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What is bond enthalpy? Explain bond enthalpy in polyatomic molecule like H_(2)O . |
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Answer» Solution :It is defined as the amount of energy required to break one mole of bonds of a particular type between two atoms in gaseous state. The UNIT of bond enthalpy is kJ`mol^(-1)` . e.g.-1 : (i) The H - H bond enthalpy in hydrogen molecule is 435.8 kJ` mol^(-1)` `H_(2(g)) rarr H_((g)) + H_((g)) , Delta_(a)H^(O) = 435.8" kJ " mol^(-1)` e.g.-2 : (ii) The bond enthalpy for molecules containing multiple bonds, for example `O_(2) and N_(2)` will be as under : `O_(2) (O = O)_((g)) rarr O_((g)) + O_((g)) , Delta_(a) ` H = 408 kJ `mol^(-1)` `N_(2)(N equiv N)_((g))rarr N_((g)) , Delta_(a) H = ` 946.0 KK `mol^(-1)` Characteristic of bond enthalpy : Here bondenthalpy is energy of BROKEN the bond (dissociation) so, it endothermic, but making enthalpy is exothermic. It is important that larger the bond dissociation enthalpy, stronger will be the bond in the molecule. So enthalpy is increase in this order : single bond `rarr` double bond `rarr`Triple bond. HCl is heteronuclear diatomic molecules and its bond dissociation enthalpy is as under `HCl_((g)) rarr H_((g)) + Cl_((g)) : Delta_(a) H = 431.0 " kJmol"^(-1)` Bond dissociation enthalpy for polyatomicmolecule `H_(2)O` : In case of polyatomic molecule, the measurement ofbond strength is more complicated. Example: In case of `H_(2)`O molecule, the enthalpy needed to break the two O - H bond is not the same but different. `H_(2)O_((g)) rarr H_((g)) + OH_((g)) , Delta_(a) H_(1)^(Ө) = 502 " kJ mol"^(-1)` ` OH_((g)) rarr H_((g))+ O_((g)): Delta_(a)H_(2)^(Ө) = ` 427 kJ `mol^(-1)` The difference in the `Delta H^(Ө)` value shows that the SECOND O - H bond undergoes some change because of changed chemical environoment . This is the reason for some different in energy of the same O - H bond. In different molecules like `C_(2)H_(5)` OH (ethanol ) and water. Average Bond enthalpy: The mean or average bond enthalpy is used in more than one same bond containing polyatomic molecules. It is obtained by dividing total bond dissociation enthalpy by the NUMBER of bonds broken. Average bond enthalpy = `("Total enthalpy of broken bond")/("Number of all broken bond") ` `({:("Average enghalpy of"),("O - H bond in"H_(2)O):}) = (Delta_(a) H_(1)^(Ө) + Delta_(a) H_(2)^(Ө))/(2)` ` = (502 + 427)/(2)` = `464.5 kJ mol^(-1)` |
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