A5.00l reaction vessel is filled with 1.00mol of h2, 1.00mol of i2, and 2.50mol of hi. if equilibrium constant for the reaction is 129 at 500k, what are the equilibrium concentrations of all species?

[I₂] ; [H₂]  = 0.067 mol/L

[HI] = 0.765 mol/L

Explanation:

This is the reaction:

                  I₂(g)    +    H₂(g)    ⇄    2HI (g)

Initially       1mol          1mol            2.5mol

I have the initially amount of each gas at first. Some amount (x), has reacted during the reaction.

                  I₂(g)    +    H₂(g)    ⇄    2HI (g)

React            x               x                  2x

In the equilibrium I have to subtract, what I had initially and the amount that has reacted, and then, as I had 2.5 mol of HI at the begining, I have to sum, the amount of the reaction. As I have to find out the concentrations, I have to /5L, which is the volume of the vessel.

Eq.           (1-x)/5         (1-x)/5         (2.5+2x)/5

Let's make the expression for Kc

Kc = [HI]² / ( [I₂] . [H₂])

129 =  ( (2.5+2x) /5 )² / ( (1-x) /5) . ( (1-x) /5) )

129 =   ( (2.5+2x) /5 )² / ( (1-x) /5)²

129 = (2.5+2x)² / 25  /  (1-x)² / 25

129 = (2.5 + 2x)² / (1-x)²

129 = 2.5² + 2 . 2.5 .2x + 4x² / 1 - 2x + x²

129 (1 - 2x + x²) = 2.5² + 2 . 2.5 .2x + 4x²

129 - 258x + 129x² = 6.25 + 10x + 4x²

129 - 6.25 -258x -10x + 129x²-4x² = 0

122.75 - 268x + 125x² = 0 (a quadratic function)

a = 125

b = -268

c = 122.75

(-b +- (√b² - 4ac)) / 2a

x₁ = 1.48

x₂ = 0.66

We take the x₂ value, cause the x₁, will get a negative concentration and that is impossible.

[I₂] =  ( 1-0.66 ) /5 = 0.067

[H₂] = ( 1-0.66 ) /5 = 0.067

[HI] = (2.5+ 2. 0.66) /5 = 0.765

they both have a y in them

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.