At 25°C, the standard enthalpy of formation of KCl(s)is -435.87 kJ/mol. When one mole of KCl(s)is formed by reacting potassium vapor and chlorine gas at 25°C, the standard enthalpy of reaction is -525.86 kJ/mol. Find $\Delta$H° for the sublimation of potassium, K(s)$\rarr$K(g), at 25°C.

Question

Quizplus · Accepted Answer

The enthalpy change for the sublimation of potassium can be found using Hess's law. The enthalpy of formation of KCl(s) from K(g) and Cl2(g) is -525.86 kJ/mol, and the enthalpy of formation from K(s) and Cl2(g) is -435.87 kJ/mol. The difference between these values gives the enthalpy of sublimation of potassium: -435.87 kJ/mol - (-525.86 kJ/mol) = 89.99 kJ/mol.

At 25°C, the Standard Enthalpy of Formation of KCl(s)is -435 Δ\DeltaΔ

At 25°C, the Standard Enthalpy of Formation of KCl(s)is -435 $\Delta$