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!

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.