Question

Calculate standard enthalpy of solution of KCl in large excess of water at 298 K. Given that standard enthalpy of formation of K+(aq), Cl⁻(aq) and KCl(s) are - 251.2 kJ mol⁻¹, - 167.08 kJ mol⁻¹ and - 437.6 kJ mol⁻¹ respectively.​

Answer 1

The standard enthalpy of solution of KCl at 298 K is calculated using the standard enthalpies of formation for KCl(s), K+(aq), and Cl−(aq), resulting in an enthalpy of solution of 19.32 kJ/mol.

Step-by-step explanation:

The standard enthalpy of solution for KCl can be calculated using the standard enthalpies of formation for KCl(s), K+(aq), and Cl−(aq). To find the enthalpy of solution, which is the change in enthalpy when one mole of a substance dissolves in water, we use the formula:

ΔHsol = [ΣΔHf(products)] − [ΣΔHf(reactants)]

In this case, the dissolution reaction is:

KCl(s) → K+(aq) + Cl−(aq)

Applying the given standard enthalpies of formation:

ΔHsol = [ΔHf(K+(aq)) + ΔHf(Cl−(aq))] − [ΔHf(KCl(s))]

ΔHsol = [(−251.2 kJ/mol) + (−167.08 kJ/mol)] − [−437.6 kJ/mol]

ΔHsol = −418.28 kJ/mol + 437.6 kJ/mol

ΔHsol = 19.32 kJ/mol

Therefore, the standard enthalpy of solution of KCl in water at 298 K is 19.32 kJ/mol.