¹/273.15 K. = 0.061J K⁻¹.  ΔS =​.0.061J K⁻¹

Q: Given: reaction with products total entropy =2×130.7 J K⁻¹ mol⁻¹and reactants total entropy =205.0 J K⁻¹ mol⁻¹.Compute ΔSrxn∘?

Soln: ΔSrxn∘=∑Sproducts∘−∑Sreactants∘

Products= 261.42×130.7=261.4. Reactants: 205.0 So, ΔS∘=261.4−205.0=56.4 J K⁻¹ mol⁻¹ 

 ΔSrxn∘=56.4 J K⁻¹ mol⁻¹

Q: Calculate the entropy change (ΔS) for the fusion of 1 mole of ice at 273 K, given that the enthalpy of fusion (ΔH) is 6.01 kJ/mol.

Soln:   ΔS = nΔH / T  = 6.01 kJ/mol / 273 K→ 0.022 kJ/mol·K = 22 J/mol·K

Q: The entropy change (ΔS) for a reaction is 0.05 kJ/mol·K. If the reaction is spontaneous at 298 K, what can be concluded about the enthalpy change (ΔH)?

Soln: ΔG = ΔH – TΔS, Given Values ΔS = 0.05 kJ/mol·K, T = 298 K. For a reaction to be spontaneous, ΔG must be negative. Now, putting the values in equation, ΔG = ΔH – TΔS → ΔG = ΔH – 298×0.05 =  ΔG = ΔH – 14.9. for the reaction to be spontaneous, ΔG must be negative. Therefore, the enthalpy change (ΔH) must be less than 14.9 kJ/mol for the reaction to be spontaneous at 298 K.

Q: Calculate the entropy change (ΔS) for the vaporization of 1 mole of water at 373 K, given that the enthalpy of vaporization (ΔH) is 40.66 kJ/mol.

Soln:  ΔS = nΔH / T  → 1×40.66 kJ/mol / 373 K = 0.109 kJ/mol·K = 109 J/mol·K

Q: The standard entropy change (ΔS°) for a reaction is 0.1 kJ/mol·K. If the standard enthalpy change (ΔH°) is 50 kJ/mol, calculate the standard Gibbs free energy change (ΔG°) at 298 K.

Soln:  ΔG° = ΔH° – TΔS° = 50 kJ/mol – (298 K) (0.1 kJ/mol·K) = 50 kJ/mol – 29.8 kJ/mol

= 20.2 kJ/mol

 GIBBS FREE ENERGY

This concept was introduced in the 1870s by Josiah Willard Gibbs, who called it “available energy”, the energy a system could deliver as a work under constant temperature and pressure. Gibbs free energy in fact gives the answers of the following questions: will the reaction be possible or not? Will the reaction be feasible or not? Or will this reaction be spontaneous or not? Or at what temperature will this reaction start? The mathematical expression used for the Gibbs free energy is G=H-TS, and the change in Gibbs free energy is ΔG=ΔH-TΔS. From this equation, it is obtained that all parameters changed except T, which is constant. ΔH is the heat or energy exchange between the system and its surroundings, i.e., ΔH is positive when heat or energy is absorbed by the system from the surroundings. So, the reaction is endothermic. Similarly, ΔH is negative when heat is released by the system to the surroundings. So, the reaction is exothermic. In the same way, ΔS has a positive value when the entropy of the system is increased and ΔS has a negative value when the entropy of the system is decreased. Thus, the Gibbs free energy is nothing only it is the simple combination result of enthalpy and entropy. It is helpful to determine the spontaneity of reaction, calculating the equilibrium constant, and determine the maximum amount of energy that can be extracted from the system.  The ΔG may obtain any of the following values, from which we decide whether the reaction is spontaneous or not.

  If ΔG is negative, the reaction will be feasible and will be spontaneous. “

If ΔG is positive, the reaction will not be feasible and will not be spontaneous and the reaction needs outside energy to occur.”

If ΔG is zero, the reaction will be at equilibrium.

DEFINITION: It is the fraction of total energy that is isothermally available for converting into useful work. The amount of energy available to a system for doing work is known as Gibbs’ free energy. The unit of Gibbs free energy is joules or kilojoules.

CHANGE IN GIBBS FREE ENERGY: The change in Gibbs free energy at constant temperature can be written as:  ΔG(G₂-G₁) = ΔH(H₂-H₁)-TΔS(S₂-S₁)

The key criteria to determine the spontaneity of a reaction is:

• ΔG < 0: Process is spontaneous (exergonic).
• ΔG > 0: Process is non-spontaneous (endergonic).
• ΔG=0: System is at equilibrium

RELATIONSHIPS:

• Basic definition ΔG=ΔH−T ΔS
•  Standard Conditions ΔG∘=ΔH∘−T ΔS∘ 
•   Equilibrium Constant ΔG∘=−RTlnK
•    Pressure Dependence (Ideal Gas)   ΔG=ΔG∘+nRTln(P/P∘)
•    Concentration/Activity Dependence ΔG=ΔG∘+nRTln([c]/[c]∘) or ΔG=ΔG∘+nRTlna
• Electrochemical Work ΔG=−nFEcell

IMPORTANT NUMERICALS RELATED TO GIBBS FREE ENERGY

Q: Calculate ΔG for a reaction with ΔH = 20 kJ/mol and ΔS = 0.05 kJ/mol·K at 298 K.

Soln:  ΔG = ΔH – TΔS. = 20 kJ/mol – (298 K)(0.05 kJ/mol·K). = 20 kJ/mol – 14.9 kJ/mol

= 5.1 kJ/mol.

Q: Determine if a reaction is spontaneous with ΔG = -15 kJ/mol.

Soln:   Yes, the reaction is spontaneous because ΔG is negative.

Q: Calculate the equilibrium constant (K) for a reaction with ΔG = -5 kJ/mol at 298 K.

Soln:    ΔG = -RT ln(K) =-5,000 J/mol = -(8.314 J/mol·K)(298 K) ln(K)

ln(K) = 2.01→ K = e^^2.01  →K ≈ 7.48

Q: For a reaction, ΔH = 50 kJ/mol and ΔS = 0.1 kJ/mol·K. At what temperature will the reaction be spontaneous?

Soln: ΔG = ΔH – TΔS → T > ΔH / ΔS → T > 50 kJ/mol / (0.1 kJ/mol·K). T > 500 K.

Q: The ΔG for a reaction is -50 kJ/mol at 300 K. What is the equilibrium constant (K) for the reaction?

Soln:  ΔG = -RT ln(K) →ΔG= -50,000 J/mol = -(8.314 J/mol·K)(300 K) ln(K). →ln(K) = 20.06→ K = e^20.06→ K ≈ 4.95 × 10^8.

Q:  The standard Gibbs free energy change (ΔG°) for a reaction is -100 kJ/mol at 298 K. Calculate the equilibrium constant (K) for the reaction.

Soln:   ΔG° = -RT ln(K). → -100,000 J/mol = -(8.314 J/mol·K)(298 K) ln(K)

ln(K) = 40.37 → K = e^40.37 →K ≈ 3.24 × 10^17