Question

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.

Answer 1

HNO₂ ⇄ H⁺ + NO₂⁻

ΔG° = 19.102 kJ/mol

Step-by-step 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