Question

Bond dissociation enthalpy is used to defining enthalpy change of a reaction as
$\left( 1 \right)$ $\Delta {H_r} = \sum {{{\left( {Bond\,dissociation\,enthalpy} \right)}_{{\text{Reactant}}}} - \sum {(Bond\,dissociation} } \,enthalpy{)_{{\text{Product}}}}$
$\left( 2 \right)$ $\Delta {H_r} = \sum {{{\left( {Bond\,dissociation\,enthalpy} \right)}_{{\text{Product}}}} - \sum {(Bond\,dissociation} } \,enthalpy{)_{{\text{Reactant}}}}$
$(3)$ $\Delta H = \sum {{{\left( {Bond\,dissociation\,enthalpy} \right)}_{{\text{Reactant}}}} + \sum {(Bond\,dissociation} } \,enthalpy{)_{{\text{Product}}}}$
$\left( 4 \right)$ None of these

Answer 1

Hint: The Bond dissociation enthalpy is an enthalpy used to break (Different atoms like A-B) one mole of the bond to give separate two gases atom (A+B). If the energy is used to break homolysis bonds (Same atom like A-A) to give free radicals ( ${A^ \bullet } + {A^ \bullet }$).

Complete step by step answer:
As we know the bond enthalpy is defined as the change in the bond dissociation enthalpy bond broken of reactants) and bond dissociation enthalpy (bond formation of reactant).
In $\left( 1 \right)$ , Bond enthalpy is defined as the change in the bond dissociation enthalpy (bond broken of reaction) and the bond dissociation enthalpy (bond formation of reactant).
$\Delta {H_r} = \sum {{{\left( {Bond\,dissociation\,enthalpy} \right)}_{{\text{Reactant}}}} - \sum {(Bond\,dissociation} } \,enthalpy{)_{{\text{Product}}}}$
In $\left( 2 \right)$ , Bond enthalpy is defined as the change in the bond dissociation enthalpy (bond formation of reaction) and the bond dissociation enthalpy (bond broken of reactant).
$\Delta {H_r} = \sum {{{\left( {Bond\,dissociation\,enthalpy} \right)}_{{\text{Product}}}} - \sum {(Bond\,dissociation} } \,enthalpy{)_{{\text{Reactant}}}}$
In $\left( 3 \right)$, , Bond enthalpy is defined as the sum of the bond dissociation enthalpy (bond broken of reaction) and the bond dissociation enthalpy (bond formation of reactant).
$\Delta H = \sum {{{\left( {Bond\,dissociation\,enthalpy} \right)}_{{\text{Reactant}}}} + \sum {(Bond\,dissociation} } \,enthalpy{)_{{\text{Product}}}}$
As we, discussed above the definition and the equation $\left( 1 \right)$ are mention the same condition for the bond dissociation enthalpy
Hence, the correct option is $\left( 1 \right)$ .

Note:
Hess’s law is defined as the sum of the changes in enthalpy for a series of intermediate reaction steps to find the overall change in enthalpy for a reaction.
$1.$Enthalpy change for a reaction is independent of the number of ways a product can be obtained. If the initial and final conditions are the same.
$2.$Negative enthalpy change for a reaction indicates exothermic process, while positive enthalpy change corresponds to endothermic process.