Question

Constants | Periodic Table

For the reaction
I2(g)+Br2(g)←−→2IBr(g),
Kc=280 at 150 ∘C. Suppose that 0.500 mol IBr in a 2.00-L flask is allowed to reach equilibrium at 150 ∘C.

Part A

Part complete

What is the equilibrium concentration of IBr?

0.223

M

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Correct

Part B

What is the equilibrium concentration of I2?

nothing

  M  

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Incorrect; Try Again

Part C

What is the equilibrium concentration of Br2?

nothing

  M  

Answer 1

Initial concentration of IBr = mol of IBr / volume in L

= 0.500 mol / 2.00 L

= 0.250 M

ICE Table:

[I2] [Br2] [IBr]

initial 0 0 0.25

change +1x +1x -2x

equilibrium +1x +1x 0.25-2x

Equilibrium constant expression is

kc = [IBr]^2/[I2]*[Br2]

280.0 = (0.25-2*x)^2/(1*x)^2

sqrt(280.0) = (0.25-2*x)/(1*x)

16.73320053068151 = (0.25-2*x)/(1*x)

16.73*x = 0.25-2*x

-0.25 + 18.73*x = 0

x = 0.01335

At equilibrium:

[I2] = +1x = +1*0.01335 = 0.01335 M

[Br2] = +1x = +1*0.01335 = 0.01335 M

[IBr] = 0.25-2x = 0.25-2*0.01335 = 0.22331 M

A)0.223 M

B)0.0134 M

C)0.0134 M