Question

If the enthalpy of atomization for Br₂(l) is x kJ/mol and the bond enthalpy for Br₂ is y kJ/mol, the relation between them:

• A. is x > y
• B. is x < y
• C. is x = y
• D. does not exist

Answer 1

The correct option is A

is x > y

Enthalpy of atomization, $\mathrm{\Delta }{}_{\mathrm{a}}\mathrm{H}^0$, is the change in enthalpy when one mole of bonds is completely broken to obtain atoms in the gas phase.

The explanation for the correct option:

${\mathrm{Br}}_2\left(\mathrm{l}\right)\stackrel{∆{\mathrm{H}}_{\mathrm{vap}}}{\to }{\mathrm{Br}}_2\left(\mathrm{g}\right)\stackrel{∆{\mathrm{H}}_{\mathrm{BE}}}{\to }2\mathrm{Br}\left(\mathrm{g}\right)$

Hence, $∆{\mathrm{H}}_{\mathrm{atom}}=∆{\mathrm{H}}_{\mathrm{vap}}+∆{\mathrm{H}}_{\mathrm{BE}}$ [vap= vapourisation, BE= Bond enthalpy]

A. x > y

Now,According to question

$$\begin{gathered} \mathrm{Enthalpy}\mathrm{of}\mathrm{atomization}=\mathrm{X}\raisebox{1ex}{$\mathrm{KJ}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \\ \mathrm{Bond}\mathrm{dissociation}\mathrm{enthalpy}\mathrm{of}\mathrm{bromine}=\mathrm{Y}\raisebox{1ex}{$\mathrm{KJ}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \end{gathered}$$

So by Hess law

So, $\mathrm{x}>\mathrm{y}$

The explanation for the incorrect option:

• B. x < y: This is not the correct option as according to the Enthalpy of atomization, $∆{\mathrm{H}}_{\mathrm{atom}}=∆{\mathrm{H}}_{\mathrm{vap}}+∆{\mathrm{H}}_{\mathrm{BE}}$, So, it y cannot be greater than x.
• C. x = y: This is not the correct option as according to the Enthalpy of atomization, $∆{\mathrm{H}}_{\mathrm{atom}}=∆{\mathrm{H}}_{\mathrm{vap}}+∆{\mathrm{H}}_{\mathrm{BE}}$, So, x is not equal to y.
• D. Does not exist- This is the incorrect option as according to the Enthalpy of atomization, $∆{\mathrm{H}}_{\mathrm{atom}}=∆{\mathrm{H}}_{\mathrm{vap}}+∆{\mathrm{H}}_{\mathrm{BE}}$,

Hence, A is the correct option.