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For the equilibrium 2IBr(g)⇌I2(g)+Br2(g) Kp=8.5×10−3 at 150 ∘C. a.)If 2.7×10−2 atm of IBr is placed in...

For the equilibrium 2IBr(g)⇌I2(g)+Br2(g) Kp=8.5×10−3 at 150 ∘C. a.)If 2.7×10−2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of IBr after equilibrium is reached? b.)If 2.7×10−2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of I2 after equilibrium is reached? c.)If 2.7×10−2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of Br2 after equilibrium is reached?

a.)If 2.7×10−2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of IBr after equilibrium is reached?

B.)If 2.7×10−2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of I2 after equilibrium is reached?

C.) If 2.7×10−2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of Br2 after equilibrium is reached?

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Answer #1

The reaction is 2IBR(g) ßà I2(g)+ Br2(g),

Kp= [I2] [Br2]/[IBr]2,

Writing the ICE table

Component                             Initial partial pressure (atm)             change (atm)      equilibrium (atm)

IBr                                                   0.027                                                   -2x                                0.027-2x

I2                                                        0                                                         x                                       x

Br2                                                     0                                                        x                                       x

Kp= x2/(0.027-2x)2= 8.5*10-3

Taking square root, x/(0.027-2x)= 0.092, x=0.092*(0.027-2x)

Hence x+2x*0.092= 0.092*0.027

Hence x*(1+0.092*2)= 0.092*0.027, x= 0.092*0.027/1.184 =0.0021

Hence at equilibrium, partial pressures ( atm)

IBR= 0.027-2*0.0021= 0.0228, I2= Br2=0.0021

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