Question

Consider the following chemical equation:

NH4NO3(s)⟶NH+4(aq)+NO−3(aq)

What is the standard change in free energy in  kJmol at 298.15K?

The heat of formation data are as follows:

ΔH∘f,NH4NO3(s)=-365.6kJmolΔH∘f,NH+4(aq)=-132.5kJmolΔH∘f,NO−3(aq)=-205.0kJmol

The standard entropy data are as follows:

S∘NH4NO3(s)=151.1Jmol KS∘NH+4(aq)=113.4Jmol KS∘NO−3(aq)=146.4Jmol K

Your answer should include two significant figures.

Answer 1

Given:

Gof(NH4NO3(s)) = 51.3 KJ/mol

Gof(NH4+(aq)) = -110.9 KJ/mol

Gof(NO3-(aq)) = 87.6 KJ/mol

Balanced chemical equation is:

NH4NO3(s) ---> NH4+(aq) + NO3-(aq)

ΔGo rxn = 1*Gof(NH4+(aq)) + 1*Gof(NO3-(aq)) - 1*Gof( NH4NO3(s))

ΔGo rxn = 1*(-110.9) + 1*(87.6) - 1*(51.3)

ΔGo rxn = -74.6 KJ

Given:

Hof(NH4NO3(s)) = -365.6 KJ/mol

Hof(NH4+(aq)) = -132.5 KJ/mol

Hof(NO3-(aq)) = -205.0 KJ/mol

Balanced chemical equation is:

NH4NO3(s) ---> NH4+(aq) + NO3-(aq)

ΔHo rxn = 1*Hof(NH4+(aq)) + 1*Hof(NO3-(aq)) - 1*Hof( NH4NO3(s))

ΔHo rxn = 1*(-132.5) + 1*(-205.0) - 1*(-365.6)

ΔHo rxn = 28.1 KJ

Given:

Sof(NH4NO3(s)) = 151.1 J/mol.K

Sof(NH4+(aq)) = 113.4 J/mol.K

Sof(NO3-(aq)) = 146.4 J/mol.K

Balanced chemical equation is:

NH4NO3(s) ---> NH4+(aq) + NO3-(aq)

ΔSo rxn = 1*Sof(NH4+(aq)) + 1*Sof(NO3-(aq)) - 1*Sof( NH4NO3(s))

ΔSo rxn = 1*(113.4) + 1*(146.4) - 1*(151.1)

ΔSo rxn = 108.7 J/K

Now we have:

ΔHo = 28.1 KJ/mol

ΔSo = 108.7 J/mol.K

= 0.1087 KJ/mol.K

T = 298.15 K

use:

ΔGo = ΔHo - T*ΔSo

ΔGo = 28.1 - 298.15 * 0.1087

ΔGo = -4.3089 KJ/mol

Answer: -4.3 KJ/mol