CLASS 11 chemistry ch 6

Class 11 chemistry chapter 6

Thermodynamics Laws in Pdf Notes 

Zeroth Law Second Law 3rd Law

 Thermodynamics :

Thermodynamics Laws in Pdf Notes 

Zeroth Law Second Law 3rd Law

 Thermodynamics :

Science which deals with study of different forms of energy and quantitative relationship.

System & Surroundings :

The part of universe for study is called system and remaining portion is surroundings.

State of system & state function :

State of system is described in terms of T, P, V etc. The property which depends only on state of system not upon path is called state function eg. P, V, T, E, H, S etc.

Extensive & Intensive Properties :

Properties which depends on quantity of matter called extensive prop. eg. mass, volume, heat capacity, enthalpy, entropy  etc.  The  properties  which do  not  depends  on  matter  present depends  upon  nature  of substance called  Intensive properties. eg.  T,P, density, refractive index, viscosity, bp, pH, mole fraction etc.

Internal energy :

The total energy with a system. i.e. U =  Ee En + Ec + Ep + Ek + ------ ∆ U = U2 – U1 or UP – UR & U is state function and extensive properly. If U1 >U2 energy is released.

 Heat (q) :

It I a form of energy which is exchanged between system and surrounding due to difference of temperature. Unit is Joule (J) or Calorie (1 Calorie = 4.18    J).

First Law of Thermodynamics :

It is law of conservation energy. Energy can neither be created not destroyed, it may be converted from one from into another. Mathematically   ∆U = q + w, w = –p.   V∆ (work of expansion) ∆U = q – p. ∆ V or q = ∆ U + p. ∆V, q,w are not state function. ∆But   U is state function.

Enthalpy (H):

At constant volume ∆V = 0,qv =∆So H = U + p.   ∆V, qp = H2 H1 = ∆H ➱ ∆H =  ∆ U + P.∆V.

Relationship between qp, qv i.e.   ∆H&  ∆U  It is ∆ H=  ∆U+  ∆ng.RT or qp = qv +  ∆    ng.RT

Exothermic and Endothermic reactions :

 ∆H = –Ve for exothermic and ∆H = +Ve for endothermic reaction i.e. evolution and absorption of heat.

Eg C+O2      → CO2 + 393.5 KJ,   H = –393.5 KJ (exothermic) N2 + O2       → 2NO – 180.7 KJ,   H = 180.7 KJ (Endothermic)

Enthalpy of reaction (  ∆rH) :

The amount of heat evolved or absorbed when the reaction is completed.

Standard Enthalpy of reaction ( ∆ rHo) at 1 bar pressure and specific temp. (290K) i.e. standard state.

 Different types of Enthalpies of reactions:

( i ) Enthalpy of combustion (∆cH) ( ii) Enthalpy of formation (∆fH)

( iii ) Enthalpy of neutralization ( iv ) Enthalpy of solution

( v ) Enthalpy of atomization (∆aH) ( vi ) Enthalpy of Ionisation (∆iH)

( vii ) Enthalpy of Hydration (∆hyolH) ( viii ) Enthalpy of fusion (∆fusH)

( ix ) Enthalpy of vaporization (∆vapH) ( x ) Enthalpy of sublimation (∆subH)

(∆subH) = ∆fus(H) - ∆vapH) --------

Bond  enthalpy :  

It  is  amount  of  energy  released  when  gaseous  atoms combines to form one mole of bonds between them or heat absorbed when one mole of bonds between them are broken to give free gaseous atoms. Further ∆ rH = ∑B.E. (Reactants) - ∑B.E. (Products)

Spontaneous and non-spontaneous operations:

 The process that can happen by itself is called a spontaneous process.  A process that cannot occur alone or through initiation is called non-spontaneous.

 Driving forces of the automatic process:

 (I) Tilt of the lower energy state.

 (2) Direction for maximum randomness.

 Entropy:

 It is a measure of randomness or disorder in the system.  Gas> Liquid> Solid.

 Spontaneity in terms of (∆S) ∆S (total) = ∆S (universe) = ∆S (system) + ∆S (circumference) If ∆S (total) + ve, the process is spontaneous.  If (S (total) –ve, the operation is not automatic.

 The second law of thermodynamics:

 In any spontaneous process, the entropy of the universe always increases.  Spontaneous operation cannot be reversed.

 Gibbs Free Energy (G):

 It is defined as G = H - T.S & ∆G = ∆H - T.∆S (Helbholtz pocket equation) It is equally useful work.  - =G = W (useful) = W (max).  If G = ve, the process is automatic.

 Effects of T on spontaneous operation:

 ∆G = ∆H - T. ∆ S.

 (I) For endothermic processes, they may not be automatic at legal temperature.

 (2) For exothermic processes it may not be automatic at high temperature.  And spontaneously by voting law.

 Calculation (∆ rGo)

 ∆rGo = ∑∆fGo (p) - ∆fGo (r)

 The relationship between (∆rGo) and the equilibrium constant (k)

 ∆G = ∆Go + RTlnQ & ∆Go = –2.303RT record.

 Calculation of the change of the universe:

  ∆rSo = ∑∆ permission (p) - So (r)