Question

Aluminum nitride (AlN), is a wide band-gap semiconductor material and has use in opto- electronics. It can be formed by reacting metallic aluminum with pure, gaseous nitrogen (N2).

(a) Write the chemical reaction for the formation of one mole of Aluminum Nitride.

(b) What is the change in enthalpy (∆H) if all the reactants and products are at 298 K and 1 atm pressure?

(c) What is the change in entropy (∆S) for the reaction if all the reactants and products are at 298 K and 1 am pressure?

(d) Repeat parts (b) and (c) when the reactants and products are at 298 K and 10 atm pressure.

Data for this problem: ∆H₂₉₈ (AlN) = −76,470 cal/mol.

Answer 1

a) Al + 0.5N2 ----- AlN (Balanced Reaction for the formation of one mole of Aluminium Nitirde)

b) Enthalpy of the reaction - Enthalpy(AlN) -Enthalpy(Al) - 0.5 * Enthalpy of N2

=> -76.470 Kcal/mol

since the enthalpy of Al(s) and enthalpy of N2 is equal to zero

c) The value of entropy must be present for AlN, the data not give

Entropy of the reaction = Entropy of AlN - Entropy of Al(s) -0.5*Entropy of N2

Entropy of N2(g) = 191.5 J/K

Entropy of Al(s) = 28.3 J/K

substitute this data