Given the following thermodynamic data, determine K at 298K for the reaction of NO3 and NO2 to produce N2O5.

Question

NO₃(g) + NO₂(g) ⇌ N₂O₅(g)                       

∆H°_Rxn     = -45.2 kJ

∆S°_Rxn      = -75.5 J/K

∆G°_Rxn,298  = -32.7 kJ

Answer 1

We can use the expression ∆Gº = -RT lnK to answer this problem.

∆G = -32.7 kJ (given, but see below)

R = 8.314 J/Kmol = 0.008314 kJ/Kmol (note the change to get units to agree with those for ∆G)

T = 298K

Solving for K we have...

-32.7 kJ = -(0.008314 kJ/Kmol)(298K) ln K

ln K = 13.20

K = 5.4x10⁵

The value of ∆G is given as -32.7 kJ, but if we calculate it from the data provided, it turns out to be -27.7 kJ.

∆G = ∆H - T∆S

∆G = -45.2 kJ - (298K)(-0.0755 kJ/K)

∆G = -45.2 kJ + 22.5 kJ

∆G = -22.7 kJ

If you use this value of ∆G, you can re-calculate K