Chemistry, 30.05.2020 03:02 3giez4

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?

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Answer from: hey840

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:

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

Se cumple la relación de Keq = 280 cuando las presiones están en equilibrio

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

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

Siendo las presiones en equilibrio:

P IBr: 17.3 - 2X

P I₂: X

P Br₂: X

Donde X representa el avance de reacción.

Remplazando en Keq:

280 = (17.3 - 2X)² / X²

280X² = 4X² - 69.2X + 299.29

0 = -276X² - 69.2X + 299.29

Resolviendo para X:

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 = 15.454atm

P I₂: X = 0.923atm

P Br₂: X = 0.923atm

Answer from: noahmoritz2004

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

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 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.

Answer from: Quest

lol idk man i guess u gotta ask bill nye the science guy

Answer from: Quest

All the elements are arranged by melting points on the periodic table. in row they go from metals to gasses. in the row from sodium to argon, sodium is more reactive than the one on its right, magnesium.

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

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