F y b. sc. ii. chemical energetics

Need of thermodynamics and the Laws of Thermodynamics.
Important principles and definitions of thermochemistry.
Concept of standard state and standard enthalpies of formations,
Integral and differential enthalpies of solution and dilution.
Calculation of bond energy, bond dissociation energy and resonance energy from thermochemical data.
Variation of enthalpy of a reaction with temperature – Kirchhoff’s equation.
Statement of Third Law of thermodynamics and calculation of absolute entropies of substances.
JFI -just for information

1. F. Y. B. Sc. SEMESTER- II CORE COURSE: DSC-2B (Physical Chemistry & Organic Chemistry) CHEMICAL ENERGETICS Dr. Mithil S. Fal Desai Assistant Professor in Chemistry SHREE MALLIKARJUN & Shri Chetan Manju Desai College Canacona Goa 1

2. Chemical Energetics Need of thermodynamics and the Laws of Thermodynamics . Important principles and definitions of thermochemistry. C oncept of standard state and standard enthalpies of formations, Integral and differential enthalpies of solution and dilution. Calculation of bond energy, bond dissociation energy and resonance energy from thermochemical data. Variation of enthalpy of a reaction with temperature – Kirchhoff’s equation. Statement of Third Law of thermodynamics and calculation of absolute entropies of substances. 2

3. What is thermodynamics? The study that deals with the relations between heat and other forms of energy as it undergoes physical or chemical transformation . 3

4. Need of T hermodynamics This study is of everyday importance as it improves our understanding about chemical and physical process in which there is change in the energy in the form of heat into other forms. Scope : 1 ) Most physical chemistry laws are derived from thermodynamics. eg . Van’t Hoff law, phase rule and distribution law. 2) Can predict whether a chemical change is possible in given set of conditions. (T, P and C, mainly) 3) Can predict extend of chemical change. (chemical equilibrium) Limitations : 1 ) Applicable to macroscopic system. (energy of bulk is considered) 2) It concern with only initial and final state. (no rate at which energy change is considered) 4

5. Laws of Thermodynamics. First Law of Thermodynamics Second Law of Thermodynamics Third Law of Thermodynamics 5

6. Important terms in thermodynamics. System + Surrounding =Universe Open/close/isolated system Internal energy (E) Heat (q) Heat capacity (c) Work (w) Pressure (p) kJ/ mol Enthalpy(H=E+PV) Entropy(S) Exothermic/Endothermic Chemical potential ( μ ) Helmholtz free energy(A) Gibbs free energy (G) Adiabatic process( dq =0) Isothermal process ( dT =0) Isobaric process( dP =)) Isochoric process ( dV =0) Standard state Reference state STP 6

7. Heat capacity (c) Capacity to absorb heat! In a system if calories of heat (q) absorbed by a mass (m) and temperature changes from T1 to T2 the heat capacity is given by Heat capacity is defined as heat absorbed by unit mass raising the temperature by 1 ◦C or 1 K.   7

8. Joule-Thomson effect (JFI) The phenomenon of lowering of temperature when a gas is made to expand adiabatically from a region of high pressure to region of low pressure. 8 One mole N2 at 10 atm in V1 at T1 Half mole N2 at 1 at in V2 at T2 T2 < T1 accordingly

9. First Law of Thermodynamics Total energy of an isolated system remains constant .(may change from one form to other) Or Total energy in the universe is constant . Consider a system is changed from A to B consider the energy change from E a to E b ΔE = E b - E a This energy change can occur by evolution or absorption of heat (q) and with and without work (w) being done on the system. ΔE = q – w (as, w = P X ΔV) ΔE = q - PΔV 9

10. Some easy considerations and calculation C yclic isothermal process (ΔE=0) and w=q Isochoric process (ΔV=0) and ΔE=q v Adiabatic process (q=0) and ΔE = -w Simple problem : A gas cylinder fitted with a piston expands against a constant pressure of 1 atm from volume of 5 L to 10 L. This system absorbs 400 J thermal energy from its surrounding determine change in internal energy of system. w= -P(V2-V1)= 1 atm (10L – 5L) = 5 L atm (as, 1 L atm = 101.2 J) w= -506 J ΔE= q - w ΔE= 400 J - (-506 J) ΔE= 906 J 10

11. Second Law of Thermodynamics A spontaneous change is accompanied by an increase in the total the total entropy of the system ad its surrounding ΔS system + ΔS surrounding > 0 Some other well known e xplanations Lord Kelvin ‘ It is impossible to take heat from a hotter reservoir and convert it completely into work by a cyclic process without transferring a part of heat to cooler reservoir.’ Clausius ‘It is impossible for a cyclic process to transfer heat from a body at a lower temperature to one at higher temperature without at the same time converting some work to heat.’   11

12. Calculation for understanding An engine operating between 423 K and 298 K takes 500 J heat from a high temperature reservoir. Assuming that there is negligible friction loss calculate the work that can be done by this engine, 147.75 J   12

13. Third Law of Thermodynamics At absolute zero , the entropy of a pure crystal is zero (at T= 0 K, S=0) Clausius introduced numerical definition of entropy ‘entropy of a system not undergoing any changes is a constant quantity .’ Thus, for a change in entropy ( Δ S) is equal to heat (q) energy absorbed or evolved divided by the temperature (T). or   13

14. Calculation for understanding Calculate the entropy change when one mole of ethanol is evaporated at 351 K. The molar heat of vaporisation of ethanol is 39840 J mol -1 . = 113.5 J K -1 mol -1   14

15. Concept of Standard State and reference state The standard state of a substance at specified temperature and 1 bar pressure is its pure form. The reference state is most stable state of a substance at specified temperature and 1 bar pressure. 15

16. S tandard Enthalpy of formation Standard Enthalpy of formation is change in enthalpy that takes place when one mole of compound is formed from all substances being there in standard states(298 K and 1 atm ). H 2(g) + ½ O 2(g) H 2 O (g) ( Δ H◦ f = -57.84) Calculate Δ H ◦ for the reaction CO 2(g ) + H 2(g)  CO (g ) + H 2 O (g) Given that Δ H◦ f for CO 2(g ) , CO (g ) and H 2 O (g) are -393.5, -111.31 and 241.80 kJ K -1 mol -1 , respectively. Δ H ◦ = Δ H◦ f (product) - Δ H◦ f (reactants) Δ H◦ =[( Δ H◦ f CO (g ) ) + ( Δ H◦ f H 2 O (g ) )] - [( Δ H◦ f ( CO 2(g ) ) + ( Δ H◦ f ( H 2(g ) )] Δ H◦ =[(-111.3)+ (-241.8)] - [(-393.5) + (0)] Δ H◦ = 40.4 kJ K -1 mol -1 16

17. Integral enthalpies of solution and dilution Solution A + ( A + B ) = AB ( Results in concentrated solution ) Dilution B + ( A + B ) = AB (Results in concentrated solution dilute solution) A - Solute B - S olvent ( A + B )- Solution 17 A A + B B A + B

18. Integral enthalpy of solution )   H eat absorbed or released when a particular amount solute is dissolved in a definite amount of solvent OR Enthalpy change accompanying with pure solvent of particular volume with addition of definite amount of solute . 18 HCl (g) + 10 H 2 O (l )  [ HCl+10 H 2 O ] ( aq ) ; = – 69.0 kJ mol -1 HCl (g) + 25 H 2 O (l)  [ HCl+25H 2 O ] ( aq ) ; = – 72.0 kJ mol -1 HCl (g) + 200 H 2 O (l)  [HCl+25H 2 O] ( aq ) ; = – 74.0 kJ mol -1  

19. Integral enthalpy of dilution If a infinite amount of solvent is added to a solution with a known concentration of solute, the corresponding change of enthalpy is called as integral heat of dilution on infinite dilution. 19 Mathematically, Enthalpy change associated with preparation of solution Eg . Calculate the integral heat of dilution at 298 K for the addition of 195 moles of water to 1 mole of HCl in 5 moles of water. The enthalpy of formation of 1 mole of HCl in 5 moles of water is – 37.37 kcal and enthalpy of formation in 200 moles of water is -39.79 kcal. HCl (g) + 5H 2 O (l)  [HCl+5H 2 O] ( aq ) ; = – 37.37 kcal [HCl+5H 2 O ] ( aq ) + 195H 2 O (l)  [ HCl+195H 2 O ] ( aq ) ; = – 39.79 kcal  

20. To summarize Integral enthalpies of solution and dilution 20 Consider one mole acid is diluted with different moles of water and enthalpy change is plotted against the moles of water added.

21. Differential enthalpies of solution and dilution 21

22. Differential enthalpy of solution ( Δ H sol ) It is the partial derivative of the total heat of solution with respect to the molar concentration of one component of the solution, when the concentration of the other component or components, temperature, and pressure are constant. Δ H sol -change in heat when solute is added -change in number of moles of solute   22

23. Differential enthalpy of dilution( Δ H dil ) Heat evolved or absorbed when a unit weight (one mole) of solvent is added to an infinite weight of solution at a constant temperature and concentration . Δ H dil = [ ] T,p,nB,nC …. =H A - H A* H A -Partial molar enthalpy of solvent in the solution H A* - molar enthalpy of solvent = infinitesimal change of mole number of dilution   23 Y . C. Wu and T. F. Young, JOURNAL OF RESEARCH of the National Bureau of Standards, 85 , 1980 http ://www.personal.psu.edu/mrh318/Wu-Young-JRNBS-1980.pdf

24. Just for information y or f(x) y ; integrating f(x) y y= 2(2) 2 - 2(0) 2 y= 8   24 Area = 8 x y= 2x 0.5 1 1 2 1.5 3 2 4 Integration: finding area under the curve Differentiation: finding slope of the curve Slope Slope Slope  

25. Calculation of bond energy, bond dissociation energy and resonance energy from thermochemical data 25

26. Bond energy and bond dissociation energy Bond energy is defined as average amount of energy required to break all bonds of a particular type in one mole of substance. Bond dissociation energy is the amount of energy needed to break down a particular bond. 26

27. Problems on bond energy Given that energies for H-H, O=O and O-H are 104, 118 and 111 kcal mol -1 , respectively. Calculate heat of the following reaction. H 2 (g) + ½ O 2 (g)  H 2 O (g) Solution: In above reaction, one H-H and half O=O bond are broken and two O-H bonds are formed. - ) - [1 ) + ½ )] - [ 1 + ½ ] - [ + 59] The heat of reaction is 59 kcal mol -1   27

28. Resonance energy 28 The difference in energy for dehydrogenation of cyclohexene to cyclohexa -1,3,5- triene in calculated and observed value is Resonance Energy. * Not to scale

29. Variation of enthalpy of a reaction with temperature - Kirchhoff’s equation Kirchoff’s equation is can be stated as d( or Where are differential heat content (enthalpy) of reactant and product, respectively. is difference in heat capacity of product and reactant at constant pressure. T1 and T2 are two different temperatures. Change in heat of reaction at constant pressure per degree change of temperature is equal to difference in heat capacities of product and reactants at constant pressure.   29

30. Problem on Kirchhoff’s equation The heat of reaction below reaction at 27 ◦C is -22 kcal. Calculate the heat of reaction at 77 ◦C. The molar heat capacities at constant pressure at 27 ◦ C for hydrogen, chlorine and hydrochloric acid are 6.82, 7.70 and 6.80 cal mol -1 , respectively. ½ H 2 + ½ Cl 2  HCl = Heat capacity of product – heat capacity of reactant ] = -22.123 Heat of reaction at 77 ◦ C is -22.123 kcal   30

31. Statement of Third Law of Thermodynamics Refer slide 13 31

32. Calculation of absolute entropies of substances 32

33. 33 Eg 1. Calculate the standard entropy of formation, of CO 2 gas. Given the standard entropies of CO 2(g) , C (g) and O 2 (g ) are 213.6, 5.740 and 205.0 JK -1 , respectively. Solution: C (g) + O 2 (g)  CO 2(g) ΔS◦ f = S◦ product – S◦ reactants ΔS◦ f = S ◦(CO 2 ) – [S◦( C ) + S◦ (O 2 )] ΔS ◦ f = 213.6– [ 5.740 + 205.0] ΔS◦ f = 2.86 JK -1

34. Eg 2 . Calculate the entropy change in the surroundings when 1.0 mol H 2 O (l) is formed from its element under standard conditions at 298 K. (Given Δ H◦ f H 2 O (l) =-286 kJ mol -1 ) Solution: As enthalpy change is negative the energy is released from the system in the form of heat is 286 kJ.   34