Question

Integral enthalpy of solution of \[KCI\], when \[1{\text{ }}mole\] of it is dissolved in \[20{\text{ }}mole\] water is \[ + 15.90{\text{ }}kJ\]. When \[1{\text{ }}mole\] is dissolved in \[200{\text{ }}mol\] water. Delta H is \[18.58{\text{ }}kJ\]. Calculate enthalpy of dilution (or hydration).

Answer 1

Hint: We need to understand the concept of enthalpy of dilution or dilution and learn to calculate its value using certain given parameters. The enthalpy shift associated with the dilution phase of a product in a solution at constant pressure is referred to as the heat of dilution, or enthalpy of dilution. The dilution process is equal to the dissolution process if the initial state of the component is a pure liquid (assuming the solution is liquid), and the heat of dilution is the same as the heat of solution.

Complete answer:
The heat of dilution is usually normalised by the solution's mole number, and its dimensional units are energy per unit mass or volume of material, which is generally expressed in \[kJ/mol{\text{ }}\left( {or{\text{ }}J/mol} \right).\] When a solution containing 1 mole of solute is diluted from one concentration to another, the enthalpy of hydration or dilution changes. The change in enthalpy of hydration or dilution is given as:
\[\Delta H\] enthalpy = \[18.58{\text{ }} - {\text{ }}15.90\]
∴ \[\Delta H\] enthalpy = $2.68kJ$

Note:
We must note that the differential and integral heats of dilution are two ways to describe the heat of dilution. On a micro scale, the differential heat of dilution is correlated with the process of adding a small volume of liquid to a large quantity of solution. The enthalpy change caused by adding a mole of solvent to a very large amount of solution at a constant temperature and pressure is thus known as the molar differential heat of dilution. The integral heat of dilution, on the other hand, is considered on a larger scale. The molar integral heat of dilution is defined as the enthalpy change in this phase, normalised by the mole number of solute.