Respuesta :
Answer:
[H₂] = 0.178 M
[I₂] = 0.184 M
[HI] = 1.415 M
Explanation:
For the equilibrium:
H₂(g) + I₂(g) ⇄ 2 HI(g)
the equilibrium constant is given by the equation:
Kc = [ HI]² / [H₂][I₂]
Lets use first the reaction quotient which has the same expression as the equilibrium constant to predict the direction the reaction will take, i.e towards reactants or product side.
Q =( 0.895)²/(0.438)(0.444) = 4.12
Q is less than Kc so the reaction will favor the product side.
We can set up the following table to account for all the species at equilibrium:
H₂ I₂ HI
initial 0.438 0.444 0.895
change -x -x +2x
equilibrium 0.438 - x 0.444 - x 0.895 + 2x
Now we are in position to express these concentrations in terms of the equilibrium conctant, Kc
54.3 = (0.895 + 2x)² / (0.438 -x)(0.444 - x)
performing the calculatiopns will result in a quadratic equation:
0.801 + 3.580x +4x² = (0.194 - 0.882x + x²)x 54.3
Upon rearrangement and some algebra, we have
0.801 + 3.580 x + 4x² = 10.534 - 47.893x + 54.3 x²
0 = 9.733 - 51.473 x + 54.3 x²
This equation has two roots X₁ = 0.687 and X₂ = 0.26
The first is physically impossible since it will imply that more 0.687 will make the quantity at equilibrium for both H₂ and I₂ negative.
Therefore the concentrations at equilibrium of each gas are:
[H₂] = (0.438 - 0.260) = 0.178 M
[I₂] = (0.444 - 0.260) M = 0.184 M
[HI] = [0.895 + 2x(0.260)] M = 1.415 M
Note if we plug these values into the equilibrium expression we get 61 which is due to the rounding errors propagating in the quadratic equation.