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At 984 °C the equilibrium constant for the reaction: 2 IBr(g) = 12(g) + Br2(g) is...

Question

At 984 °C the equilibrium constant for the reaction: 2 IBr(g) = 12(g) + Br2(g) is...

At 984 °C the equilibrium constant for the reaction: 2 IBr(g) = 12(g) + Br2(g) is Kp = 1.52. If the initial pressure of IBr i

At 984 °C the equilibrium constant for the reaction: 2 IBr(g) = 12(g) + Br2(g) is Kp = 1.52. If the initial pressure of IBr is 0.00596 atm, what are the equilibrium partial pressures of IBr, I2, and Brz? P(IBr) = P(12) = p(Br2) =

Answers

ICE Table:

p (IBr) p (I2) p (Br2) initial 0.00596 change -2x +1x +1x equilibrium 0.00596-2x +1x +1x

Equilibrium constant expression is

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

1.52 = (1*x)^2/(5.96*10^-3-2*x)^2

sqrt(1.52) = (1*x)/(5.96*10^-3-2*x)

1.233 = (1*x)/(5.96*10^-3-2*x)

7.348*10^-3-2.466*x = 1*x

7.348*10^-3-3.466*x = 0

x = 2.12*10^-3

At equilibrium:

p(IBr) = 0.00596-2x = 0.00596-2*0.00212 = 0.00172 atm

p(I2) = +1x = +1*0.00212 = 0.00212 atm

p(Br2) = +1x = +1*0.00212 = 0.00212 atm

Answer:

p(IBr) = 0.00172 atm

p(I2) = 0.00212 atm

p(Br2) = 0.00212 atm


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