A mixture of 0.80 M of bromine and 0.97 M of iodine is placed into a...

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A mixture of 0.80 M of bromine and 0.97 M of iodine is placed into a...

A mixture of 0.80 M of bromine and 0.97 M of iodine is placed into a rigid 1.00 L container at 240 °C: Br2(g) + I2(g) ⇔ 2IBr(g).
When the mixture has reached equilibrium, the concentration of iodine monobromide is 1.50 M. What is the equilibrium constant for this reaction?

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Answer 1

Answer #1

ICE Table:
[Br2] [12] [IBr] initial 0.8 0.97 change - 1x - 1x +2x equilibrium 0.8–1x 0.97–1x +2x

Given at equilibrium,
[IBr] = 1.5
+2x = 1.5
x = 0.75
Equilibrium constant expression is
Kc = [IBr]^2/[Br2][I2]
Kc = (+2x)^2/(0.8-1x)(0.97-1x)
Kc = (+2*0.75)^2/(0.8-1*0.75)(0.97-1*0.75)
Kc = 205
Answer: 205

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Answer 2

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