Question

Why is ∆Hf ° 62.4 kJ/mol for I2 (
g., but 0.0 kJ/mol for I2 (s)?

Answer 1

Opposite the ordinal equation, the reaction can become energy-absorbing,
 N2 (g) + four liquid (l) ---> a pair of NO2 (g) + four H2 (g),

delta H = 212.5 kJ/mol a pair of NH3 (g) ----------------> N2 (g) + three H2 (g)
 delta H = 100 and fifteen kJ/mol transfer those a pair of equations at constant time.

You get the equation needed and also the succeeding delta H is that the total of the several 
delta H: 4H2O (l) + 2NH3 (g) ------> 2NO2 (g) + 7H2 (g) , delta H = 327.5 kJ/mo

Answer 2

Answer: Because enthalpy of the substances in their standard state is always 0.

Explanation:

Standard enthalpy of formation or standard heat of formation is defined as the change in enthalpy during the formation of 1 mole of a substance from its constituent elements when all the substances are present in their standard states.

Iodine is the 53rd element which is present in Group 17 and Period 5 of the period table.

Standard enthalpy formation of the substances present in their standard state is always equal to 0.

As, iodine occurs in solid state only. So, its standard formation enthalpy will always be 0 in solid state.