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

/ Exams & Certification / By

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 equilibrium constant KaK_aKa​ for nitrous acid (HNO₂) represents the dissociation of HNO₂ into its ions. The correct equilibrium equation for the dissociation of nitrous acid is:

HNO2(aq)⇌H+(aq)+NO2−(aq)\text{HNO}_2(aq) \rightleftharpoons \text{H}^+(aq) + \text{NO}_2^-(aq)HNO2​(aq)⇌H+(aq)+NO2−​(aq)

So, the correct option is:

2. HNO₂(aq) ⇌ H⁺(aq) + NO₂⁻(aq)

b. Calculation of ΔG° for the Dissociation of Nitrous Acid

To calculate the standard Gibbs free energy change (ΔG°) for the dissociation of nitrous acid, we can use the following relationship between KaK_aKa​ and ΔG∘\Delta G^\circΔG∘:ΔG∘=−RTln⁡Ka\Delta G^\circ = -RT \ln K_aΔG∘=−RTlnKa​

Where:

  • RRR is the gas constant = 8.314 J/(mol·K)
  • TTT is the temperature in Kelvin (given as 25°C, so T=298 KT = 298 \, \text{K}T=298K)
  • KaK_aKa​ is the equilibrium constant for the dissociation of nitrous acid, given as 4.5×10−44.5 \times 10^{-4}4.5×10−4

Substituting the values:ΔG∘=−(8.314)×(298)×ln⁡(4.5×10−4)\Delta G^\circ = – (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)=ln⁡(4.5)+ln⁡(10−4)=1.5041−9.2103=−7.7062\ln(4.5 \times 10^{-4}) = \ln(4.5) + \ln(10^{-4}) = 1.5041 – 9.2103 = -7.7062ln(4.5×10−4)=ln(4.5)+ln(10−4)=1.5041−9.2103=−7.7062

Now calculate ΔG∘\Delta G^\circΔG∘:ΔG∘=−(8.314)×(298)×(−7.7062)\Delta G^\circ = – (8.314) \times (298) \times (-7.7062)ΔG∘=−(8.314)×(298)×(−7.7062)ΔG∘=19104.5 J/mol≈19.1 kJ/mol\Delta G^\circ = 19104.5 \, \text{J/mol} \approx 19.1 \, \text{kJ/mol}ΔG∘=19104.5J/mol≈19.1kJ/mol

So, ΔG∘≈19.1 kJ/mol\Delta G^\circ \approx 19.1 \, \text{kJ/mol}ΔG∘≈19.1kJ/mol.

c. Value of ΔG at Equilibrium

At equilibrium, the Gibbs free energy change ΔG\Delta GΔG is zero because the system has reached a state of balance between the forward and reverse reactions. Therefore:ΔG=0 kJ/mol\Delta G = 0 \, \text{kJ/mol}ΔG=0kJ/mol

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

To calculate the Gibbs free energy change at non-standard conditions, we use the equation:ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT \ln QΔG=ΔG∘+RTlnQ

Where:

  • ΔG∘\Delta G^\circΔG∘ is the standard Gibbs free energy change (19.1 kJ/mol)
  • RRR is the gas constant (8.314 J/mol·K)
  • TTT is the temperature in Kelvin (298 K)
  • QQQ is the reaction quotient, calculated as:

Q=[H+][NO2−][HNO2]Q = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]}Q=[HNO2​][H+][NO2−​]​

Substituting the given concentrations:Q=(5.9×10−2)(6.7×10−4)0.21=3.953×10−50.21=1.88×10−4Q = \frac{(5.9 \times 10^{-2})(6.7 \times 10^{-4})}{0.21} = \frac{3.953 \times 10^{-5}}{0.21} = 1.88 \times 10^{-4}Q=0.21(5.9×10−2)(6.7×10−4)​=0.213.953×10−5​=1.88×10−4

Now calculate ΔG\Delta GΔG:ΔG=19104.5+(8.314)(298)ln⁡(1.88×10−4)\Delta G = 19104.5 + (8.314)(298) \ln(1.88 \times 10^{-4})ΔG=19104.5+(8.314)(298)ln(1.88×10−4)

First, calculate the natural logarithm:ln⁡(1.88×10−4)=ln⁡(1.88)+ln⁡(10−4)=0.6290−9.6439=−9.0149\ln(1.88 \times 10^{-4}) = \ln(1.88) + \ln(10^{-4}) = 0.6290 – 9.6439 = -9.0149ln(1.88×10−4)=ln(1.88)+ln(10−4)=0.6290−9.6439=−9.0149

Now calculate ΔG\Delta GΔG:ΔG=19104.5+(8.314)(298)(−9.0149)=19104.5−22229.7=−3125.2 J/mol\Delta G = 19104.5 + (8.314)(298)(-9.0149) = 19104.5 – 22229.7 = -3125.2 \, \text{J/mol}ΔG=19104.5+(8.314)(298)(−9.0149)=19104.5−22229.7=−3125.2J/mol

So, ΔG≈−3.13 kJ/mol\Delta G \approx -3.13 \, \text{kJ/mol}ΔG≈−3.13kJ/mol.

Final Answers:

  • a. The correct chemical equation is HNO₂(aq) ⇌ H⁺(aq) + NO₂⁻(aq).
  • b. The standard Gibbs free energy change ΔG∘\Delta G^\circΔG∘ is approximately 19.1 kJ/mol.
  • c. The Gibbs free energy at equilibrium is 0 kJ/mol.
  • d. The Gibbs free energy at the given concentrations is approximately -3.13 kJ/mol.

Related Posts

Scroll to Top