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The dissociation of molecular iodine into iodine atoms is represented as I2(g) ⇌ 2I(g) At 1000...

The dissociation of molecular iodine into iodine atoms is represented as I2(g) ⇌ 2I(g) At 1000 K, the equilibrium constant Kc for the reaction is 3.80 × 10−5. Suppose you start with 0.0454 mol of I2 in a 2.28−L flask at 1000 K. What are the concentrations of the gases at equilibrium?

What is the equilibrium concentration of I2? M

What is the equilibrium concentration of I? M

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Answer #1

concentration of I2 = 0.0454 / 2.28 = 0.0199 M

I2(g) <------------------> 2I(g)

0.0199                            0

0.0199 - x                       2x

Kc = (2x)^2 / 0.0199 - x

3.80 x 10^-5 = 4x^2 / 0.0199 - x

x = 4.30 x 10^-4

equilibrium concentration of I2 = 0.0195 M

equilibrium concentration of I = 8.60 x 10^-4 M

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