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Question: \(\text{AB}\), \({{\text{A}}_{2}}\) and \({{\text{B}}_{\text{2}}}\) are diatomic molecules. If the b...

AB\text{AB}, A2{{\text{A}}_{2}} and B2{{\text{B}}_{\text{2}}} are diatomic molecules. If the bond enthalpies of A2{{\text{A}}_{2}},AB\text{AB} and B2{{\text{B}}_{\text{2}}} are in the ratio 1:1:0.51:1:0.5 and enthalpy of formation of AB\text{AB} from A2{{\text{A}}_{2}} and B2{{\text{B}}_{\text{2}}} is -100kJmol-1\text{-100kJmo}{{\text{l}}^{\text{-1}}}. What is the bond energy of A2{{\text{A}}_{2}}?
(A) 200 kJmol-1\text{200 kJmo}{{\text{l}}^{\text{-1}}}
(B) 100 kJmol-1\text{100 kJmo}{{\text{l}}^{\text{-1}}}
(C) 300kJmol-1\text{300kJmo}{{\text{l}}^{\text{-1}}}
(D) 400 kJmol-1\text{400 kJmo}{{\text{l}}^{\text{-1}}}

Explanation

Solution

Enthalpy of formation is the enthalpy change when one mole of substance is formed from its elements in their most abundant form or their standard and stable form. While bond energy is defined as the average amount of energy required to dissociate one mole gaseous bond into separate atoms.

Complete answer:
Bond energy of various bond present in the reactant and products are given then  !!Δ!! H\text{ }\\!\\!\Delta\\!\\!\text{ H}of that reaction can be calculated by following formula-
 !!Δ!! H=∑(B.E)R-∑(B.E)P\text{ }\\!\\!\Delta\\!\\!\text{ H=}\sum{{{\text{(B}\text{.E)}}_{\text{R}}}}\text{-}\sum{{{\text{(B}\text{.E)}}_{\text{P}}}}
The bond energies of A2{{\text{A}}_{2}},AB\text{AB} and B2{{\text{B}}_{\text{2}}} are in the ratio of 1:1:0.51:1:0.5
And enthalpy of formation of AB\text{AB} is known. So from the reaction
A2 + B2→ 2 AB{{\text{A}}_{\text{2}}}\,\text{+}\,{{\text{B}}_{\text{2}}}\to \,\text{2 AB}
So  !!Δ!! Hf= (BE)A-A + (BE)B-B- 2(BE)A-B.........(i)\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{\text{f}}}\text{=}\,{{\text{(BE)}}_{\text{A-A}}}\,\text{+}\,{{\text{(BE)}}_{\text{B-B}}}\text{- 2(BE}{{\text{)}}_{\text{A-B}}}.........(i)
From the question if (BE) of A-A =  a\text{(BE)}\,\text{of}\,\text{A-A = }\,\text{a}
So, (BE) of A-B =  a\text{(BE)}\,\text{of}\,\text{A-B = }\,\text{a}
And (BE) of B-B =  a2\text{(BE)}\,\text{of}\,\text{B-B = }\,\dfrac{a}{2}
Since  !!Δ!! Hf(A-B) is =  -100 kJmole-1........(ii)\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{\text{f}}}\text{(A-B)}\,\text{is = }\,\text{-100}\,\text{kJmol}{{\text{e}}^{\text{-1}}}........(ii)

After putting the value of  !!Δ!! Hf (A-B)\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{\text{f}}}\,\text{(A-B)}in equation (ii) from equation (i)
 !!Δ!! Hf(A-B) is =  -100 kJmole-1for one molecule........(ii) for two molecule of (A-B)       !!Δ!! Hf(A-B) is  =  −200 kJmole-1   a+ a2 −2a  =  −200 kJmole-1 −a2     =  −200 kJmole-1 a     =    400 kJmole-1\begin{aligned} & \text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{\text{f}}}\text{(A-B)}\,\text{is = }\,\text{-100}\,\text{kJmol}{{\text{e}}^{\text{-1}}}\text{for}\,\text{one}\,\text{molecule}........(ii) \\\ & \text{for}\,\text{two}\,\text{molecule}\,\text{of (A-B)}\,\,\,\, \\\ & \,\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{\text{f}}}\text{(A-B)}\,\text{is}\,\text{ =}\,\,-\text{200}\,\text{kJmol}{{\text{e}}^{\text{-1}}} \\\ & \,\,a+\,\dfrac{a}{2}\,-2a\,\,=\,\,-200\,\text{kJmol}{{\text{e}}^{\text{-1}}} \\\ & -\dfrac{a}{2}\,\,\,\,\,=\,\,-200\,\text{kJmol}{{\text{e}}^{\text{-1}}} \\\ & a\,\,\,\,\,=\,\,\,\,400\,\text{kJmol}{{\text{e}}^{\text{-1}}} \end{aligned}
The bond dissociation energy of A2{{\text{A}}_{2}} will be 400 kJmol-1\text{400 kJmo}{{\text{l}}^{\text{-1}}}

So, option (D) will be the correct answer.

Note:
Bond dissociation energy will be higher if the bond will be stronger. Bond dissociation is an endothermic process (a process which requires external energy), so it will be represented by positive signs.
The formation reaction could be exothermic or endothermic. If the enthalpy of formation of a reaction is an exothermic (energy releasing) process, it is represented by a negative sign.
 !!Δ!! Hf\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{\text{f}}} Data can be used to compare the stability of isomers and allotropes.