The value of Ka for nitrous acid (HNO2) at 25 ∘C is 4.5*10-4. Part APart complete Write the chemical equation for the equilibrium that corresponds to Ka. H + (aq) + NO-2 (aq) ⇌HNO2 (aq) HNO2 (aq) ⇌H + (aq) + NO-2 (aq) HNO2 (aq) ⇌H - (aq) + NO+2 (aq) HNO2 (aq) + H + (aq) ⇌H2NO+2 (aq) HNO2 (aq) + H - (aq) ⇌H2NO+2 (aq) By using the value of Ka, calculate ΔG∘ for the dissociation of nitrous acid in aqueous solution.

HNO₂ ⇄ H⁺ + NO₂⁻

ΔG° = 19.102 kJ/mol

Explanation:

Dissociation constant, Ka, for a weak acid that has as reaction:

HX ⇄ H⁺ + X⁻

Is defined as:

Ka = [H⁺][X⁻] / [HX]

Thus, for the weak acid, HNO₂, the chemical equation is:

HNO₂ ⇄ H⁺ + NO₂⁻

In a reaction, ΔG° is defined as:

ΔG° = - RT lnK

Where R is gas constant (8.314 J/molK), T is temperature (273.15 + 25°C = 298.15K) and K is equilibrium constant (4.5x10⁻⁴).

Replacing:

ΔG° = - 8.314J/molK*298.15K ln4.5x10⁻⁴

ΔG° = 19102 J/mol

ΔG° = 19.102 kJ/mol