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![Standard free energy change:
The standard free energy change is defined as the free energy change
for a process at 298 K and 1 atmospheric pressure in which the
reactants in their standard states are converted to the products in their
standard states. Thus,
Δ G° = Σ Δ G °(products) - Σ Δ G °(reactants)
=[sum of the standard free energy in the formation of products ] –
[ sum of standard energy in the formation of reactants]](https://image.slidesharecdn.com/thermodynamicsii-180130054401/75/Thermodynamics-ii-37-2048.jpg)
Thermodynamics ii
This document discusses spontaneous processes and the driving forces behind them in thermodynamics. It explains that spontaneous processes are driven by a decrease in enthalpy or an increase in entropy. While enthalpy change alone cannot predict spontaneity, the introduction of entropy and Gibbs free energy allows better determination of spontaneous processes. The document also discusses how temperature, entropy change, and the relationship between Gibbs free energy and equilibrium constant can be used to analyze spontaneity.
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Thermodynamics ii
- 1.
- 2. Spontaneous Process: In Thermodynamicssense, a process that occurs by itself without the help of external agent is a spontaneous process. All naturally occurring process is spontaneous process. Examples: 1. Rolling down of a stone downhill 2. Melting of ice 3. Flowing of water from higher to lower level 4. Vaporization of water 5. Burning of fuel 6. Mixing of gases 7. Natural radioactivity, etc.
- 3. Characteristics of SpontaneousProcess: 1. All spontaneous process are unidirectional in nature and proceed till the process is completed or a state of equilibrium is reached 2. Some spontaneous process can be reversed by applying external agency. 3. Some spontaneous process requires initiations.
- 4. What is thedriving force for spontaneous of a process? In the early development of thermodynamics energy change was considered as criteria for spontaneity a process occurs with decrease in energy is spontaneous. Example. flowing of water from higher to lower level, rolling down of stone downhill etc. In case of chemical reaction, the reaction that occurs with decrease in enthalpy i.e. exothermic is spontaneous like burning of Methane, neutralization reaction, etc. On the basis of enthalpy change, ∆H = -ve = exothermic = spontaneous ∆H = +ve = endothermic = non-spontaneous ∆ H = O = equilibrium
- 5. However many endothermicreaction were found to be spontaneous. Example, i. Melting of ice; H2O (s) → H2O (l) ∆H = +6 KJ ii. Vaporization of water H2O (l) → H2O (g) ∆H = +40.8 KJ iii. Dissociation of NH4NO3 in water, NH4NO3 (s) + H2O (l) → NH4 + (aq) + NO3 – (aq) ∆H = +25KJ These examples show that enthalpy change alone is not an enough criteria for spontaneity. These lead to necessity of another thermodynamic term. In all above examples, the molecular disorderness is increased, so the new thermodynamic function should measure disorderness. This function is called entropy.
- 6. Entropy: Entropy of asystem is the measurement of disorderness or randomness of the system. The greater the randomness, the higher is the entropy. For a given substance, the crystalline solid state has lowest entropy i.e. in solid, the molecules are arranged orderly so it has least entropy but in liquid the disorderness is more than solid while in gas, the disorderness is maximum. So the order of entropy of a substance in different physical states, is, Ssolid < Sliquid << Sgas
- 8. Entropy is astate function and depends only upon initial and final states, thus change in entropy ( ∆S) will be Δ S = S2 – S1 Where S2 = Final state entropy S1 = Entropy of Initial state
- 9. In chemical reaction,the change in entropy is, Δ Sreaction = Sproducts – Sreactants = qrev, iso/T…………………..(i) ( simply Δ Sreaction = q/T) i.e. entropy change during the process is defined as the amount of heat (q) absorbed isothermally and reversibly (infinitesimally slowly) divided by the absolute temperature (T)
- 10. Justifications: 1. When asystem absorbs heat, the molecule start moving faster because kinetic energy increases. Hence the disorder increases. More the heat absorbed, greater is the disorder. 2. For the same amount of heat absorbed at low temperature, the disorder is more than at higher temperature. Thus the presence of T in the denominator is justified. Unit Its unit is Jmol-1K-1
- 11. Entropy change duringphase transformation: i. Entropy of fusion: It is change in entropy when 1 mole of solid change into liquid at melting temperature. Mathematically, ∆Sfusion = Sliquid - Ssolid = qrev Tmelt = ∆Hfusion Tmelting where, Sliquid = molar entropy of liquid Ssolid = molar entropy of solid ∆Hfusion = enthaly of fusion Tmelting = melting temperature
- 12. i. Entropy ofvaporisation: It is change in entropy when 1 mole of liquid change into vapours at boiling temperature. Mathematically, ∆Svap = Svapour - Sliquid = ∆Hvap Tboiling where, Sliquid = molar entropy of liquid Svapour = molar entropy of vapour ∆Hvap = enthaly of vaporisation Tboiling = boiling temperature
- 13. i. Entropy ofsublimation: It is change in entropy when 1 mole of solid change into vapour at particular temperature. Mathematically, ∆Ssublimation = Svapour - Ssolid = ∆H 𝑠𝑢𝑏𝑙𝑖𝑚𝑎𝑡𝑖𝑜𝑛 T where, Sliquid = molar entropy of liquid Ssolid = molar entropy of solid T = particular temperature
- 14. Limitation of thefirst law of thermodynamics. 1. No restriction on the direction of the flow of heat: the first law establishes definite relationship between the heat absorbed and the work performed by a system. The first law does not indicate whether heat can flow from a cold end to a hot end or not. 2. The information provided by the first law of thermodynamics are not enough to predict the spontaneity or feasibility of a process. To deal with spontaneity of the chemical process some new terms viz., entropy and Gibbs free energy were introduced in thermodynamics which lead to the formulation of second law of the thermodynamics.
- 15. Second law ofthermodynamics: There are limitation of first law of thermodynamics. To overcome those limitation and to explain the spontaneity of a process, second law of thermodynamics is stated. It state that, “ Whenever a spontaneous process takes place, it is accompanied by an increase in the total entropy of universe.” Thus for a spontaneous process ∆Suniverse = ∆Ssys + ∆Ssurr > 0 According to second law, when an irreversible spontaneous process occurs, the entropy of the system and surrounding increases. i.e. ∆Suniverse>0. As we consider entire universe as a system, the second law can be stated as, “ The entropy of the universe is constantly increasing.”
- 16. Explanation of secondlaw of thermodynamics: Second law of thermodynamics is explained here in the light of entropy change. For this purpose consider spontaneous flow of heat from a system at higher temperature to the surrounding at lower temperature. Let Tsystem and Tsurrounding are the temperature of the system and surrounding respectively where Tsystem > Tsurrounding.
- 17. Let system evolvesQ quantity of heat spontaneously to the surrounding. Here, the system decreases its entropy which is given as ∆Ssystem = - Q Tsystem On the other hand, the surrounding increases its entropy which is given as ∆Ssurrounding = + Q Tsurrounding The net entropy of universe in the spontaneous flow of heat is ∆Suniverse = ∆Ssystem + ∆Ssurrounding = - 𝑄 Tsystem + +Q Tsurrounding
- 18. Since Tsystem >Tsurrounding, the numerical value of Q Tsystem < Q Tsurrounding Hence ∆Suniverse = positive or ∆Suniverse >0 Here the net entropy of the universe is greater than zero. Further the total change of the system and surrounding = -Q + Q = 0 Hence, the net entropy of the universe is increasing whereas the total energy of the universe is constant. This is second law of thermodynamics.
- 19. Class work How canyou say that melting of ice is spontaneous process? How can you say that solidification of ice is spontaneous process?
- 20. Entropy and equilibriumstate: Let us apply the entropy criteria to decide the feasibility of the process of conversion of water to ice at 1 atmospheric pressure. H20(l) → H2O(s) The entropy changes for the three different temperature 272, 273 and 274 K are given in the table below. 1. At 272 K ;freezing is spontaneous because ∆Stotal is positive. 2. At 273 K ;∆Stotal is zero. Thus ice and water are in equilibrium and no net change is observed. 3. At 274 K ;freezing is non-spontaneous because ∆Stotal is –ve, but the reverse change i.e. melting is spontaneous.
- 21. Thus, from above,entropy criterion can be summarised up as follows: ∆Stotal > 0 process is spontaneous. ∆Stotal = 0 process is in equilibrium state. ∆Stotal < 0 process is non-spontaneous
- 22. Free energy ofthe system: One of the defect of entropy, for the prediction of spontaneity is to calculate the entropy of surrounding also. Our concern aspect is the system and not the surrounding. So, a new thermodynamic term has been introduced for the prediction of spontaneity based on the system only and this new term is free energy of the system.
- 23. Free energy: This isanother thermodynamic quantity that helps in predicting the spontaneity of the process. It is usually denoted by ‘G’ and designed mathematically by the equation G = H - TS Where, H = enthalpy of the system S = entropy of the system T = absolute temperature
- 24. Consider a systemundergoing a change from state 1 to state 2 at isothermal process i.e. G1 = H1 - TS1 for initial state……………(i) G2 = H2 - TS2 for final state……………..(ii) Therefore, G2 - G1 = (H2 - H1) - T( S2 - S1) ∆G = ∆H − T∆S … … … … … (iii) Equation(iii) is known as Gibbs-Helmoltz equation or simply Gibbs energy equation.
- 25. Spontaneity in termsof free energy change: It has already been seen that the total entropy change, ΔStotal determines the spontaneity of a process. The total entropy change during a process is given by ΔStotal = ΔSsystem + ΔSsurrounding ……………………………(1) Consider a process being carried out at constant temperature and constant pressure. Suppose heat equal to q is lost by surrounding at temperature T. The entropy change of the surrounding is given by
- 26. ∆Ssurrounding = q T …………………………………………..(2) Now heat(q) gained by system at constant temperature and pressure represents its enthalpy change (∆H).Therefore , ∆Ssurrounding = q T = −∆Hsystem T (since qp= ∆Hsystem) Substituting this value in equation 1, we get ΔStotal =ΔSsystem - ΔHsystem T …………………(3)
- 27. Multiplying equation (3)by T we get, TΔStotal = TΔSsystem - ΔHsystem -TΔStotal = ΔHsystem - TΔSsystem -TΔStotal = ΔGsystem……………………………(4) We know, If ΔStotal = positive, the process is spontaneous If ΔStotal = negative, the process is non-spontaneous If ΔStotal = 0 , the process is in equilibrium
- 28. Putting the resultsin (4), we get If ΔG = negative, the process is spontaneous If ΔG = positive, the process is non-spontaneous If ΔG = 0, the process is in equilibrium
- 29. Free- energy changeand useful work: Gibb’s free energy is related with useful work. The work other than mechanical work (PV-work) is called useful work or net work done. It is denoted by Wuseful or Wnet. From first law of thermodynamics, we have q = ∆E + W There are two types of work done by/on a system. One is mechanical work i.e. expansion or contraction while the other is useful work i.e. W = Wmechanical + Wuseful or, W = P ∆V + Wuseful Or, q = ∆E + P ∆V + Wuseful
- 30. Then, q = ∆H+ Wuseful …………..(i) since (∆H = ∆E + P ∆V) Now, For a change taking place under isothermal and reversible conditions, the entropy change ∆S is given by ∆S= q T Or, q = T ∆S…………………..(ii)
- 31. Substituting this valueof q in equation(i) we have, T ∆S = ∆H + Wuseful - ∆H - T ∆S = Wuseful - ∆G = Wuseful Gibbs free energy is that thermodynamic quantity of a system that decrease in whose value during the process is equal to the maximum possible useful work done that can be obtained from the system.
- 32. Effect of temperatureon spontaneous process: The spontaneity of a process is explained on the basis of enthalpy, entropy and free energy change. These are related to each other by means of Gibb’s Helmholtz equation i.e. ∆G = ∆H − T∆S The value of ∆G at any temperature is calculated from the values of ∆H and ∆S, which in turn is used to explain the spontaneity of a process. Depending upon the sign of ∆H and ∆S, the effect of temperature on spontaneity of a process is explained as follows:
- 33. 1. If ∆𝐇𝐢𝐬 − 𝐯𝐞 𝐚𝐧𝐝 ∆S is +ve. The free energy change is always –ve . Hence, the process is spontaneous at any temperature. 2. If ∆H and ∆S are positive, then a. At low temperature, T ∆S is less than ∆H. So, ∆G will have positive value. Therefore the reaction will be non spontaneous. b. At high temperature, T ∆S is greater than ∆H. So, ∆G is negative. Therefore the reaction is spontaneous. 3. When both ∆H and ∆S are negative, then a. At low temperature, T ∆S is less than ∆H. Therefore ∆H term will predominate and the value of ∆G is negative. Therefore the change will be spontaneous. b. At higher temperature, T ∆S is greater than ∆H. Under this condition, ∆G are positive. Therefore, the reaction is non spomtaneous.
- 34. 4. If ∆His positive and ∆𝐒 𝐢𝐬 𝐧𝐞𝐠𝐚𝐭𝐢𝐯𝐞. Then ∆G is positive and the reaction is non-spontaneous. 5. If ∆H= T ∆S, then ∆G=0, the reaction is at equilibrium state.
- 36. Comment on thefollowing statements: a. An exothermic reaction is always thermodynamically spontaneous. Ans: It is due to following reasons: 1. They are accompanied by decrease of energy. 2. The heat evolved is absorbed by surrounding so that entropy of surrounding increases and hence ∆Stotal is positive. However according to Gibbs energy equation ∆G = ∆H - T∆S, if ∆Ssystem is negative and T is high. ∆G may be positive and hence the reaction may be non spontaneous.
- 37. Standard free energychange: The standard free energy change is defined as the free energy change for a process at 298 K and 1 atmospheric pressure in which the reactants in their standard states are converted to the products in their standard states. Thus, Δ G° = Σ Δ G °(products) - Σ Δ G °(reactants) =[sum of the standard free energy in the formation of products ] – [ sum of standard energy in the formation of reactants]
- 38. Relation between standardfree energy change and equilibrium constant: The standard free energy change, Δ Go is related to the equilibrium constant K by the relation, Δ Go = - RT ln K or ΔGo = –2.303 RT log K Where, R= molar gas constant T= kelvin temperature K = equilibrium constant.