Question

The bond dissociation energies of ${X_2},{Y_2}$ and $XY$ are in the ratio $1:0.5:1$. $\Delta H$ for the formation of $XY$ is $ - 200kJmo{l^{ - 1}}$ . The bond dissociation energy of ${X_2}$ will be:
A.$800kJmo{l^{ - 1}}$
B.$200kJmo{l^{ - 1}}$
C.$400kJmo{l^{ - 1}}$
D.$100kJmo{l^{ - 1}}$

Answer 1

Hint:Bond dissociation energy is the energy required to break a bond into molecular fragments. It is an endothermic reaction. Enthalpy of a reaction is the heat content of the system and change in enthalpy refers to the heat released or absorbed in a reaction at constant pressure. Consider the ratios of bond dissociation energies given and substitute them in enthalpy change formula.
Formula used: $\Delta {H_{(system)}} = \Delta {H_{(reac\tan t)}} - \Delta {H_{(product)}}$

Complete step by step answer:
Bond dissociation energy is the energy required to break a bond into molecular fragments. It is an endothermic reaction as energy is required externally for breaking a bond. It measures the strength in a bond. Greater will be the bond dissociation energy, more stable will the compound be. As the carbon-carbon bonds have high bond dissociation energy it is difficult to break them. Bond dissociation energy is calculated according to the difference in the electronegativity of the atoms and the bond dissociation energy increases if the difference in the electronegativities of the atoms increases.
Enthalpy of a reaction is the heat content of the system and change in enthalpy refers to the heat released or absorbed in a reaction at constant pressure. The change in enthalpy of the reaction or the system is calculated as the difference between the enthalpy change of the reactants and the enthalpy change in the products.
In the above question,
Given
$\Delta H$ of formation of $XY = - 200kJmo{l^{ - 1}}$
Ratio of bond dissociation energies of ${X_2},{Y_2}$ and $XY = 1:0.5:1$
Therefore, bond dissociation energy of ${X_2} = nkJ/mol$
bond dissociation energy of ${Y_2} = 0.5nkJ/mol$
bond dissociation energy of $XY = nkJ/mol$
We know that,
$\Delta {H_{(system)}} = \Delta {H_{(reac\tan t)}} - \Delta {H_{(product)}}$
$\dfrac{1}{2}{X_2} + \dfrac{1}{2}{Y_2} = XY$
$ \Rightarrow \Delta {H_{(reaction)}} = \dfrac{1}{2}\Delta H({X_2}) + \dfrac{1}{2}\Delta H({Y_2}) - \Delta H(XY)$
$ \Rightarrow - 200 = \dfrac{1}{2}(n) + \dfrac{1}{2}(0.5n) - (n)$
$ \Rightarrow - 200 = - 0.25(n)$
$ \Rightarrow n = 800$
Thus, the bond dissociation energy of ${X_2} = 800kJ/mol$

So, the correct option is (A) $800kJ/mol$.

Note:
Electronegativity of an atom is the measure of the ability of an atom to attract shared electrons. It increases as we move from left to right across a period of a periodic table and it decreases as we move down the group. Fluorine has the highest electronegativity which states that it has the highest tendency to gain electrons from another element.