Chemistry Question (problem???) 14.42?

Question

The dissociation of molecular iodine atoms is represented as

I2 (g) > 2I (g)

At 1000 K, the equilibrium constant Kc for the reaction is 3.80 x 10 (raised to -5). Suppose you start with 0.0456 mole of I2 in 2.30 L flask at 1000 K. What are the concentrations of the gases at equilibirum?


THANK YOU!

Answer 1

The initial concentration of I2 is 0.0198 M. As the reaction proceeds, you will lose an x amount of I2, but add 2x of the individual atoms.

Kc = [I]^2/[I2]
3.80*10^-5 = [2x]^2/0.0198-x

Here you can assume that since 3.8*10^-5 is much less than 0.0198, that x will be insignificant compared to 0.0198, therefore you can drop the x term from the denominator. This simplifies the equation to:

3.8*10^-5 = 4x^2/0.0198
7.5*10^-7 = 4x^2
1.88 * 10^-7 = x^2
x = 4.3 * 10^-4, so [I] = 8.6 * 10-4

If you wish, you can keep it in there and solve for x using the quadratic.