The standard molar free energies of formation of NO₂(g)and N₂O₄(g)at 25°C are 51.840 and 98.035 kJ/mol,respectively.What is the value of K_p(in atm)for the reaction written as follows at 25°C? 2NO₂ N₂O₄

Question

Quizplus · Accepted Answer

The standard Gibbs free energy change for the given reaction can be calculated by using the standard molar free energies of formation of the reactants and products:

ΔG° = ΣG°f(products) - ΣG°f(reactants)
ΔG° = [98.035 kJ/mol - (2 x 51.840 kJ/mol)] = 51.355 kJ/mol

At 25°C, the relationship between the equilibrium constant (Kp) and ΔG° is given by:

ΔG° = -RTlnKp

where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (298 K). Rearranging this equation, we get:

Kp = e^(-ΔG°/RT)

Substituting the values, we get:

Kp = e^(-51355 J/mol / (8.314 J/mol·K x 298 K))
Kp = 9.76 atm

Therefore, the best choice is C, with Kp = 9.76 atm.

The Standard Molar Free Energies of Formation of NO2(g)and N2O4(g)at ×\times×

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