Q.18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol^-1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalphy of vapourisation for water?

Concept: Molar Enthalpy of Vaporisation

The enthalpy of vaporisation is an intensive property — it depends only on the substance and conditions (temperature, pressure), not on the amount. Given the molar enthalpy change, we scale it by the number of moles to find the total enthalpy change for any quantity.

Step 1. The problem states that the enthalpy change is $40.79 \, \text{kJ mol}^{-1}$ at 100°C and 1 bar. This is already the molar enthalpy of vaporisation: $\Delta_{\text{vap}} H = 40.79 \, \text{kJ mol}^{-1}$.

Step 2. For two moles of water vaporising under the same conditions:

$\Delta H = n \times \Delta_{\text{vap}} H = 2 \, \text{mol} \times 40.79 \, \text{kJ mol}^{-1} = 81.58 \, \text{kJ}.$ …

Why enthalpy of vaporisation is intensive

When we talk about the enthalpy change "in the process," we need to distinguish between the total energy absorbed and the energy per mole. The question tells us that vaporising water at $100°\text{C}$ and $1 \, \text{bar}$ requires $40.79 \, \text{kJ mol}^{-1}$. This is already expressed as a molar quantity—an intensive property that characterises the substance itself, not the sample size.

The molar mass of water is $18.0 \, \text{g mol}^{-1}$, so $18.0 \, \text{g}$ corresponds to exactly one mole. The given enthalpy change is therefore the energy needed to convert one mole of liquid water into one mole of water vapour under these conditions.

Working through the two questions

1. Enthalpy change for two moles

Since the molar enthalpy of vaporisation is $\Delta_{\text{vap}} H = 40.79 \, \text{kJ mol}^{-1}$, vaporising $n = 2$ moles requires

The enthalpy change scales linearly with the amount of substance because we are simply repeating the same molecular process twice as many times.

2. Standard enthalpy of vaporisation

The standard enthalpy of vaporisation, $\Delta_{\text{vap}} H^\circ$, is defined as the enthalpy change when one mole of a substance vaporises at a specified temperature and the standard pressure of $1 \, \text{bar}$. …