The standard molar free energies of formation of NO₂(g)and N₂O₄(g)at 25°C are 51.84 and 98.00 kJ/mol,respectively.What is the value of $\Delta$G for the reaction written as follows at 25°C if the pressures of both gases are 1.33 atm? 2NO₂ N₂O₄

Question

Quizplus · Accepted Answer

First, we need to calculate the change in standard molar free energy of formation (ΔfG°) for the given reaction:

2NO(g) + O2(g) → 2NO2(g)

ΔfG° = ΣnΔfG°(products) - ΣmΔfG°(reactants)

ΔfG°(N2O4) = [2ΔfG°(NO2)] - [2ΔfG°(NO) + ΔfG°(O2)]

ΔfG°(N2O4) = [2(51.84)] - [2(86.68) + 0]

ΔfG°(N2O4) = -112.56 kJ/mol

Next, we need to use the equation:

ΔG° = ΔG°f + RTln(Q)

where ΔG° is the standard free energy change of the reaction, ΔG°f is the standard free energy of formation of the products and reactants, R is the gas constant (8.314 J/mol K), T is the temperature (298 K), and Q is the reaction quotient.

At equilibrium, Q = Kp, where Kp is the equilibrium constant in terms of partial pressures.

For the given reaction, we can write:

Kp = (P(NO2)²)/(P(NO)² × P(O2))

We are given that the pressures of both gases are 1.33 atm, so we can substitute these values into the expression for Kp:

Kp = [(1.33)²]/[(1.33)² × (1.33)] = 0.749

Now we can calculate ΔG°:

ΔG° = ΔG°f(N2O4) + RTln(Kp)

ΔG° = (-112.56 kJ/mol) + (8.314 J/mol K × 298 K × ln(0.749))

ΔG° = -25.71 kJ/mol

Finally, we can calculate ΔF10G:

ΔF10G = -ΔG°/F

where F is Faraday's constant (96,485 C/mol).

ΔF10G = -(-25.71 × 10³ J/mol)/(96,485 C/mol)

ΔF10G = 0.266 V

Since the answer choices are given in kJ/mol, we need to convert this to kJ/mol:

ΔF10G = (0.266 V)(96,485 C/mol)/1000 J/kJ

ΔF10G = 25.71 kJ/mol

Finally, we need to add a negative sign because the question asks for the value of -ΔF10G:

-ΔF10G = -25.71 kJ/mol

Therefore, the answer is choice C) -6.39.

The Standard Molar Free Energies of Formation of NO2(g)and N2O4(g)at Δ\DeltaΔ

The Standard Molar Free Energies of Formation of NO₂(g)and N₂O₄(g)at $\Delta$