1. Meaning of bond enthalpy of $Br_2$

Bond enthalpy of $Br_2$ is the enthalpy required to break $Br_2(g) \rightarrow 2Br(g)$ If this is $y\,\text{kJ mol}^{-1}$, then $\Delta H = y$

2. Meaning of enthalpy of atomisation of $Br_2(l)$

Enthalpy of atomisation means converting 1 mole of substance in its standard state into gaseous atoms.

For bromine, standard state is liquid, so $Br_2(l) \rightarrow 2Br(g)$ Let this be $x\,\text{kJ mol}^{-1}$.

3. Relating the two processes

To convert $Br_2(l)$ into $2Br(g)$, we can imagine two steps:

• First vaporise bromine: $Br_2(l) \rightarrow Br_2(g) \quad (\Delta H_{vap} > 0)$
• Then break the bond: $Br_2(g) \rightarrow 2Br(g) \quad (\Delta H = y)$

Therefore, $x = \Delta H_{vap} + y$

Since $\Delta H_{vap} > 0$ we get $x > y$

4. Checking options

• A: does not exist — false
• B: $x<y$ — false
• C: $x>y$ — true
• D: $x=y$ — false

Hence, the correct option is C.