Equilibrium concentrations

Question

The equilibrium constant, K_c, for the following reaction is 55.6 at 698 K.

H₂ (g) + I₂ (g)  2 HI (g)

Calculate the equilibrium concentrations of reactants and product when 0.333 moles of H₂ and 0.333 moles of I₂ are introduced into a 1.00 L vessel at 698 K.

[ H2 ]	=	M
[ I2 ]	=	M
[ HI ]	=	M

im pretty confused on these concentrations, if you could include steps that would be appreciated.

Answer 1

The equilibrium equation for this reaction is

K_c = [HI]² / [H₂] [I₂]

The balanced equation says that one mole of H₂ reacts with one mole of I₂ to produce 2 moles of HI.

We start out with 0.333 moles of both H₂ and I₂, but in going to equilibrium, some of each reactant is used up to produce HI. If X amount of reactant is used up, at equilibrium the amount left of each reactant is 0.333 - X = [H₂] and [I₂]. Now the reaction equation tells us that if X moles of reactant reacts, then 2X moles of HI is produced.

As the reaction is carried out in a liter container, the concentrations are equal to 0.333 and X.

The equilibrium expression can be written

55.6 = [2X]² / [0.333 - X] [0.333 - X] This equation can be solved for X using the quadratic formula. Once X is determined, calculate the equilibrium concentrations using the following

[HI] = 2X [H₂] = 0.333 - X [I₂] = 0.333 - X

I got X = 0.136 moles / L.

As always, check the math.

Answer 2

It's not that tough. Do a few, and they get easier. I think you've asked questions like this before, but here goes. Hopefully this will make sense to you so you can do others yourself. When given some of the concentrations or pressures, set up an ICE table.

H₂(g) + I₂(g) <==> 2HI

0.333.....0.333.........0.......Initial

-x...........-x.............+2x.....Change

0.333-x...0.333-x....2x......Equilibrium

We are given the value of K = 55.6, so set up the equilibrium expression:

K = [HI]² / [H₂] [I₂]

55.6 = (2x)² / (0.333-x)(0.333-x)

Now the rest is algebra and solving for x. You'll have to use the quadratic equation. If you can't do the algebra, then perhaps you should bone up on that aspect as well,