Answer to Question #146953 in General Chemistry for Jenny

Question

Here are thermodynamic data for the fusion of NH3:

NH3(s) → NH3(l)        ΔH° = 5.65 kJ/mol

ΔS° = 28.9 J mol-1 K-1

(a) Calculate ΔG° for the melting of 1.00 mole of NH3 at 298 K.

ΔG° = kJ/mol

(b) Calculate the freezing point of NH3.

T = K

Answer 1

(a) ΔG° = ΔH°– TΔS°

ΔH° = 5.65 kJ/mol

= 5650 J/mol

ΔS° = 28.9 J mol-1 K-1

T = 298 K

So,

ΔG° = (5650 –298×28.9) J/mol

= –2962.1 J/mol

= –2.9621 kJ/mol

So, ΔG° = –2.9621 kJ/mol

(b) given reaction is,

NH3(s) → NH3(l)

This type of transportation occurs at freezing point of NH3 by absorbing latent heat , ∆H°

If the freezing point is = Tf

As,. ∆S° = ∆H°/Tf

Or, Tf = ∆H°/∆S°

Or, Tf

= (5650 J mol-1)/(28.9 J mol-1 K-1)

= 195.5 K

Hence, freezing point of NH3 = 195.5 K