High School

The equilibrium constant for the reaction below is \(7.2 \times 10^{-4}\) at 298 K and 1 atm.

\[ \text{HNO}_2(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NO}_2^-(aq) + \text{H}_3\text{O}^+(aq) \]

When \([\text{HNO}_2(aq)] = 1.0 \, \text{M}\) and \([\text{NO}_2^-(aq)] = [\text{H}_3\text{O}^+(aq)] = 1.0 \times 10^{-5} \, \text{M}\), calculate \(\Delta G\).

A) -39.1 kJ
B) -17.9 kJ
C) +17.9 kJ
D) +39.1 kJ

Answer :

Final answer:

The Gibbs free energy (ΔG) for the reaction is calculated using the reaction quotient (Q), the equilibrium constant (K), and the standard free energy change (ΔG°) formula. The correct answer for the reaction under the given conditions is +17.9 kJ, implying that energy is required for the reaction to proceed.

Explanation:

The student's question involves calculating the Gibbs free energy (ΔG) for the reaction:

HNO2(aq) + H2O(l) → NO2⁻(aq) + H3O⁺(aq)

To calculate ΔG, we use the equation:

ΔG = ΔG° + RTlnQ

Where ΔG° is the standard free energy change, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Q is the reaction quotient.

Firstly, we need to calculate ΔG° using the given equilibrium constant (K₂ε) and the formula:

ΔG° = -RTlnK₂ε

ΔG° = -(8.314 J/mol·K)(298 K)ln(7.2 x 10⁻´)

Now, we calculate Q using the provided concentrations and then plug into the main equation:

Q = [NO2⁻][H3O⁺] / [HNO2]

Q = (1.0 x 10⁻⁵)² / (1.0 M)

Using the above values, we calculate ΔG for the system.

The correct answer is C) +17.9 kJ, indicating that the reaction requires an input of energy to proceed under the given conditions.

Final answer:

To find the Gibbs free energy change (ΔG) for the given reaction, we calculate the standard free energy change from the equilibrium constant and then add the product of the gas constant, temperature, and the natural logarithm of the reaction quotient.

Explanation:

To calculate the Gibbs free energy change (ΔG) for the reaction HNO2(aq) + H2O(l) ⇌ NO2-(aq) + H3O+(aq) at equilibrium, we use the equation ΔG = ΔG° + RTlnQ, where ΔG° is the standard free energy change, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Q is the reaction quotient.

The standard free energy change (ΔG°) can be calculated from the equilibrium constant (K) using the relation ΔG° = -RTlnK. At 298 K, with K = 7.2 x 10-4, and R = 8.314 J/mol·K, we get:

ΔG° = -8.314 × 298 × ln(7.2 x 10-4)

To find Q, we use the concentrations given:

Q = [NO2-][H3O+]/[HNO2] = (1.0 x 10-5)2/(1.0) = 1.0 x 10-10

Substituting back into the ΔG equation:

ΔG = (-8.314 × 298 × ln(7.2 x 10-4)) + (8.314 × 298 × ln(1.0 x 10-10))

Solving the equation gives us the value of ΔG, which we can then compare to the provided options to determine the correct one.