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

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. Write the chemical equation for the equilibrium that corresponds to Ka:

The correct equation for the dissociation of nitrous acid (HNO2) is:

2. HNO2(aq) ⇌ H+(aq) + NO2-(aq)

This represents the dissociation of nitrous acid into hydrogen ions (H+) and nitrite ions (NO2-), which is the equilibrium described by the acid dissociation constant (Ka).

b. Calculate ΔG° for the dissociation of nitrous acid in aqueous solution:

The relationship between the acid dissociation constant (Ka) and the standard Gibbs free energy change (ΔG°) is given by the equation:ΔG∘=−RTln⁡Ka\Delta G^\circ = -RT \ln K_aΔG∘=−RTlnKa​

Where:

• RRR = 8.314 J/mol·K (universal gas constant),
• TTT = 298 K (standard temperature in Kelvin),
• KaK_aKa​ = 4.5 × 10⁻⁴ (given Ka for nitrous acid).

Now, calculate ΔG°:ΔG∘=−(8.314 J/mol\cdotpK)×(298 K)×ln⁡(4.5×10−4)\Delta G^\circ = – (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)

First, calculate the natural logarithm of Ka:ln⁡(4.5×10−4)≈−7.81\ln(4.5 \times 10^{-4}) \approx -7.81ln(4.5×10−4)≈−7.81

Now calculate ΔG°:ΔG∘=−(8.314×298)×(−7.81)≈19321.5 J/mol≈19.3 kJ/mol\Delta G^\circ = – (8.314 \times 298) \times (-7.81) \approx 19321.5 \, \text{J/mol} \approx 19.3 \, \text{kJ/mol}ΔG∘=−(8.314×298)×(−7.81)≈19321.5J/mol≈19.3kJ/mol

So, the standard Gibbs free energy change (ΔG°) for the dissociation of nitrous acid is approximately +19.3 kJ/mol.

c. What is the value of ΔG at equilibrium?

At equilibrium, ΔG = 0, as the system has reached a state where the forward and reverse reactions occur at the same rate, and no net change in concentrations happens.

So, the value of ΔG at equilibrium is 0.

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?

To calculate the value of ΔG under non-standard conditions, we use the equation:ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT \ln QΔG=ΔG∘+RTlnQ

Where:

• QQQ is the reaction quotient, given by:

Q=[H+][NO2−][HNO2]Q = \frac{[H^+][NO_2^-]}{[HNO_2]}Q=[HNO2​][H+][NO2−​]​

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

Now, calculate ΔG:ΔG=19.3 kJ/mol+(8.314 J/mol\cdotpK)×(298 K)×ln⁡(1.88×10−4)\Delta G = 19.3 \, \text{kJ/mol} + (8.314 \, \text{J/mol·K}) \times (298 \, \text{K}) \times \ln(1.88 \times 10^{-4})ΔG=19.3kJ/mol+(8.314J/mol\cdotpK)×(298K)×ln(1.88×10−4)

First, calculate the natural logarithm of Q:ln⁡(1.88×10−4)≈−8.97\ln(1.88 \times 10^{-4}) \approx -8.97ln(1.88×10−4)≈−8.97

Now calculate ΔG:ΔG=19321.5 J/mol+(8.314×298)×(−8.97)\Delta G = 19321.5 \, \text{J/mol} + (8.314 \times 298) \times (-8.97)ΔG=19321.5J/mol+(8.314×298)×(−8.97)ΔG=19321.5 J/mol−22140.5 J/mol≈−2820 J/mol≈−2.82 kJ/mol\Delta G = 19321.5 \, \text{J/mol} – 22140.5 \, \text{J/mol} \approx -2820 \, \text{J/mol} \approx -2.82 \, \text{kJ/mol}ΔG=19321.5J/mol−22140.5J/mol≈−2820J/mol≈−2.82kJ/mol

So, the value of ΔG when the concentrations are given is approximately -2.82 kJ/mol.

Summary:

• a. The correct equilibrium equation is HNO2(aq) ⇌ H+(aq) + NO2-(aq).
• b. ΔG° = +19.3 kJ/mol.
• c. ΔG at equilibrium = 0.
• d. ΔG at the given concentrations = -2.82 kJ/mol.