At 298 K, the enthalpy of fusion of a solid (X) is 2.8 kJ mol^-1 and the enthalpy of vaporisation of

Q: Why can we just add the enthalpies together like simple numbers?

Because enthalpy is a 'state function'. This means the total change in enthalpy only depends on the initial state (solid) and the final state (gas), not the path taken. Going directly (sublimation) takes the exact same energy as going via the liquid phase (fusion + vaporisation).

Q: Does the temperature of 298 K matter for the calculation?

The temperature is important context because enthalpy values change with temperature. However, since all the given values and the requested value are at the same temperature (298 K), we don't need to perform any temperature corrections (like using Kirchhoff's equations).

Q: What if the question asked for the enthalpy of condensation?

Condensation is the reverse of vaporisation (gas to liquid). The enthalpy of condensation would simply be the negative of the enthalpy of vaporisation, so it would be $-98.2 \text{ kJ mol}^{-1}$.

Q: Why is the enthalpy of vaporisation so much larger than the enthalpy of fusion?

Melting a solid (fusion) only requires partially breaking the intermolecular forces to allow molecules to slide past each other. Boiling a liquid (vaporisation) requires completely overcoming all intermolecular forces to separate the molecules into a gas, which takes significantly more energy.

Q: Is sublimation always an endothermic process?

Yes! Sublimation involves breaking intermolecular bonds to go from a highly ordered solid to a highly disordered gas. Breaking bonds always requires an input of energy, making $\Delta H$ positive (endothermic).

Question

At $298	ext{ K}$, the enthalpy of fusion of a solid (X) is $2.8	ext{ kJ mol}^{-1}$ and the enthalpy of vaporisation of the liquid (X) is $98.2	ext{ kJ mol}^{-1}$. The enthalpy of sublimation of the substance (X) in $	ext{kJ mol}^{-1}$ is ...... (Nearest integer)

Sigma Percentile · Accepted Answer

### The Magic of State Functions

Imagine you are standing at the base of a mountain and you want to reach the peak. You could take a long, winding trail that goes up and down before finally reaching the top, or you could take a helicopter and fly straight up. Regardless of how you get there, your final altitude is exactly the same.

In thermodynamics, properties that behave like your altitude are called **state functions**. They only care about where you start and where you finish, completely ignoring the path you took to get there. Enthalpy, denoted by $H$, is one of the most important state functions in chemistry.

### Breaking Down the Phase Changes

In this problem, we are dealing with a substance $X$ that can exist as a solid, a liquid, or a gas. We are given two pieces of information:

1.  **Enthalpy of Fusion ($\Delta H_{	ext{fus}}$):** The heat required to melt the solid into a liquid. For substance $X$, this is $2.8 	ext{ kJ mol}^{-1}$.
2.  **Enthalpy of Vaporisation ($\Delta H_{	ext{vap}}$):** The heat required to boil the liquid into a gas. For substance $X$, this is $98.2 	ext{ kJ mol}^{-1}$.

But what if we want to skip the liquid phase entirely? The direct transition from a solid to a gas is called **sublimation**, and the heat required for this is the **Enthalpy of Sublimation ($\Delta H_{	ext{sub}}$)**.

### Applying Hess's Law

Because enthalpy is a state function, we can use a powerful tool called **Hess's Law**. Hess's Law states that the total enthalpy change for a chemical reaction or physical process is the same, regardless of whether it occurs in one step or multiple steps.

Think of it like this: going directly from Solid $ightarrow$ Gas must require the exact same amount of energy as going from Solid $ightarrow$ Liquid $ightarrow$ Gas.

Mathematically, we can express this beautiful relationship as:

$$\Delta H_{	ext{sub}} = \Delta H_{	ext{fus}} + \Delta H_{	ext{vap}}$$

### The Final Calculation

Now, the problem becomes incredibly simple. We just need to substitute the values given to us into our master equation.

$$\Delta H_{	ext{sub}} = 2.8 	ext{ kJ mol}^{-1} + 98.2 	ext{ kJ mol}^{-1}$$

Adding these together, we get:

$$\Delta H_{	ext{sub}} = 101 	ext{ kJ mol}^{-1}$$

And there we have it! The enthalpy of sublimation is exactly **$101$**. This elegant problem reminds us that in the world of thermodynamics, complex journeys can always be broken down into simpler, manageable steps.

Enter Numerical Value:

Visualized Solution

$Visualizing the Phase Transitions$

$Hess’s Law and Sublimation$

$Formulating the Equation$

$Substituting the Values$

$Final Calculation$

$The Power of State Functions$

The Sigma Insight: Enthalpy and Hess's Law

The Magic of State Functions

Breaking Down the Phase Changes

Applying Hess's Law

The Final Calculation

Similar Questions

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If the bond dissociation energies of $X Y$ , $X_{2}$ and $Y_{2}$ (all diatomic molecules) are in the ratio of $1 : 1 : 0.5$ and $Δ H_{f}$ for the formation of $X Y$ is $- 200 kJ mol^{- 1}$ . The bond dissociation energy of $X_{2}$ will be

Lattice enthalpy and enthalpy of solution of NaCl are $788 kJ mol^{- 1}$ and $4 kJ mol^{- 1}$ , respectively. The hydration enthalpy of NaCl is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Phase Transitions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hess’s Law and Sublimation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Formulating the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting the Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Power of State Functions