For the equilibrium 2IBr(g) I_2(g) + Br_2(g) K_p = 8.5 times 10^-3 at 150 degree

Question

For the equilibrium 2IBr(g) I_2(g) + Br_2(g) K_p = 8.5 times 10^-3 at 150 degree C. If 2.8 times 10^-2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of IBr after equilibrium is reached? Express your answer to two significant figures and include the appropriate units. If 2.8 times 10^-2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of I_2 after equilibrium is reached? Express your answer to two significant figures and include the appropriate units. If 2.8 times 10^-2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of Br_2 after equilibrium is reached? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

2IBr <==>I2 + Br2

Kp = P(I2) * P(Br2)/P^2(IBr)

8.5*10^-3 = (0.5x)^2/(0.028-x)

x = 0.018 atm

Partial pressure of IBr = 0.028-0.018 = 0.01 atm = 10*10^-2 atm

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partial pressure of I2 = 0.5 x = 9.0*10^-3 atm

Partial pressure of Br2 = 0.5x = 9.0*10^-3 atm

IBr (1/2)I2 (1/2)Br2 initial 0.028 0 0 change -x +0.5x +0.5x equilibrium 0.028-x 0.5x 0.5x

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