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Write the notes on diatomic molecules and polyatomic molecules. OR Explain the bond enthalpy. |
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Answer» Solution :DIATOMIC Molecules : `H_(2(g)) to 2H_((g)) ,Delta_(H-H)H^( Theta )=435.0 "kj/mol"` The enthalpy change INVOLVED in this process is the bond DISSOCIATION enthalpy of H-H bond. It is the same as the enthalpy of atomization of dihydrogen. This is true for all diatomic molecules. `CI_(2(g)) to 2CI_((g)) , Delta_(CI - CI) H^( Theta ) = 242 "kj/mol"` `O_(2(g)) to 2O_((g)) , Delta_(O=O) H^( Theta ) = 428 "kj/mol"` The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bond of a gaseous covalent compound is broken to form products in the gaseous phase. "It is change in enthalpy to break one mole of covalent bond and to get product in its vapour phase." Polyatomic Molecules : In the case of polyatomic molecules, bond dissociation enthalpy is different for different bonds within the same molecules. e.g., `CH_(4(g)) to C_((g)) + 4H_((g)) , Delta_(a) H^( Theta ) = 1665 "kj mol"` In methane, all the four C - H bonds are identical in bond length and energy. However, the energies required to break the individual C - H bonds in each successive step differ : `CH_(4(g)) to CH_(3(g)) + H_((g)) , Delta_("bond") H^( Theta ) = + 427 "kj mol"^(-1)` `CH_(3(g)) to CH_(2(g)) + H_((g)) , Delta_("bond")H^( Theta ) = +439 "kj mol"^(-1)` `CH_(2(g)) to CH_((g)) + H_((g)) , Delta_("bond")H^( Theta ) = +452 "kj mol"^(-1)` `CH_((g)) to C_((g)) + H_((g)) , Delta_("bond")H^( Theta ) = +347 "kj mol"^(-1)` Therefore, `CH_(4(g)) to C_((g)) + 4H_((g)), Delta_(a) H^( Theta )= 1665 "kj mol"^(-1)` In such cases we use mean bond enthalpy of C-H bond. for example in `CH_(4) , Delta_(C-H) H^( Theta )` is calculated as, `Delta_(C-H) H^( Theta ) = (1)/(4) (Delta_(a) H^( Theta ) )= (1)/(4) (1665) = 416 "kj/mol"` That means C-H bond enthalpy in methane is 416 kJ/mol. It has been found that mean C-H bond enthalpies differ slightly from compound to compound. `Delta_(r) H^( Theta ) "bond enthalpies"_("reactants") - sum "bond enthalpies"_("products")` THUS the net enthalpy change of a reaction is the amount of energy required to break all the bonds in the reactant molecules minus the amount of energy required to break all the bonds in the product molecules. This relationship is approximate and is valid when all substances in the reaction are in gaseous state. Table - (a) : Some Mean Single Bond Enthalpies in kJ `"mol"^(-1)` at 298 K 
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