Notes On Bond Enthalpy - CBSE Class 11 Chemistry
When a bond is formed between atoms energy is released. When bonds dissociates energy required for dissociation. The energy released when a bond is formed or the energy required to dissociate the bonds is called Bond enthalpy. Definition:  The amount of energy released when 1 mole of bonds are formed from isolated atoms in the gaseous state or the amount of energy required to dissociate 1 mole of bonds present between the atoms in the gaseous molecules. Representation: ∆bH or ∆bond H Enthalpy changes are associated with chemical bonds. two different terms are used to express enthalpy changes associated with chemical bonds. They are Bond dissociation enthalpy and mean or average bond enthalpy The bond dissociation enthalpy is applies to diatomic molecules where as mean or average bond energy is applies to poly atomic molecules. The bond dissociation enthalpy for the di atomic molecules is the same as their bond enthalpy. Example:  Hydrogen gas dissociates to give hydrogen atoms. the bond dissociation energy of Hydrogen molecule is equal to enthalpy of atomisation of hydrogen. H2 (g) → H(g)+ H(g)                           ∆H-H H0 =435.0kJ/mol The bond enthalpy for polyatomic molecules is the average value of bond dissociation enthalpies. Example:  Methane When methane bonds dissociates it requires 1665 kJ/mol energy In methane all C-H bonds are equal in bond length and bond energy. But energy required to break the bonds in successive steps are differ. STEP 1: CH4(g) → CH3(g) + H(g) ; Δbond H° = +427 KJmol-1 STEP 2: CH3(g) → CH2(g) + H(g) ; Δbond H° = +439 KJmol-1 STEP 3: CH2(g) → CH(g) + H(g) ; Δbond H° = +452 KJmol-1 STEP 4: CH(g) → C(g) + H(g) ; Δbond H° = +347 KJmol-1                         ΔC-HH° = ¼ (ΔaH°)                                    = ¼ (1665 KJmol-1)                                    = 416  KJmol-1 Bond enthalpy is mainly used to calculate enthalpy changes in  reaction. Enthalpy of reaction = Sum of enthalpies of reactants - Sum of bond enthalpies of products. Example:Combustion of propane CH3 CH2 CH3 +5O2 → 3CO2 +4H2O Bond enthalpies: C-C =347 kJ/mol C-H = 414 kJ/mol C=O = 741 kJ/mol O=O = 498 kJ/mol O-H = 464 kJ/mol Number of bonds broken on reactants 8 C-H bonds, 2 C-C bonds and 5 O=O are present. Number of bonds formed in products 6 C=O & 8O-H bonds. Enthalpy of combustion of propane = Sum of enthalpies of reactants - Sum of bond enthalpies of products. ∆rH0= [(8*414)+(2*347)+(5*498)-(6*741)+(8*464)]         = -1662 kJ/mol

#### Summary

When a bond is formed between atoms energy is released. When bonds dissociates energy required for dissociation. The energy released when a bond is formed or the energy required to dissociate the bonds is called Bond enthalpy. Definition:  The amount of energy released when 1 mole of bonds are formed from isolated atoms in the gaseous state or the amount of energy required to dissociate 1 mole of bonds present between the atoms in the gaseous molecules. Representation: ∆bH or ∆bond H Enthalpy changes are associated with chemical bonds. two different terms are used to express enthalpy changes associated with chemical bonds. They are Bond dissociation enthalpy and mean or average bond enthalpy The bond dissociation enthalpy is applies to diatomic molecules where as mean or average bond energy is applies to poly atomic molecules. The bond dissociation enthalpy for the di atomic molecules is the same as their bond enthalpy. Example:  Hydrogen gas dissociates to give hydrogen atoms. the bond dissociation energy of Hydrogen molecule is equal to enthalpy of atomisation of hydrogen. H2 (g) → H(g)+ H(g)                           ∆H-H H0 =435.0kJ/mol The bond enthalpy for polyatomic molecules is the average value of bond dissociation enthalpies. Example:  Methane When methane bonds dissociates it requires 1665 kJ/mol energy In methane all C-H bonds are equal in bond length and bond energy. But energy required to break the bonds in successive steps are differ. STEP 1: CH4(g) → CH3(g) + H(g) ; Δbond H° = +427 KJmol-1 STEP 2: CH3(g) → CH2(g) + H(g) ; Δbond H° = +439 KJmol-1 STEP 3: CH2(g) → CH(g) + H(g) ; Δbond H° = +452 KJmol-1 STEP 4: CH(g) → C(g) + H(g) ; Δbond H° = +347 KJmol-1                         ΔC-HH° = ¼ (ΔaH°)                                    = ¼ (1665 KJmol-1)                                    = 416  KJmol-1 Bond enthalpy is mainly used to calculate enthalpy changes in  reaction. Enthalpy of reaction = Sum of enthalpies of reactants - Sum of bond enthalpies of products. Example:Combustion of propane CH3 CH2 CH3 +5O2 → 3CO2 +4H2O Bond enthalpies: C-C =347 kJ/mol C-H = 414 kJ/mol C=O = 741 kJ/mol O=O = 498 kJ/mol O-H = 464 kJ/mol Number of bonds broken on reactants 8 C-H bonds, 2 C-C bonds and 5 O=O are present. Number of bonds formed in products 6 C=O & 8O-H bonds. Enthalpy of combustion of propane = Sum of enthalpies of reactants - Sum of bond enthalpies of products. ∆rH0= [(8*414)+(2*347)+(5*498)-(6*741)+(8*464)]         = -1662 kJ/mol

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