Question

Use the thermodynamic data provided below to estimate the boiling point (in K) of ICl. Report your answer to zero decimal places in standard notation (i.e. 123. kJ *For numbers ending in zero, be sure to include the decimal!*).

Substance		ΔH°f (kJ/mol)		S° (J mol-1K-1)
ICl (l)		-23.89		135.1
ICl (g)		17.78		247.6

Answer 1

Answer – Given, thermodynamic data of ∆H^of anf S^of and we need to calculated the ∆H^of and ∆S^of for the reaction as follow –

ICl(l) -----> ICl(g)

∆H^of = Sum of the ∆H^of product – sum of the ∆H^of of reactant

          = 17.98 – (-23.89)

          = 41.87 kJ/mol

∆S^of = Sum of the S^of product – sum of the S^of of reactant

          = 247.6 – 135.1

          = 112.5 J/mol

          = 0.1125 kJ/mol. K

We know the formula

T = ∆H^of / ∆S^of

   = 41.87 kJ.mol^-1 / 0.1125 kJ.mol^-1.K^-1

   = 372 K

the boiling point (in K) of ICl is 372 K