Question

Find deltaHr, deltaSr, and deltaGr for 2NO2 --> N2O4 at 298K. Find deltaUr as well.

Answer 1

For above reaction, 2NO₂(g)​N₂O₄(g)

ENTHALPY OF REACTION

ΔH_reaction= [1ΔH_f(N2O4 (g))] -[2ΔH_f(NO2 (g))]

[1(9.16)] -[2(33.18)] = -57.2 kJ
-57.20 kJ     (exothermic)

ENTROPY CHANGE

ΔS_reaction=ΔS_f,product -ΔS_f,reactant

[1ΔS_f(N2O4 (g))] - [2ΔS_f(NO2 (g))]
[1(304.18)] - [2(239.95)] = -175.72 J/K
-175.72 J/K     (decrease in entropy)

FREE ENERGY OF REACTION (AT 298.15 K)

From ΔGf° values:

ΔG_reaction= ΔG_f,product  - ΔG_f,reactant
[1ΔG_f(N2O4 (g))] - [2ΔG_f(NO2 (g))]
[1(97.82)] - [2(51.3)] = -4.78 kJ
-4.78 kJ (spontaneous due to negative value)

For ΔU_reaction, we have to use following formula-

ΔU= PΔV - ΔH

For constant volume,ΔV = 0,

ΔU=- ΔH= 57.2 KJ

Hope, you get it what you want.