To solve this problem, we need to use the concept of equilibrium and the expression for the equilibrium constant. We'll start by setting up an ICE table to determine the equilibrium concentrations.
Given:
- Initial concentration of HI (H2I) = 1 mole
- Volume of the container = 3.0 L
Let's assume the equilibrium concentration of HI is x, which means the concentrations of H2 and I2 will also be x because the stoichiometric coefficient for all three species is 1.
The ICE table is as follows:
H2(g) + I2 (g) <=> 2 HI(g)
Initial: x x 1
Change: -x -x +2x
Equilibrium: 1-x 1-x 1 + 2x
Now, let's substitute these equilibrium concentrations into the Kc expression:
Kc = [HI]^2 / [H2][I2]
Kc = (1 + 2x)^2 / (1 - x)(1 - x)
Since we are given that Kc = 54, we can write the equation:
54 = (1 + 2x)^2 / (1 - x)(1 - x)
Now, let's solve this equation.
Take the square root of both sides:
√54 = 1 + 2x / (1 - x)(1 - x)
Simplify:
√54 = (1 + 2x) / (1 - x)^2
Now, let's cross-multiply:
√54 * (1 - x)^2 = 1 + 2x
Expand and simplify:
54(1 - 2x + x^2) = 1 + 2x
54 - 108x + 54x^2 = 1 + 2x
Rearrange and simplify:
54x^2 - 110x + 53 = 0
Now, we have a quadratic equation. To solve for x, we can either factor or use the quadratic formula. In this case, factoring is more difficult, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from the quadratic equation:
x = (-(-110) ± √((-110)^2 - 4(54)(53))) / (2(54))
Simplifying:
x = (110 ± √(12100 - 11664)) / 108
x = (110 ± √436) / 108
Now, we can calculate the two possible values of x:
x = (110 + √436) / 108 ≈ 0.331
or
x = (110 - √436) / 108 ≈ 0.162
Since we are looking for the equilibrium concentration of HI, we choose the positive value:
x ≈ 0.331
Therefore, the equilibrium concentration of HI is approximately 0.331 M.
The closest answer choice is e. 0.33, so that would be the correct option.