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Alina [70]
3 years ago
15

En la reacción I2(g) + Br2(g) « 2 IBr(g), Keq = 280 a 150°C. Suponga que se permite que 0.500 mol de IBr en un matraz de 1.00 L

alcancen el equilibrio a 150°C. ¿Cuáles son las presiones parciales de equilibrio de IBr, I2 y Br2?
Chemistry
2 answers:
user100 [1]3 years ago
5 0

Answer:

p_{I_2}=0.926atm\\p_{Br_2}=0.926atm\\p_{IBr}=15.5atm

Explanation:

Hello,

In this case, given the initial load of 0.500 mol of IBr in the 1.00-L, we compute its initial concentration:

[IBr]_0=\frac{0.500mol}{1.00L}=0.500M

Hence, by knowing the original reaction, we should invert it as IBr will produce iodine and bromine considering the initial load:

2IBr(g)\rightleftharpoons I_2(g) + Br_2(g)

Therefore, the equilibrium constant should be inverted:

K'=\frac{1}{Keq}=\frac{1}{280}=3.57x10^{-3}

So we write the law of mass action:

K'=\frac{[I_2][Br_2]}{[IBr]^2}

That in terms of the change x due to the reaction extent turns out:3.57x10^{-3}=\frac{(x)(x)}{(0.500-2x)^2}

In such a way, solving by using solver or quadratic equation we obtain:

x_1=-0.0339M\\x_2=0.0267M

Clearly, the solution is 0.0267M, thus, the equilibrium concentrations are:

[I_2]=x=0.0267M

[Br_2]=x=0.0267M

[IBr]=0.5M-2x=0.5M-2*0.0267M=0.447M

Thus, with the given temperature (150+273.15=423.15K), we compute the partial pressures by using the ideal gas equation:

p_{I_2}=[I_2]RT=0.0267\frac{mol}{L} *0.082\frac{atm*L}{mol*K}*423.15K\\\\p_{I_2}=0.926atm\\\\p_{Br_2}=[Br_2]RT=0.0267\frac{mol}{L} *0.082\frac{atm*L}{mol*K}*423.15K\\\\p_{Br_2}=0.926atm\\\\p_{IBr}=[IBr]RT=0.447\frac{mol}{L} *0.082\frac{atm*L}{mol*K}*423.15K\\\\p_{IBr}=15.5atm

Best regards.

love history [14]3 years ago
3 0

Answer:

P IBr: 15.454atm

I₂: 0.923 atm

P Br₂: 0.923atm

Explanation:

Basados en la reacción:

I₂(g) + Br₂(g) ⇄ 2 IBr(g)

La constante de equilibrio, Keq, es definida como:

Keq = \frac{P_{IBr}^2}{P_{I_2}P_{Br_2}}

<em>Se cumple la relación de Keq = 280 cuando las presiones están en equilibrio</em>

Usando PV = nRT, la presión inicial de IBr es:

P = nRT / V; 0.500mol*0.082atmL/molK*423.15K / 1.00L = <em>17.3 atm</em>

<em />

Siendo las presiones en equilibrio:

P IBr: 17.3 - 2X

P I₂: X

P Br₂: X

<em>Donde X representa el avance de reacción.</em>

Remplazando en Keq:

280 = (17.3 - 2X)² / X²

280X² = 4X² - 69.2X + 299.29

0 = -276X² - 69.2X + 299.29

<em>Resolviendo para X:</em>

X = -1.174 → Solución falsa. No existen presiones negativas

X = 0.923 → Solución real

Así, las presiones parciales en equilibrio de cada compuesto son:

P IBr: 17.3 - 2X = <em>15.454atm</em>

P I₂: X =<em> 0.923atm</em>

P Br₂: X = <em>0.923atm</em>

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