The standard enthalpy of formation of Mg ²⁺ (aq) from Mg(s) is -460 kJ mol ⁻¹ . Given the following thermodynamic data: Enthalpy of sublimation of Mg(s) = 150 kJ mol ⁻¹ First ionization enthalpy of Mg(g) = 740 kJ mol ⁻¹ Second ionization enthalpy of Mg(g) = 1450 kJ mol ⁻¹ The magnitude of the hydration enthalpy of Mg ²⁺ (g) is ________ kJ mol ⁻¹ . (Nearest integer)

Step-by-step solution

The formation of Mg ²⁺ (aq) from Mg(s) can be represented by the following thermodynamic cycle: Mg(s) Mg(g) Mg ^+ (g) Mg ²⁺ (g) Mg ²⁺ (aq) According to Hess's law, the total enthalpy change is the sum of the enthalpies of the individual steps: _ f H ^ = _ sub H ^ + IE ₁ + IE ₂ + _ hyd H ^ Substituting the given values: -460 = 150 + 740 + 1450 + _ hyd H ^ -460 = 2340 + _ hyd H ^ _ hyd H ^ = -460 - 2340 = -2800 kJ mol ⁻¹ The magnitude of the hydration enthalpy is 2800 kJ mol ⁻¹ . Answer: 2800