If enthalpy of atomisation for Br2(l) is x kJ/mol and bond enthalpy for Br2 is y kJ/mol, the relatio

Q: Why is bond enthalpy only defined for gases?

If we measured bond enthalpy in a liquid or solid, the energy required would also include the energy to overcome intermolecular forces (like hydrogen bonding or van der Waals forces). By defining it only for gases, we isolate the pure energy of the covalent bond itself.

Q: Is vaporisation always endothermic?

Yes. To turn a liquid into a gas, you must pull the molecules apart against their attractive intermolecular forces. This always requires an input of energy, making $\Delta H_{\text{vap}}$ positive.

Q: What if the question asked about Iodine?

Iodine is a solid at room temperature. The logic remains the same, but instead of vaporisation, you would have the enthalpy of sublimation ($\Delta H_{\text{sub}}$). The atomisation energy would still be greater than the bond enthalpy.

Q: When would $x$ be equal to $y$?

$x$ equals $y$ only when the element is already a gas in its standard state at room temperature, such as $\text{H}_2$, $\text{N}_2$, $\text{O}_2$, $\text{F}_2$, or $\text{Cl}_2$.

Q: Can bond enthalpy ever be negative?

No. Breaking a stable chemical bond always requires an input of energy. Therefore, bond dissociation enthalpies are always positive (endothermic).

Question

If enthalpy of atomisation for $	ext{Br}_2(l)$ is $x 	ext{ kJ/mol}$ and bond enthalpy for $	ext{Br}_2$ is $y 	ext{ kJ/mol}$, the relation between them

Options:

(A) is $x < y$

(B) does not exist

(C) is $x > y$

(D) is $x = y$

Sigma Percentile · Accepted Answer

### The Subtle Difference Between Atomisation and Bond Enthalpy

When we study thermochemistry, it is incredibly easy to confuse the terms **Enthalpy of Atomisation** and **Bond Enthalpy**. At first glance, they both seem to describe the exact same process: breaking a molecule apart into its constituent atoms. However, there is a profound physical distinction hidden in their standard definitions, and this question tests exactly that.

### Analyzing the Setup

Let's visualize the physical reality of Bromine. At room temperature, Bromine is a heavy, reddish-brown liquid, denoted as $	ext{Br}_2(l)$.

The **Enthalpy of Atomisation** ($x$) is defined as the total energy required to completely break one mole of a substance in its standard state into gaseous atoms. For liquid bromine, this means taking $	ext{Br}_2(l)$ and ripping it apart into $2	ext{Br}(g)$.

On the other hand, **Bond Enthalpy** ($y$) is strictly defined for breaking bonds in *gaseous* molecules. It is the energy required to take $	ext{Br}_2(g)$ and break it into $2	ext{Br}(g)$.

### The Master Equation: Hess's Law

To understand the relationship between $x$ and $y$, we can construct a thermodynamic cycle using **Hess's Law**. Instead of jumping directly from liquid bromine to gaseous atoms, let's take a two-step detour:

1. **Vaporisation:** First, we must boil the liquid bromine to turn it into a gas. This requires overcoming the intermolecular van der Waals forces. 
   $$	ext{Br}_2(l) \longrightarrow 	ext{Br}_2(g) \quad (\Delta H_{	ext{vap}} > 0)$$

2. **Bond Dissociation:** Now that we have gaseous bromine molecules, we can break the covalent $	ext{Br}-	ext{Br}$ bonds. This is our bond enthalpy, $y$.
   $$	ext{Br}_2(g) \longrightarrow 2	ext{Br}(g) \quad (\Delta H = y)$$

According to Hess's Law, the total energy of the direct path (atomisation, $x$) must equal the sum of the energies of our two-step path:

$$x = \Delta H_{	ext{vap}} + y$$

### Final Calculation

Because vaporising a liquid always requires an input of energy (it is an endothermic process), $\Delta H_{	ext{vap}}$ is strictly a positive number.

Therefore, if we add a positive number to $y$ to get $x$, it mathematically guarantees that:

**$$x > y$$**

The enthalpy of atomisation for a liquid will always be greater than its bond enthalpy because you have to pay the "energy tax" of vaporisation first. If the question had asked about Chlorine ($	ext{Cl}_2$), which is already a gas at room temperature, then the vaporisation step wouldn't exist, and $x$ would exactly equal $y$.

Animated Solution for Chemistry - Chemical Thermodynamics: If enthalpy of atomisation for $Br_{2} (l)$ is $x kJ/mol$ and bond enthalpy for $Br_{2}$ is $y kJ/mol$ , the relation between them

Select Answer:

Visualized Solution

\text{Initial State of Bromine}

\text{Enthalpy of Atomisation } (x)

\text{Bond Enthalpy } (y)

\text{Hess's Law Cycle}

\text{Relating } x \text{ and } y

\text{Final Conclusion}

\text{The Way Forward}

The Sigma Insight: Enthalpy and Hess's Law

The Subtle Difference Between Atomisation and Bond Enthalpy

Analyzing the Setup

The Master Equation

Final Calculation

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