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.

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation: