For the equilibrium 2IBr(g)⇌I2(g)+Br2(g) Kp=8.5×10−3 at 150 ∘C.
Part A If 2.1×10−2 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. SubmitMy AnswersGive Up
Part B If 2.1×10−2 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 2.1×10−2 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.
For the equilibrium 2IBr(g)⇌I2(g)+Br2(g) Kp=8.5×10−3 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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