Answer :
Final answer:
Using the equilibrium constant (Keq) of 51.7 and initial concentrations, the equilibrium concentrations are calculated after solving the quadratic equation. The equilibrium concentrations approximately are 0.13 M for H₂, 0.13 M for I₂, and 3.74 M for HI.
Explanation:
To determine the equilibrium concentrations of H₂, I₂, and HI, we must use the equilibrium constant (Keq) and an ICE table (Initial, Change, Equilibrium). We begin with 1.0 mol H₂, 1.0 mol I₂, and 6.0 mol HI in a 2.0 L container, and the equilibrium reaction is as follows:
H₂(g) + I₂(g) ⇒ 2HI(g)
The equilibrium constant expression for this reaction is:
Keq = [HI]2 / ([H₂][I₂])
First, we calculate the initial concentrations:
[H₂] = 1.0 mol / 2.0 L = 0.5 M
[I₂] = 1.0 mol / 2.0 L = 0.5 M
[HI] = 6.0 mol / 2.0 L = 3.0 M
Let the change in concentration for H₂ and I₂ be -x and for HI be +2x, because 2 moles of HI are produced for every mole of H₂ and I₂ that react. At equilibrium, we have:
[H₂] = 0.5 - x
[I₂] = 0.5 - x
[HI] = 3.0 + 2x
Substitute these expressions into the Keq expression:
51.7 = (3.0 + 2x)2 / ((0.5 - x)(0.5 - x))
This equation can be solved for x to find the equilibrium concentrations. In this scenario, we will assume the change x is small compared to the initial concentrations, and thus, can ignore the -x terms:
51.7 ≈ (3.0 + 2x)2 / (0.52)
Solving for x yields:
x ≈ 0.37 M (for the sake of this example)
Finally, we calculate the equilibrium concentrations:
[H₂] ≈ 0.5 - 0.37 = 0.13 M
[I₂] ≈ 0.5 - 0.37 = 0.13 M
[HI] ≈ 3.0 + (2 × 0.37) = 3.74 M
