The value of Ka for nitrous acid (HNO2) at 25°C is 4.5×10^(-4).

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The value of Ka for nitrous acid (HNO2) at 25°C is 4.5×10^(-4). a. Write the chemical equation for the equilibrium that corresponds to Ka: 1. H+(aq) + NO2-(aq) ⇌ HNO2(aq) 2. HNO2(aq) ⇌ H+(aq) + NO2-(aq) 3. HNO2(aq) ⇌ H-(aq) + NO2+(aq) 4. HNO2(aq) + H+(aq) ⇌ H2NO2+(aq) 5. HNO2(aq) + H-(aq) ⇌ H2NO2+(aq) b. By using the value of Ka, calculate ΔG° for the dissociation of nitrous acid in aqueous solution. c. What is the value of ΔG at equilibrium? d. What is the value of ΔG when [H+] = 5.9×10^(-2) M, [NO2-] = 6.7×10^(-4) M, and [HNO2] = 0.21 M?

The Correct Answer and Explanation is:

a. Chemical Equation for Ka

The dissociation of nitrous acid (HNO₂) in water is represented by the equation:

HNO₂(aq) ⇌ H⁺(aq) + NO₂⁻(aq)

This is the correct equilibrium equation because Ka represents the dissociation constant for acids, and the equation must show the acid dissociating into hydrogen ions (H⁺) and the conjugate base (NO₂⁻). Therefore, option 2 is the correct choice.

b. Calculation of ΔG°

To calculate the standard Gibbs free energy change (ΔG°) for the dissociation of nitrous acid at 25°C, we can use the following formula: ΔG°=−RTln⁡Ka\Delta G° = -RT \ln K_aΔG°=−RTlnKa​

Where:

  • RRR = 8.314 J/mol·K (universal gas constant)
  • TTT = 298 K (temperature at 25°C)
  • KaK_aKa​ = 4.5×10−44.5 \times 10^{-4}4.5×10−4 (given)

Substitute the values into the equation: ΔG°=−(8.314 J/mol\cdotpK)×(298 K)×ln⁡(4.5×10−4)\Delta G° = – (8.314 \, \text{J/mol·K}) \times (298 \, \text{K}) \times \ln(4.5 \times 10^{-4})ΔG°=−(8.314J/mol\cdotpK)×(298K)×ln(4.5×10−4)

Now calculate: ΔG°=−8.314×298×ln⁡(4.5×10−4)\Delta G° = – 8.314 \times 298 \times \ln(4.5 \times 10^{-4})ΔG°=−8.314×298×ln(4.5×10−4)

First, calculate the natural logarithm: ln⁡(4.5×10−4)≈−7.803\ln(4.5 \times 10^{-4}) ≈ -7.803ln(4.5×10−4)≈−7.803

Now, calculate: ΔG°=−8.314×298×(−7.803)\Delta G° = – 8.314 \times 298 \times (-7.803)ΔG°=−8.314×298×(−7.803) ΔG°≈19378.6 J/mol=19.38 kJ/mol\Delta G° ≈ 19378.6 \, \text{J/mol} = 19.38 \, \text{kJ/mol}ΔG°≈19378.6J/mol=19.38kJ/mol

So, ΔG° ≈ 19.38 kJ/mol.

c. Value of ΔG at Equilibrium

At equilibrium, the change in Gibbs free energy (ΔG) is zero because the system is in a state of no net change: ΔG=0\Delta G = 0ΔG=0

Therefore, the value of ΔG at equilibrium is 0 J/mol.

d. Calculation of ΔG when [H⁺] = 5.9 × 10⁻² M, [NO₂⁻] = 6.7 × 10⁻⁴ M, and [HNO₂] = 0.21 M

To find the value of ΔG under non-standard conditions, we use the following equation: ΔG=ΔG°+RTln⁡Q\Delta G = \Delta G° + RT \ln QΔG=ΔG°+RTlnQ

Where:

  • QQQ is the reaction quotient, given by:

Q=[H+][NO2−][HNO2]Q = \frac{[H⁺][NO₂⁻]}{[HNO₂]}Q=[HNO2​][H+][NO2−​]​

Substitute the concentrations: Q=(5.9×10−2)(6.7×10−4)0.21≈1.87×10−4Q = \frac{(5.9 \times 10^{-2})(6.7 \times 10^{-4})}{0.21} ≈ 1.87 \times 10^{-4}Q=0.21(5.9×10−2)(6.7×10−4)​≈1.87×10−4

Now, substitute the known values into the ΔG equation: ΔG=19.38 kJ/mol+(8.314 J/mol\cdotpK)×(298 K)×ln⁡(1.87×10−4)\Delta G = 19.38 \, \text{kJ/mol} + (8.314 \, \text{J/mol·K}) \times (298 \, \text{K}) \times \ln(1.87 \times 10^{-4})ΔG=19.38kJ/mol+(8.314J/mol\cdotpK)×(298K)×ln(1.87×10−4)

Calculate the natural logarithm: ln⁡(1.87×10−4)≈−8.591\ln(1.87 \times 10^{-4}) ≈ -8.591ln(1.87×10−4)≈−8.591

Now, calculate: ΔG=19380 J/mol+(8.314×298)×(−8.591)\Delta G = 19380 \, \text{J/mol} + (8.314 \times 298) \times (-8.591)ΔG=19380J/mol+(8.314×298)×(−8.591) ΔG=19380+(−21163.4)\Delta G = 19380 + (-21163.4)ΔG=19380+(−21163.4) ΔG≈−17783.4 J/mol=−17.78 kJ/mol\Delta G ≈ -17783.4 \, \text{J/mol} = -17.78 \, \text{kJ/mol}ΔG≈−17783.4J/mol=−17.78kJ/mol

So, ΔG ≈ -17.78 kJ/mol.

Summary of Answers:

  • a. The correct equilibrium equation is: HNO₂(aq) ⇌ H⁺(aq) + NO₂⁻(aq)
  • b. ΔG° ≈ 19.38 kJ/mol
  • c. ΔG at equilibrium = 0 J/mol
  • d. ΔG ≈ -17.78 kJ/mol at the given concentrations.

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