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 for nitrous acid when [no−2]=[h3o+]=7.73×10−5 m and [hno2]=0.663 m .

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

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

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!

catya your not at all which answer i've seen you do this on another question!