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?
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
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...
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