Respuesta :
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?
Answer: The equilibrium partial pressure of [tex]NO_2\text{ and }N_2O_4[/tex] 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,
[tex]PV=\frac{w}{M}RT[/tex]
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 = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = Temperature of the gas = [tex]45^oC=[45+273]K=318K[/tex]
Putting values in above equation, we get:
[tex]P\times 2.12L=\frac{50.0g}{92g/mol}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 318\\\\P=\frac{50.0\times 0.0821\times 318}{2.12\times 92}=6.693atm[/tex]
Relation of [tex]K_p[/tex] with [tex]K_c[/tex] is given by the formula:
[tex]K_p=K_c(RT)^{\Delta ng}[/tex]
where,
[tex]K_p[/tex] = equilibrium constant in terms of partial pressure
[tex]K_c[/tex] = equilibrium constant in terms of concentration = 0.619
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature = [tex]45^oC=[45+273]K=318K[/tex]
[tex]\Delta n_g[/tex] = change in number of moles of gas particles = [tex]n_{products}-n_{reactants}=(2-1)=1[/tex]
Putting values in above equation, we get:
[tex]K_p=0.619\times (0.0821\times 318)^{1}\\\\K_p=16.16[/tex]
For the given chemical equation:
[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]
Initial: 6.693
At eqllm: 6.693-x 2x
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{(p_{NO_2})^2}{p_{N_2O_4}}[/tex]
Putting values in above expression, we get:
[tex]16.16=\frac{(2x)^2}{6.693-x}\\\\x=-7.59,3.56[/tex]
So, equilibrium partial pressure of [tex]NO_2=2x=(2\times 3.56)=7.12atm[/tex]
Equilibrium partial pressure of [tex]N_2O_4=(6.693-x)=(6.693-3.56)=3.133atm[/tex]
Hence, the equilibrium partial pressure of [tex]NO_2\text{ and }N_2O_4[/tex] is 7.12 atm and 3.133 atm respectively.
