Part A If 50.0 gg of N2O4N2O4 is introduced into an empty 2.12 LL container, what are the partial pressures of NO2NO2 and N2O4N2O4 after equilibrium has been achieved at 45∘C∘C?

The question is incomplete, here is the complete question:

At 45°C, Kc = 0.619 for the reaction N₂O₄(g) ⇌ 2 NO₂(g).

If 50.0 g of N₂O₄ is introduced into an empty 2.12 L container, what are the partial pressures of NO₂ and N₂O₄ after equilibrium has been achieved at 45°C?

The equilibrium partial pressure of is 7.12 atm and 3.133 atm respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

Or,

where,

P = Pressure of the gas  = ?

V = Volume of the gas  = 2.12 L

w = Weight of the gas  = 50.0 g

M = Molar mass of gas  = 92 g/mol

R = Gas constant =

T = Temperature of the gas =

Putting values in above equation, we get:

Relation of with is given by the formula:

where,

= equilibrium constant in terms of partial pressure

= equilibrium constant in terms of concentration = 0.619

R = Gas constant =

T = temperature  =

= change in number of moles of gas particles =

Putting values in above equation, we get:

For the given chemical equation:

Initial:            6.693

At eqllm:    6.693-x           2x

The expression of for above equation follows:

Putting values in above expression, we get:

So, equilibrium partial pressure of

Equilibrium partial pressure of

Hence, the equilibrium partial pressure of is 7.12 atm and 3.133 atm respectively.