Question

At the normal boiling temperature of iron, TB = 3330 K, the rate of change of the vapor pressure of liquid iron with temperature is 3.72 x 10-3 atm/K. Calculate the molar latent enthalpy of boiling of iron at 3330 K:

Answer 1

The molar latent enthalpy of boiling of iron at 3330 K is  ΔH = 342 $\times$ 10^3 J.

Explanation:

Molar enthalpy of fusion is the amount of energy needed to change one mole of a substance from the solid phase to the liquid phase at constant temperature and pressure.

                      d ln p = (ΔH / RT^2) dt

                   (1/p) dp = (ΔH / RT^2) dt

                    dp / dt = p (ΔH / RT^2) = 3.72 $\times$ 10^-3

                  (p) (ΔH) / (8.31) (3330)^2 = 3.72 $\times$ 10^-3

                          ΔH = 342 $\times$ 10^3 J.

Answer 2

H = 102,94