Question

The ΔH∘f of TiI3(s) is -328 kJ/mol and the ΔH∘ for the reaction 2Ti(s)+3I2(g)→2TiI3(s) is -839 kJ Calculate the ΔH of sublimation of I2(s), which is a solid at 25 ∘C.

Answer 1

The enthalpy of sublimation (ΔHsub) is the energy required to convert one mole of a substance from the solid to the gaseous state. It can be estimated as the sum of the enthalpies of fusion and vaporization. To calculate the ΔHsub of I2, we first calculate the ΔHf of I2 using the given ΔHf of TiI3 and ΔHr for the reaction. Then, we can calculate ΔHsub of I2 using the equation ΔHsub = ΔHf + ΔHvap.

Step-by-step explanation:

The enthalpy of sublimation (ΔHsub) is the energy required to convert one mole of a substance from the solid to the gaseous state. It can be estimated as the sum of the enthalpies of fusion and vaporization. In this case, we can use the enthalpy of formation (ΔHf) of TiI3 to calculate the ΔHsub of I2.

We are given the ΔHf of TiI3(s) as -328 kJ/mol and the ΔHr for the reaction 2Ti(s) + 3I2(g) → 2TiI3(s) as -839 kJ/mol. To calculate the ΔHsub of I2(s), we need to first calculate the ΔHf of I2(s). Since the reaction involves 3 moles of I2, we can use the equation:

ΔHf = ΔHr / (moles of product)

By substituting the given values, we can solve for ΔHf of I2:

ΔHf = -839 kJ/mol / (2 moles x 3)

ΔHf = -139.83 kJ/mol

Finally, we can calculate ΔHsub of I2 using the equation:

ΔHsub = ΔHf + ΔHvap