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?
<u>Answer:</u> The equilibrium partial pressure of is 7.12 atm and 3.133 atm respectively.
<u>Explanation:</u>
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:
<u>Initial:</u> 6.693
<u>At eqllm:</u> 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.