Equilibrium Constant (Kp) Calculation:

The given chemical reaction is:

N2O4(g) ⇌ 2NO2(g)

The equilibrium constant (Kp) of the reaction is given as 0.141 at 25°C and 1 bar.

Spontaneity of the Reaction:

A reaction is spontaneous if it occurs without any external intervention. The spontaneity of the reaction can be determined by comparing the Gibbs free energy (ΔG) of the system with the standard Gibbs free energy (ΔG°) of the system.

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If ΔG > 0, the reaction is non-spontaneous.

If ΔG = 0, the reaction is at equilibrium.

The standard Gibbs free energy change (ΔG°) for the given reaction is:

ΔG° = -RTlnKp

Where,

R = Universal gas constant = 8.314 J/mol-K

T = Temperature in Kelvin = 25 + 273 = 298 K

Kp = Equilibrium constant = 0.141

ΔG° = - (8.314 J/mol-K) x 298 K x ln(0.141)

ΔG° = - 14.44 kJ/mol

Since ΔG° is negative, the reaction is spontaneous at 25°C.

Partial Pressure of NO2 and N2O4:

The partial pressure of a gas in a mixture is the pressure that the gas would exert if it alone occupied the same volume as the mixture at the same temperature. The partial pressure of NO2 and N2O4 in the given reaction can be calculated using the following formula:

PNO2 = 2x/(x+1) x Ptotal

PN2O4 = 1/(x+1) x Ptotal

Where,

x = degree of dissociation

Ptotal = Total pressure of the mixture = 1 bar

At equilibrium, the degree of dissociation (x) can be calculated using the equilibrium constant (Kp) as:

Kp = (PNO2)2/PN2O4

0.141 = (2x/1+x)2/ (1/1+x)

x = 0.211

Using the above formulae, we can calculate the partial pressures of NO2 and N2O4 as:

PNO2 = 2x/(x+1) x Ptotal = 0.449 bar

PN2O4 = 1/(x+1) x Ptotal = 0.551 bar

Therefore, the partial pressure of NO2 and N2O4 at equilibrium are 0.449 bar and 0.551 bar, respectively.