For the equilibrium:
H2 + I2 <=> 2HI
The equilibrium konstant Kc is equal to 50.5 at 448ºC. If [H2]o=[I2] o= 1 mol/l. What will the concentration of H2, I2 and HI be at equilibrium
K= [2x]^2/(1-x)(1-x)
K= 4x^2/(1-2x+x^2)
(K-4)x^2-2Kx+K=0
solve for x with the quadratic equation
concentration of H2, I2=x
concentration of HI= 2x
To determine the concentrations of H2, I2, and HI at equilibrium, we'll use the given equilibrium constant (Kc) and the initial concentrations of H2 and I2. The balanced equation is:
H2 + I2 ⇌ 2HI
Let's set up an ICE table to track the changes in concentration:
Initial: [H2] = [I2] = 1 mol/l, [HI] = 0 mol/l
Change: -x -x +2x
Equilibrium:[1-x] [1-x] [2x]
Here, x represents the change in concentration. Since one mole of H2 reacts with one mole of I2 to produce two moles of HI, we can subtract x from the initial concentration of H2 and I2, and add 2x to the initial concentration of HI at equilibrium.
The equilibrium constant expression for the reaction is given as:
Kc = [HI]^2 / ([H2] * [I2])
Substituting the equilibrium concentrations into the equation:
50.5 = ([2x]^2) / ([1-x] * [1-x])
Simplifying the equation:
50.5 = 4x^2 / (1-x)^2
Now, we can rearrange the equation to solve for x:
50.5(1-x)^2 = 4x^2
Expanding the square:
50.5(1-2x+x^2) = 4x^2
Multiplying out:
50.5 - 101x + 50.5x^2 = 4x^2
Rearranging:
46.5x^2 - 101x + 50.5 = 0
To solve this quadratic equation, you can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For the equation, a = 46.5, b = -101, and c = 50.5. Substituting these values into the formula will give you two possible values for x.
Once you find the value(s) of x, substitute it back into the expression for equilibrium concentrations to get the concentrations of H2, I2, and HI at equilibrium.