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.

Explanation:

Hello,

In this case, for the given dissociation reaction, we can compute the enthalpy of reaction considering the enthalpy of formation of each involved species (products minus reactants):

Next, the entropy of reaction considering the standard entropy for each involved species (products minus reactants):

Next, since the Gibbs free energy of reaction is computed in terms of both the enthalpy and entropy of reaction at the given temperature (298.15 K), we finally obtain (two significant figures):

Best regards.

the answer for sodium is 0.186

the answer for rubidium is 0.251