Heat Of Solution Formula: The heat of solution, also known as enthalpy of solution, is a thermodynamic parameter that represents the amount of heat energy absorbed or released when a solute dissolves in a solvent to form a solution at constant pressure. It is an important concept in chemistry, particularly in the study of solution thermodynamics.
Heat of Solution Formula:
The heat of solution (ΔHsoln) is calculated using the following formula:
ΔH soln = H final - H initial
Where:
ΔH soln is the heat of solution.
H final is the enthalpy of the solution (the total energy of the system after dissolution).
H initial is the enthalpy of the individual components (solute and solvent) before dissolution.
ΔHsoln = Hfinal - Hinitial
Where ΔHsoln is the heat of solution, Hfinal is the enthalpy of the solution, and Hinitial is the enthalpy of the individual components (solute and solvent) before dissolution.
ΔHsoln per mole = ΔHsoln / moles of solute
This formula calculates the enthalpy change for the dissolution of one mole of solute in the solvent.
Mass of solute dissolved = moles of solute × molar mass of solute
This formula is used to find the mass of solute that has dissolved in the solvent.
Moles of solvent = (Initial moles of solvent) - (moles of solute)
This formula determines the moles of solvent remaining after the solute has dissolved.
Also Check – Escape Speed Formula
ΔHdilution = ΔHsoln - ΔHsolvent
Where ΔHdilution is the heat of dilution, ΔHsoln is the heat of solution, and ΔHsolvent is the heat associated with the dissolution of solute in the solvent.
ΔHhydration = ΔHsoln - ΔHlattice
This formula calculates the heat associated with the solvation of ionic compounds, taking into account the energy required to break the lattice structure of the solid.
ΔHcrystallization = -ΔHsoln
This equation relates the heat of solution to the heat released during the crystallization of a solute from a solution.
These formulas are essential for understanding and calculating the heat changes associated with the dissolution of solutes in solvents, which is a fundamental concept in Heat Of Solution Formula.
Also Check – Amperes Law Formula
Example 1: Dissolution of Sodium Chloride in Water
Let's calculate the heat of solution when 5 grams of sodium chloride (NaCl) dissolve in 100 mL of water. Given that the enthalpy of sodium chloride is -411 kJ/mol and the enthalpy of water is -285.8 kJ/mol.
Moles of NaCl = Mass (g) / Molar mass (g/mol)
Moles of NaCl = 5 g / 58.44 g/mol (molar mass of NaCl)
Moles of NaCl ≈ 0.0857 mol
Moles of water = Volume (mL) / Molar volume (mL/mol)
Moles of water = 100 mL / 18.01528 mL/mol (molar volume of water)
Moles of water ≈ 5.54 mol
ΔHsoln = (Moles of NaCl HNaCl) + (Moles of water Hwater) - Hinitial
ΔHsoln = (0.0857 mol -411 kJ/mol) + (5.54 mol -285.8 kJ/mol) - 0 kJ (Initial enthalpy of individual components)
ΔHsoln ≈ -233.59 kJ
So, the heat of solution for dissolving 5 grams of sodium chloride in 100 mL of water is approximately -233.59 kJ.
Also Check – Terminal Velocity Formula
Example 2: Dissolution of Sugar in Water
Calculate the heat of solution when 10 grams of sugar (sucrose, C12H22O11) dissolve in 200 mL of water. Given that the enthalpy of sugar is 0 kJ/mol and the enthalpy of water is -285.8 kJ/mol.
Moles of sugar = Mass (g) / Molar mass (g/mol)
Moles of sugar = 10 g / 342.3 g/mol (molar mass of sugar)
Moles of sugar ≈ 0.0292 mol
ΔHsoln = (Moles of sugar Hsugar) + (Moles of water Hwater) - Hinitial
ΔHsoln = (0.0292 mol 0 kJ/mol) + (5.54 mol -285.8 kJ/mol) - 0 kJ
ΔHsoln ≈ -1585.33 kJ
The heat of solution for dissolving 10 grams of sugar in 200 mL of water is approximately -1585.33 kJ.