Question

NH₄NO₃(aq)N₂O(g) + 2H₂O(l)


Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15K.

ANSWER:

Answer 1

Given:

Gof(NH4NO3(aq)) = -190.56 KJ/mol

Gof(N2O(g)) = 104.2 KJ/mol

Gof(H2O(l)) = -237.129 KJ/mol

Balanced chemical equation is:

NH4NO3(aq) ---> N2O(g) + 2 H2O(l)

ΔGo rxn = 1*Gof(N2O(g)) + 2*Gof(H2O(l)) - 1*Gof( NH4NO3(aq))

ΔGo rxn = 1*(104.2) + 2*(-237.129) - 1*(-190.56)

ΔGo rxn = -179.498 KJ

We have:

T = 298.15 K

ΔGo = -179.498 KJ

ΔGo = -179498 J

use:

ΔGo = -R*T*ln Kc

-179498 = - 8.314*298.15* ln(Kc)

ln Kc = 72.4127

Kc = 2.81*10^31

Answer: 2.81*10^31