For the equilibrium 2IBr(g)I2(g)+Br2(g) Kp=8.5×103 at 150 C. Part A If 2.1×102 a

Question

For the equilibrium 2IBr(g)I2(g)+Br2(g) Kp=8.5×103 at 150 C.

Part A If 2.1×102 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. SubmitMy AnswersGive Up

Part B If 2.1×102 atm of IBr is placed in a 2.0-L container, what is the partial pressure of I2 after equilibrium is reached? Express your answer to two significant figures and include the appropriate units. SubmitMy AnswersGive Up

Part C If 2.1×102 atm of IBr is placed in a 2.0-L container, what is the partial pressure of Br2 after equilibrium is reached? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

For the equilibrium 2IBr(g)I2(g)+Br2(g) Kp=8.5×103 at 150 C.

P-IBr = (2.1*10^-2) atm, V = 2 L

From the initial conditions

P-IBr = 2.1*10^-2 atm

P-Br2 = 0

P-I2 = 0

in equilibrium

P-IBr = 2.1*10^-2 - 2x

P-Br2 = 0+ x

P-I2 = 0 + x

substittue in Kp

8.5*10^-3 = P-I2 * P-Br2 / (P-IBr)^2

substitute in P

8.5*10^-3 = P-I2 * P-Br2 / (P-IBr)^2

8.5*10^-3 = x*x / (2.1*10^-2 - 2x )^2

8.5*10^-3 = x^2 / (2.1*10^-2 - 2x )^2

sqrt(8.5*10^-3) = x / (0.021 -2x)

0.092195 = x / (0.021 -2x)

10.846x = 0.021 -2x

12.846x = 0.021

x = 0.021/12.846 = 0.001634

substitute

P-IBr = 2.1*10^-2 - 2x = 0.021 -2*0.001634 = 0.017732 atm

P-Br2 = 0+ x = 0.001634 atm

P-I2 = 0 + x = 0.001634 atm

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