Thermodynamics-Notes

                                          UNIT 6    THERMODYNAMICS

It is the branch of chemistry which deals with energy changes of macroscopic systems.

Thermodynamic  terms: SYSTEM:-It is that part of universe in which energy changes are made.

SURROUNDING:-The part of universe other than system is called surrounding.

Universe=System+surrounding   , Boundry is the wall which separates system and surrounding.

Types  of system:- 1.Open system:- In this system , there is an exchange of matter as well as energy between system and surroundings. E.g. Reactants in an open beaker.

2.Closed system:- In this system , there is an exchange of  energy but not matter  between system and surroundings. E.g. Reactants in a closed beaker.

3.Isolated system:- In this system , there is neither exchange of matter nor energy between system and surroundings. E.g. Tea in thermos flask.

State of system is the state of existence of  system which can be described in terms of state variables

State variables are those variables which describe the state of system. These are mainly macroscopic properties like P , T, V and amount of substance(n).

Internal  energy is the total amount of energycontained within a system, which is the sum of  chemical, electrical, mechanical and other types of energies. It can change when:  1. Heat passes into or out of the system.   2. Work is done on or by the system.   3. Matter enter or leaves the system.

Adiabatic process is that in which there is no transfer of heat between system and surrounding. So temperature of the system changes.

Isothermal process is that in which there is an exchange of energy with the surroundings. So temperature of the system changes.

We can change the state of system by  mechanical work or electrical work. In both of the cases, there is change in temperature(∆T) as well as internal energy(∆U). Which is given by  ∆T=T₂-T₁ , change in internal energy (∆U) = U₂-U₁ ,  also ∆U=q(heat absorbed by the system)

Sign conventions:-   Work done on the system= +w   ,  work done by the system= -w  ,  heat absorbed by the system= +q  , heat given out of system = -q

First law of thermodynamics:- Energy of an isolated system is always constant.  ∆U = q+w .      if w=0 , q=0 then ∆U =0 for an isolated system.

Derivation for mechanical work or pressure-volume work:-

Consider one mole of ideal gas filled in a cylinder fitted with frictionless piston and the work of compression is going on.

V_i = initial volume of gas.   V_f = final volume of gas.  P_ex = external pressure.  P = Total pressure.  By the application of P_ex , piston covers distance  l and volume changes from V_i to V_f. Then work done on the system is given by:

  W = force  X distance    OR   W = P_ex .A X l as ( F= PA)& ( Axl = V) ,  W = Pex .(- ∆V) or  W = -P∆V

Here negative sign is used to make w +ve , as volume change will be –ve in compression work.

Reversible process is that process which proceeds infinitely slowely by a series of equilibrium states and even a  small change can reverse the process. (while irreversible process cannot be reversed back)

Various forms  of ∆U = q+w :-

1.     For  isothermal irreversible change:  q = -w = P∆V.                                                                         

2.     For  isothermal reversible change: q = -w = 2.303 nRT log V_f/V_i

3.     For adiabatic change; q=0 , ∆U =W_ad

ENTHALPY( H) : If we consider that heat is absorbed by the system constant pressure (q_p) then expansion work will be done by the system(-w) . so      ∆U = q_p - p∆V

                                                       U2-U1 = q_p –p (V₂-V₁)   on rearranging we get

                                                            Q_p  =( U₂+pV₂) – (U₁+pV₁)  but  U+PV= H

                                                             Q_p = H₂ – H₁  or   q_p = ∆H

 so enthalpy change is heat absorbed at constant pressure. We can also write ∆H= ∆U+p∆V.   ∆H is negative for exothermic reactions and positive for endothermic reactions. At constant volume, ∆V=0, so the above equation becomes ∆H = ∆U = q_v. Here q_v is heat absorbed at constant volume.

Consider two gases with volume V₁ and V₂ , no. Of moles of gaseous reactants are n₁ and n₂.  So   pV₁ = n₁RT  and  pV₂ = n₂RT  then  p( V₂-V₁) = (n₂-n₁) RT    or we can conclude that        p∆V  =  ∆nRT  where ∆n= moles of gaseous products – moles of gaseous reactants

Hence the new equation becomes     ∆H = ∆U + ∆nRT

Extensive property is a property whose value depends on quantity of matter present in the system. E.g. mass, volume, internal energy, enthalpy and heat capacity.

Intensive property is a property which depends upon nature of matter and is independent of the quantity of matter. E.g. temperature, density, pressure.

Molar property is the value of extensive property of system for one mole of substance.

Specific heat capacity is the quantity of heat required to raise the temperature of unit mass of  substance by one degree Celsius.

Molar heat capacity is the heat capacity for one mole of substance.

We know that      Heat supplied ∞   change in temperature

                                             q    ∞  ∆T   or    q= C∆T  (C is heat capacity)

                                                                      q = c_m ∆T (c=specific heat ,m=no. of moles)

Relationship between Cp and Cv

We know that  at constant temperature  q_v = Cv∆T = ∆U

                At  constant pressure,  q_p = Cp∆T = ∆H

For one mole of ideal gas,  ∆H = ∆U + P∆V  or   ∆H = ∆U + R∆T

                                                                        Cp∆T = Cv∆T + R∆T  or   Cp – Cv = R

Measurements of  ∆H and ∆U is on page 163 and 164 of NCERT

Reaction enthalpy is the enthalpy change accompanying a reaction.

∆_rH = Total enthalpies of products – total enthalpies of reactants

Eg.   CH₄ + 2O₂ à CO₂ + 2H₂O

   ∆_rH = [H_m(CO₂) + 2H_m(H₂O)]  -  [H_m(CH₄) + 2H_m(O₂)]

Note: do the definition of following enthalpies from NCERT

Enthalpy of fusion   H₂O(s) à H₂O(l)      ∆_fus H = 6 kJ/mol

Standard enthalpy of vapourisation     H₂O(l) à H₂O(g)      ∆_vap H⁰ = 40.79 kJ/mol

Standard enthalpy of sublimation       CO₂(s) à CO₂(g)     ∆_sub H⁰ = 25.2 kJ/mol

Standard enthalpy of formation          H₂ + 1/2O₂ à H₂O      ∆_f H⁰ = -285.8 kJ/mol

Enthalpy of formation can be used to calculate enthalpy change for the reaction:

     CaCO₃  à CaO + CO₂ ,    ∆rH = [∆_fH(CaO) + ∆_fH(CO₂)]  -  ∆_fH(CaCO₃)

                                                 = [-635.1 -393.5] – (-1206.9)  = 178.3 kj/mol  (endothermic)

Thermochemical equation is the balanced chemical equation along with the value of ∆rH.

Eg. C₂H₅OH + 3O₂ à 2CO₂ + 3H₂O   ∆_rH = -1367kJ/mol

Calculations for the heat of reaction:-     Fe₂O₃ + 3H₂ à 2Fe + 3H₂O

Given : ∆_fH(H₂O) = -283.83 kJ/mol , ∆_fH(Fe₂O₃) =  -824.2 kJ/mol

 ∆_rH = 3 [∆_fH(H₂O)] – [∆_fH (Fe₂O₃)]

        = 3x -285.83  -  (-824.2) =  -33.3 kJ/mol

Hess’s law :- If a reaction takes place in several steps then its standard reaction enthalpy is the sum of standard enthalpies of all intermediate reactions, into which the overall reaction may be divided at the same temperature.

∆rH = ∆rH₁ + ∆rH₂ + ∆rH₃   (if reaction is taking place in three steps)

ENTHALPIES

1.     Std. enthalpy of combustion C₄H₁₀ + 13/2O₂ à 4CO₂ + 5H₂O  ∆cH⁰=  -2658kJ/mol

2.     Enthalpy of atomisation  CH₄  à C + 4H   ∆_aH⁰=  1665kJ/mol

3.     Bond enthalpy   a. For diatomic molecule   H₂  à H + H         ∆_H-H H= 435kJ/mol

b. For polyatomic molecules       H₂O  àH + OH     ∆_bH=  502kJ/mol

                                                 OH à O + H            ∆_bH= 427kJ/mol

                                           ∆_O-H H⁰=  ½ (∆_bH) = ½ x 929 = 464.5 kJ/mol

Relation between standard reaction enthalpy and bond enthalpy:

∆_rH⁰  = ∑Bond enthalpies of reactants - ∑Bond enthalpies of products

4.     Enthalpy of solution is the enthalpy change when one mole of substance dissolves in a specified amount of solvent.

                ∆_sol H⁰

AB( s)                                 A+(aq) + B-(aq)

∆_latticeH⁰

                                                            ∆_hydH⁰

              A+(g) + B-(g)

                                                         So       ∆_solH⁰  =     ∆_latticeH⁰   +  ∆_hydH⁰

5.     Lattice enthalpy is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.

e.g. Na⁺Cl^-   à Na⁺(g) + Cl^-(g)           ∆_latticeH⁰  = +788kJ/mol

We can calculate lattice enthalpy by Born – Haber cycle

                 Na(s)+1/2Cl₂ (g                                   

 

 _121KJ/mol      1/2 ∆_bondH⁰

 Na(s)+1/2Cl₂ (g             _-348.6   ∆_egH⁰

_495.6                 ∆_iH⁰

 Na(s)+1/2Cl₂ (g           Na(s)+1/2Cl₂ (g

_108.4                ∆_subH⁰

Na(s)+1/2Cl₂ (g)                      ∆_latticeH⁰

_-411.2               ∆_fH⁰

                       NaCl (s)

Applying Hess’s law here we get     ∆_latticeH⁰ = 411.2 + 108.4 + 121+ 496 - 348.6 = 788 kJ

                 But     ∆_solH⁰  =  ∆_latticeH⁰ +  ∆_hydH⁰   = 788 – 784(from literature) = 4 kJ/mol

Spontaneity means having the potential to proceed without getting the assistance of external agency. The deriving forces for spontaneous process are:

1.     Decrease in enthalpy    2.  Increase in entropy.

So we can say that all exothermic processes are spontaneous becoz they have tendency to get minimum energy. But here the question arises on the spontaneity of endothermic processes like  C(s) + 2S(l) à CS₂(l)          ∆_rH⁰ = +128.5kJ/mol

Now the answer can be increase in randomness. Heat added to the system at lower temperature causes greater randomness than heat added at higher temp.

So     ∆S=q_rev/T     and          ∆S_total  =  ∆S_system +  ∆S_sur

When the system is in equilibrium, entropy is maximum and ∆S=0. So we can say that entropy of spontaneous process increases till it reaches maximum and at equilibrium, entropy change is zero. So ∆S of reversible process is ∆S_sys=q_sys.rev /T

Both for reversible and irreversible processes ∆U=0 but ∆S_total is not 0 for irreversible system.

Problem: For 4Fe(s) + 3O₂(g) à 2Fe₂O₃(s)      ∆S= -549.4J/K/mol

Why is the reaction spontaneous with –ve entropy. Given ∆rH=-1648x10³ J/mol

Ans. If we consider that heat absorbed by surroundings is -∆rH then

 ∆S_sur= -∆_rH/T = -(-1648x10³)/298 = 5530J/K/mol   But     ∆S_total  =  ∆S_system +  ∆S_sur     so     ∆S=5530-549.4 = 4980.6 J/K/mol       and reaction is spontaneous.

GIBBS ENERGY AND SPONTANEITY :-  Gibbs function , an extensive property or a state function can be written as   G=H-TS    or  ∆G = ∆H - T∆S (at constant temp.)

In thermal equilibrium, temp. of system and surrounding is equal. Increase in enthalpy of surrounding is equal to decrease in enthalpy of system.

So    ∆S_sur = ∆H_sur/T  = - ∆H_sys/T

   ∆S_total  =  ∆S_system +  ∆S_sur    

∆S_total  =  ∆S_system -  ∆H_sys/T   on rearranging, we get  T∆S_tot = T∆S_sys - ∆H_sys

For a spontaneous process ∆S_tot >0  or  T∆S_sys - ∆H_sys>0  ,  -(∆H-T∆S)>0  ,  -∆G>0

Or  ∆G<0 for a spontaneous process.

∆G & EQUILIBRIUM:-  ∆G allows us to  i) predict the spontaneity of reaction    ii)predict the useful work extracted from the system.

At equilibrium, ∆G=0 . standard free energy is related to equilibrium constant(K)  as

   ∆_rG⁰ = -RT ln K  , ∆_rG⁰ = -2.303 RT log K  or    ∆_rH⁰ - T∆S⁰ = -2.303 RT log K

For endothermic reaction: ∆H⁰ large +ve ,  K<1For exothermic reaction: ∆H⁰ large -ve   ,  K>1