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

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.