Question

HNO2(aq)+H2O(l)↽−−⇀H3O+(aq)+NO−2(aq) is Ka=5.6×10−4 M at 25.0 ∘C . Calculate the value of Δ????∘rxn at 25.0 ∘C for nitrous acid when [NO−2]=[H3O+]=[HNO2]=1.00 M . Δ????∘rxn= 1.8552 kJ/mol Will the acid spontaneously dissociate under these conditions? Calculate the value of Δ????rxn for nitrous acid when [NO−2]=[H3O+]=7.73×10−5 M and [HNO2]=0.663 M .

Answer 1

ΔGrxn = 18,561 kJ/mol. The acid will not spontaneously dissociate.

Under the other conditions: ΔGrxn = -27,361 kJ/mol

Step-by-step explanation:

For the reaction:

HNO₂(aq)+H₂O(l) ⇄ H₃O⁺(aq) + NO₂⁻(aq)

To obtain the ΔG° reaction it is necessary to use:

ΔGrxn = ΔG° + RT ln Q (1)

Where:

ΔG° is 18,561 kJ/mol (Using ΔG° = -RT ln ka)

R is gas constant (8,314472x10⁻³ kJ/molK)

T is temperature (25°C = 298,15K)

And Q is reaction quotient ([NO₂⁻][H₃O⁺]/[HNO₂]) = 1,00

Replacing these values in (1) you will obtain:

ΔGrxn = 18,561 kJ/mol

As ΔG > 0 the reaction will be spontaneous in reverse direction. That means the acid will not spontaneously dissociate.

When molar concentrations of [NO₂⁻] = [H₃O⁺] = 7,73x10⁻⁵M and [HNO₂] is 0,663M the ΔGrxn is calculated also with (1) just changing Q that will be: (7,73x10⁻⁵M)² / 0,663M

Replacing in (1)

ΔGrxn = 18,561 kJ/mol - 45,922 kJ/mol = -27,361 kJ/mol

This reaction will be spontaneous in forward direction. In other words, the acid will spontaneously dissociate.

I hope it helps!