Chemistry, 03.03.2020 03:49 diyah843

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?

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Answer from: kennaklein2

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 $NO_2\text{ and }N_2O_4$ 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,

$PV=\frac{w}{M}RT$

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 = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gas = 45^oC=[45+273]K=318K

Putting values in above equation, we get:

$P\times 2.12L=\frac{50.0g}{92g/mol}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 318\\\\P=\frac{50.0\times 0.0821\times 318}{2.12\times 92}=6.693atm$

Relation of K_p with K_c is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

K_p = equilibrium constant in terms of partial pressure

K_c = equilibrium constant in terms of concentration = 0.619

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 45^oC=[45+273]K=318K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=(2-1)=1$

Putting values in above equation, we get:

$K_p=0.619\times (0.0821\times 318)^{1}\\\\K_p=16.16$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial: 6.693

At eqllm: 6.693-x 2x

The expression of K_p for above equation follows:

$K_p=\frac{(p_{NO_2})^2}{p_{N_2O_4}}$

Putting values in above expression, we get:

$16.16=\frac{(2x)^2}{6.693-x}\\\\x=-7.59,3.56$

So, equilibrium partial pressure of $NO_2=2x=(2\times 3.56)=7.12atm$

Equilibrium partial pressure of N_2O_4=(6.693-x)=(6.693-3.56)=3.133atm

Hence, the equilibrium partial pressure of $NO_2\text{ and }N_2O_4$ is 7.12 atm and 3.133 atm respectively.

Answer from: Quest

a is the correct answer

Answer from: Quest

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Part A If 50.0 gg of N2O4N2O4 is introduced into an empty 2.12 LL container, what are the partial pr...

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