Question

${\text{AB, }}{{\text{A}}_2}$ and ${{\text{B}}_{\text{2}}}$ are diatomic molecules. If the bond enthalpies of ${{\text{A}}_2}{\text{, AB}}$ and ${{\text{B}}_{\text{2}}}$ are in the ratio $1:1:0.5$ and enthalpy of formation of ${\text{AB}}$ from ${{\text{A}}_{\text{2}}}$ and ${{\text{B}}_{\text{2}}}$ is $- 100{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$ . What is the bond energy of ${{\text{A}}_{\text{2}}}$ ?
A.$200{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
B.$100{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
C.$300{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
D.$400{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$

Answer 1

Bond enthalpy or bond energy is used to define the average enthalpies required to dissociate the bond present in different gaseous compounds into free atoms in the gaseous state. For example, in methane, all the 4 carbon – hydrogen bonds are identical in bond energy and bond length. But the energies required to break up the individual carbon – hydrogen bonds in each successive step differ. Therefore, in such a case, the mean bond energy or mean bond enthalpy of the carbon – hydrogen bond is used.
-The standard enthalpy change for the formation of one mole of a compound from its element in their most stable states of aggregation or reference states is termed as ‘standard enthalpy of formation’. The reference state of an element is 1 bar pressure and a temperature of 25 degree Celsius.

Complete step by step answer:
-Given that ${\text{AB, }}{{\text{A}}_2}$ and ${{\text{B}}_{\text{2}}}$ are diatomic molecules and the bond enthalpies of ${{\text{A}}_2}{\text{, AB}}$ and ${{\text{B}}_{\text{2}}}$ are in the ratio $1:1:0.5$ .
-Also given that the enthalpy of formation of ${\text{AB}}$ from ${{\text{A}}_{\text{2}}}$ and ${{\text{B}}_{\text{2}}}$ is $- 100{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$ .
-We need to find out the bond energy of ${{\text{A}}_{\text{2}}}$.
-Let the value of bond energy of ${{\text{A}}_{\text{2}}}$ be x.
-Since, the bond enthalpies or energies f ${{\text{A}}_2}{\text{, AB}}$ and ${{\text{B}}_{\text{2}}}$ are in the ratio $1:1:0.5$ , so, the value of bond energy of ${\text{AB}}$ will also be x and the bond energy of ${{\text{B}}_{\text{2}}}$ will be $\dfrac{{\text{x}}}{{\text{2}}}$ .
-Since enthalpy of formation is for one mole of compound formed, therefore the reaction of formation of ${\text{AB}}$ from ${{\text{A}}_{\text{2}}}$ and ${{\text{B}}_{\text{2}}}$ can be written as:
$\dfrac{1}{2}{{\text{A}}_{\text{2}}} + \dfrac{1}{2}{{\text{B}}_{\text{2}}} \to {\text{AB}}$
The value of enthalpy $- 100{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$ is for this reaction.
If B.E. denotes bond energy, then from this equation we can write
$\dfrac{1}{2}{\text{B}}{\text{.E}}{{\text{.}}_{{{\text{A}}_{\text{2}}}}} + \dfrac{1}{2}{\text{B}}{\text{.E}}{{\text{.}}_{{{\text{B}}_{\text{2}}}}} - {\text{B}}{\text{.E}}{{\text{.}}_{{\text{AB}}}} = - 100$
Substitute all the values.
$\dfrac{{\text{x}}}{{\text{2}}} + \dfrac{{\text{x}}}{4} - {\text{x = - 100}} \\\ \Rightarrow \dfrac{{{\text{2x + x - 4x}}}}{4} = - 100 \\\ \Rightarrow \dfrac{{ - {\text{x}}}}{4} = - 100 \\\ \Rightarrow {\text{x = 400}} \\\$
Therefore, the value of bond energy of ${{\text{A}}_{\text{2}}}$ is $400{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$ and

Hence the correct option is D.

Note:
-When a bond is strong, it takes a very large amount of energy to break it which means that a very high bond energy is present in a strong bond. This correlates with the bond order and the bond length of the bond.
-If bond order is higher, then the bond length is shorter and shorter bond length means a high bond energy due to increased electric attraction. And if the bond length is longer, then it will mean a smaller bond energy.