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.