THERMOCHEMISTRY

Heat of Reaction:-

            Consider a general Reaction

                                    A+B   <=============>    C+D

Let        H_A is enthalpy of substance A

             H_B is enthalpy of substance B

             H_C is enthalpy of substance C

             H_D is enthalpy of substance D

We know that heat of reaction ∆H is

∆H = ∑enthalpy of product – ∑enthalpy of reactant

∆H = ∑ [ Hc + H_D  ] –∑ [H_A  + H_B]

∆H = ∑H_P  - H_R

OR                   ∆H = ∑ ∆Hproduct  - ∑∆Hreactant

Exothermic and Endothermic Reaction:-

 Endothermic   Reaction:-

                   If ∑ ∆H product is larger than ∑∆H reactant

                  ∆H = ∑∆H product - ∑∆H Reactant

                         = +ve

          The Reaction in which heat is absorbed by the system from the surrounding are called Endothermic Reaction.

Ex.-  1) Heat of formation of CS₂

               C_(S) + 2S_(s) + Heat  ------------>         CS₂               ∆H⁰ = +88K

        2)   N_{2 (g)} +2C_{2 (g)} + Heat  ---------->       2NO_{2 (g)}        ∆H⁰= +180KJ       

Exothermic Reaction:

 If ∑∆H product is less than ∑∆H reactant

                   ∆H = ∑H product - ∑H reactant

                         = -Ve

Therefore the Reaction in which heat is released from the system to the surrounding called Exothermic Reaction

Ex-> 1) Heat of combustion of carbon

          C_(s) + O_2(g)  ------------>     CO_2(g)+ Heat                 ∆H = -393.51 KJ

        2)  Heat of combination of Nitrogen

            N_2(g) +3H_{2(g) --------------->}    2NH_3(g) + Heat         ∆H= -100.4 KJ      

Standard State: - The most stable state of a substance at room temperature (298k and 1 atm) is known standard state. 

Standard enthalpy of chemical reaction:-

          Enthalpy change accompanying the reaction when all the substance involved are in their standard state.

 It is denoted by ∆H⁰

  Thermochemical equation: -

                  Balanced chemical equation which indicates exact value of change, physical states and number of moles of reactants and products is known as Thermochemical equation.

 E.g:-    C_(s) + 0_2(g) ------------>  CO_2(g)       ∆H = - 393.51kJ/mole

This balanced equation indicates that when 1 mole of solid carbon reacts with 1mole gaseous oxygen, 395kJ of heat energy is given out at a given temperature and hence, it is a thermo chemical equation.

  

Point while writing Thermochemical equation:-

1)   A Thermochemical equation should be well balanced.

2)   The physical state of the reactants and products should be mentioned by using the symbol. s (solid), l (liquid), g (gas) and aq. (aqueous)

3)   Proper sign of ∆H (-ve for exothermic reaction and +ve for endothermic reaction) should be used.

4)   The temperature at which enthalpy change takes place should be mentioned.

5)   The substances used in chemical reaction are taken in their standard state.

6)   Enthalpy of all the elements in their standard state is taken as zero.

7)   The Thermochemical equation can be reversed by reversing the sign of ∆H.

 E.g:-    C_(s) + 0_{2(g)------------->}   CO_2(g)                   ∆H = - 393.51kJ/mole

        This equation can be reversed as

            CO_2(g)   ------------->     C_(s) + 0₂              ∆H = + 393.5lkJ/mole

I) Standard enthalpy of formation (standard heat of formation/∆_fH)

          The std. enthalpy of formation of a compound is defined as the enthalpy change that accompanies a reaction in which one mole of the pure compound in its standard state is formed from its element also in their standard state.

It is denoted by ∆_fH

Ex;- H_{2 (g)} + ½ O_2(g)                            H₂O_(l) ,                  ∆ _fH⁰ =-286kj/mole

II) Standard enthalpy of combustion (standard heat of combustion/∆_cH)

The std. enthalpy of combustion of a substance is the std enthalpy change accompanying a reaction in which one mole of the substance in its std state burn completely with sufficient quantity of oxygen

CO_(g) + ½ O_2(g)   ------------>      CO_2(g)            ∆_cH⁰ =-283kj/mole

Bond enthalpy (Bond energy)

The enthalpy change necessary to break the covalent bond in 1 mole of gaseous molecule to produced atom or ion or radical is called bond enthalpy.

      H_{2 (g) --------------->}  H _(g) + H _(g)                   ∆H = 436.4KJ

           Ex.->  Cl_{2 (g)}  ------------->   Cl_(g) + Cl_(g)                    ∆H = 242.7KJ

  Ex.->  HCl_(g) --------------> H_(g) + Cl_(g)                     ∆H = 431.9KJ

In case of CH₄ molecule, four identical C-H bond’s However the four C-H bond req. diff. amount of energy

StepI ->       CH₄(g)  ----------->    CH₃ (g) + H(g)                ∆H = 427KJ

step II->      CH₃(g)  ------------>   CH₂ (g) + H(g)                ∆H = 439KJ

step III->     CH₂(g) ------------->   CH (g) + H(g)                 ∆H = 452KJ

                       CH(g) ----------->-                   C (g) + H(g)   ∆H = 347KJ

 stepIV->     

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CH₄(g) ------------->  C(g) +4H(g)                    ∆H = 1665K

Average bond enthalpy = 1665 KJ / 4 = 416 KJ

                        ∆H(C-H) = 416 KJ/mol.

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.

Second law of thermodynamics and enthalpy

Second law of thermodynamics:-

Statements:-

1)   The spontaneous flow of heat is always unidirectional from higher temperature to lower temperature.

2)   The heat cannot be completely converted into the equivalent amount of work without producing permanent changes either in the system or its surrounding.

3)   No machine has yet been made that has efficiency unity.

ENTHALPY

Definition: -

        It is defined as the sum of internal energy and pressure, volume type energy. Denoted by H.

Mathematical expression:-

                             H = U + PV

Since U, P, V is state function hence H is also state function

Therefore ∆H = H₂ – H₁

Where H₁ is enthalpy of system in initial state

              H₂ is enthalpy of system in final state

          H₁ = U₁ + P₁ V₁

H₂ =  U₂+ P₂ V₂

  ∆H = H₂ – H₁

  ∆H = U₂+ P₂ V₂  - (U₁ + P₁ V₁)

  ∆H = U₂- U₁ + P₂V_{2 -} P₁V₁

At constant pressure P = P₁ = P₂

  ∆H = ∆U + P ∆V                               -------------1

            q_P = ∆U + Pex ∆V        ( Isobaric process) 

Assuming that P_ex = P

           q_P = ∆U + P∆V                                   ------------2

Compare equation   1 & 2

           ∆H = q_P

Increase in enthalpy of system is equal to heat absorbed by it when reaction is carried out at constant pressure

Relationship between ∆H and ∆U for chemical reaction

          We known that

                   ∆H = ∆U + P ∆V

For reaction involving solid and liquid ∆V is neglected because it is not expand

 P∆V is neglected

Hence ∆H = ∆U

For reaction involving gases ∆U is not neglected 

                    ∆H = ∆U + P∆V

                        = ∆U + P(V₂  - V₁)

                        = ∆U + (PV₂ - PV₁)

Where  V₁ = volume of reactant in initial state

             V₂ = volume of reactant in final state

Assuming ideal behaviors of gas

          PV = nRT

  n₁ = No. of mole of gases Reactant

  n₂ = No. of mole of gases product

          PV₁ = n₁ RT

          PV₂ = n₂ RT

          ∆H = ∆U + n₂ RT - n₁ RT

          ∆H = ∆U + RT (n₂  - n₁)

          ∆H = ∆U + RT ∆n

Where, ∆n is difference between No. of mole of gases product and gases Reactant

 ∆H = change in enthalpy

 R = gas constant

 T= Temperature  

 ∆U = increases in internal energy.

Expression for Work done in chemical Reaction:-

          We known that

                             W = -Pex ∆V

Assuming Pex = P

                             W = -P∆V

                             W = - P (V₂  - V₁)

                                 =  PV₂  -  PV₁

                                 = - ( PV₂   +  PV₁  )

Again          PV = nRT

                             PV₁  = n₁ RT

                             PV₂  = n₂ RT

                             W = - (n₂ RT + n₁ RT)

                              W = -RT (n₂ -  n₁)

                              W = -RT ∆n -----------------1

Equation   1 gives work done by the system in chemical reaction

i)     If n₂ > n_{1 ,}W is negative & work is done by the system on surrounding

ii)    if n₂ <n₁, W is positive &work done on the system by surrounding

iii) if n₂ = n_1,W= 0, no work is done