Condition under which ∆H = ∆U

Explain a condition in which ∆H = ∆U  
Condition under which ∆H = ∆U
 1) When a reaction is carried out in a closed vessels so that the volume of the system     remain constant 
  ∆V = 0 
∆H = ∆U + P∆V 
∆H = ∆U + P x (0)
∆H = ∆U 
 2) When a reaction involves only solid and liquids, 
    Therefore (not expands) 
∆V is neglected 
∆H= ∆U + P∆V 
∆H = ∆U 
 3) In reaction in which n1=n2 i.e. number of gases Reactant is equal to number of gases         product 
∴ n2 - n1 = 0 
         ∆n = 0 
 ∆H = ∆U + RT∆n 
 ∆H = ∆U + RT x (0)
 ∆H = ∆U + 0 
∆H = ∆U 

Enthalpy of physical changes:-

A) Enthalpy of phase Transition 
 a) Enthalpy of fusion (∆fusH)          
       Enthalpy change when one mole of solid substance is converted into its liquid state    at its melting point is called enthalpy of fusion. 
                                 OR 
       The enthalpy change due to the fusion of one mole of solid without change in    temperature at constant pressure is called enthalpy of fusion. 
      It is denoted by symbol ∆fusH 
    Example -   H2O(s)    ----------->  H2O(l)    ∆fusH = + 6.01 kJ/mol 00C 

b) Enthalpy of freezing (∆freezH) 
Enthalpy change when one mole of liquid substance is converted into its solid at    its       solid at its freezing point is called enthalpy of freezing 
    Example    H2O(l)   ----------->     H2O(s)    ∆freezH = -6.01 kJ/mol at 00C 

c) Enthalpy of Vaporization (∆vapH) 
The enthalpy change due to the vaporization of one mole of liquid without changing     its temp at const pressure is called enthalpy of vaporization. 
                     H2O(l)     ----------->      H2O(g)    ∆vapH=+40.7kJmol-1at1000C                                              H2O(l)  ----------->        H2O(g)     ∆vapH=+44.0kJmol-1at 25⁰C  

d) Enthalpy of condensation ( ∆conH) 
The enthalpy change due to the condensation of one mole of gas without changing its     temp at const pressure is called enthalpy of condensation. 
                   H2O(g)    ----------->       H2O(l) ∆conH = -40.7 kJmol-1 at1000C
   
e) Enthalpy of Sublimation (∆subH) 
         The direct conversion of solid vapour without going through liquid state is called sublimation.
The enthalpy change due to the conversion of one mole of solid directly into its vapor’s at cost temp and pressure is called enthalpy of sublimation.
                  H2O(s)    ----------->     H2O(g)           ∆subH = 51.08 kJmol-1 at00C …..(i)

 QUE :  Prove that enthalpy of sublimation is equal to enthalpy of fusion and enthalpy of vaporization.

It should be denoted that whether the conversion of solid to vapour take placed in one or two step 



Case1 :-    Direct method 
The solid is directly converted into vapour 
              H2O(s)   ----------->         H2O(g)    ∆subH = 51.08 kJmol-1at00C        ......(i)
Case 2 : - Indirect method 
Step 1) The solid is converted into liquid 
        H2O(s)  ----------->          H2O(l)            ∆fusH = + 6.01 kJmol-1 at00C                                     H2O(l)      ----------->      H2O(g)            ∆vap H = + 45.07 kJmol-1 at00C
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     H2O(s)     ----------->      H2O(g)   ∆fus H + ∆vap H = ( 6.01 + 45.07)  kJmol-1       
                                                                                 = 51.08 kJmol-1 at00C … (ii)   
From equation i & ii 

                      ∆subH = ∆fusH + ∆vapH 


B)Enthalpy of atomic or molecular changes 

a)Enthalpy of Ionization:- 
The enthalpy change due to the removal of an electron from each atom or ion in 1 mole of gases atom or ion is called enthalpy of ionization. 
Ex. -> Na(g) ----------->    Na+(g) + e-   ∆ionH = 494 kjmol-1
Ex. -> Ca(g) ----------->      Ca+(g) + e-     ∆ion H = 590 kjmol-1
       Ca+(g)  ----------->       Ca+2(g) + e-     ∆ion H = 1150 kjmol-1

Note :->  The enthalpy change for the removal of one electron from atom in one mole is called First Ionization enthalpy 

The enthalpy change for the removal of second electron from each atom in one mole is called second ionization enthalpy 

QUE : Explain that second ionization enthalpy is greater than first ionization enthalpy. 

The second ionization enthalpy is greater than first ionization enthalpy due to following Reason. 

The first ionization enthalpy is the energy necessary to remove the electron from neutral atom where as the second ionization enthalpy is the energy necessary to remove the electron from positive ion, it is more difficult to remove an electron from positive ion due to the strong attraction between electron and positive ion. 

Ex. - Ca(g)  ----------->     Ca+(g) + e-                   ∆ionH = 590 kjmol-1   1st I.E

        Ca+(g)  ----------->      Ca+2(g) + e-      ∆ionH= 1150 kjmol-1         2st I.E 

b)Enthalpy of Atomization ( ∆atoH) 

The enthalpy change due to the dissociation of all molecules in 1 mole of gases phase substance into gaseous atom is called enthalpy of atomization.

 Cl2(g)           Cl(g) + Cl(g)     ∆atomH = 242 kjmol-1 

 The enthalpy atomization of Cl2 ( g) is 242 kg/mol 

CH4(g)            C(g) + 4H (g)     ∆atom H = 1660 kjmol-1  

Gaseous substance   gaseous atom 

c) Enthalpy of solution ( ∆solnH ) 

      Enthalpy change when 1 mole of solute is dissolve in specific amount of solvent known as enthalpy of solution 

OR

      Enthalpy change when one mole of solute is dissolve completely in excess of solvent known as enthalpy of solution.

Ex. -> KCl (s) + 200H2O (l) -----------> KCl (200H2O)              ∆H= +18.58 kJ

            HCl (g) + 200 H2O (l) -----------> HCl (200H2O)      ∆H = -72.90 kJ

Solute: The compound which is present in smaller proportion is called solute. 

Solvent: The compound which is present in larger proportion is called solvent.

         

       ∆solH = ∆LH + ∆hydraH 

       ∆LH = crystal lattice enthalpy

       ∆H is +ve heat absorbed (increase in enthalpy) 

       ∆H is –ve heat evolved (decrease in enthalpy) 

d)Enthalpy of dilution 

The enthalpy of dilution is defined as enthalpy change that occurs when a solution of one concentration is diluted to form solution of another concentration.