If 2.9x10-2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of I2 after equilibrium is reached?
For the equilibrium
2IBr(g) ⇌ I2:(g) + Br2(g) Kp = 8.5 x 10-3 at 150°C.
Part B
If 2.9x10-2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of I2 after equilibrium is reached?
Homework Answers
Equilibrium constant expression is
Kp = p(I2)*p(Br2)/p(IBr)^2
0.0085 = (1*x)^2/(2.9*10^-2-2*x)^2
sqrt(0.0085) = (1*x)/(2.9*10^-2-2*x)
9.22*10^-2 = (1*x)/(2.9*10^-2-2*x)
2.674*10^-3-0.1844*x = 1*x
2.674*10^-3-1.184*x = 0
x = 2.257*10^-3
At equilibrium:
p(I2) = +1x = +1*0.002257 = 0.00226 atm
Answer: 0.0023 atm
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If 2.9x10-2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of I2 after equilibrium is reached?
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