Equilibrium concentration

Question

The value of the equilibrium constant Ke for the decomposition of the gas diiodine I2 in iodine atoms is 3.6 x 10^-10 at a specific temperature. Calculate the equilibrium concentration of the iodine atoms when 0.50 mol of diiodine is heated to the same temperature in a 5.0-liter reaction vessel. the answer is 6.0. 10^-6

Please explain to me step by step and also which formula you used to solve this problem.

Thank you for your help in advance

Answer 1

K = 3.6 x 10^-10 for I₂(g) ⇔ 2I(g) .5 moles of diatomic iodine in 5 liters. I assume this is a K_C although generally K_P is more appropriate for gases. Generally, try to always attach units to any data and answers.

Let x be the moles of diatomic iodine reacted, so I₂ moles will be the initial .5 minus x at equilibrium or .5-x is the equilibrium concentration of I₂

The I produced when x moles of I₂ decomposes will be +2x from the stoichiometry. Therefore, the final concentration of I = 2x

The K expression is [I]²/[I₂] so

3.6 x 10^-10 = (2x/5)²/((.5-x)/5) = 4x²/(5(.5-x))

We can make the assumption that x << .5 because the K is very small (We can check to see if this assumption is correct at the end. Otherwise, you have to multiply out and solve the quadratic in x)

Also, we could have defined x as the concentration change of I₂ and the initial concentration of I₂ as.5moles / 5l = .1 mole/liter solving this with molarity as a surrogate for moles. This is fine if the volume is constant. the final equation would be

3.6 x 10^-10 = 4x²/(.1-x) This is a little more manageable.

Again, assuming that x << .1

3.6 x 10^-10 = 40x²

x = 3 x 10^-6 and the final I conc = 2x = 6 x 10^-6 (the error of assuming that x << 1 is in the 6th place - much more accurate than our dat.)