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

Question

Alina [70]

3 years ago

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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 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] · Answer 1 · 2022-04-29T07:21:44+03:00

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:

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