At a high temperature, the equilibrium constant for the decomposition of hydrogen iodide is 65.0. If the initial concentration of HI is 1.60 M, what is the concentration of hydrogen at equilibrium?
2Hl(g) H2(g) + I2(g)
A. 1.42 M
B. 0.753 M
C. 0.240 M
D. 0.802 M
I'm actually not sure how to work through this question at all so I don't have any ideas as to which answer it is. Can someone show me how to work through it?
Dr. Bob, your responses are not good whatsoever.
0.753
To solve this question, we will use the concept of the equilibrium constant (Kc) and the stoichiometry of the balanced chemical equation. The equilibrium constant is calculated as the concentration of the products divided by the concentration of the reactants, each raised to the power of their stoichiometric coefficients.
In this case, the equilibrium constant expression for the decomposition of hydrogen iodide is:
Kc = [H2][I2] / [HI]^2
Given: Kc = 65.0 and [HI] = 1.60 M (initial concentration)
Since the initial concentration of H2 and I2 is zero, we can assume their concentrations at equilibrium as "x."
Plugging these values into the equilibrium constant expression, we get:
65.0 = (x)(x) / (1.60 - x)^2
Simplifying the equation, we have:
65.0 = x^2 / (1.60 - x)^2
To solve for "x," we can cross-multiply and rearrange the equation:
65.0(1.60 - x)^2 = x^2
104.0 - 104x + 65x^2 = x^2
65x^2 + x^2 - 104x - 104 = 0
Combining like terms, we have a quadratic equation:
66x^2 - 104x - 104 = 0
Using the quadratic formula, we can solve for "x":
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
Plugging in the values for this equation, with a = 66, b = -104, and c = -104, we can solve for "x."
......2HI ==> H2 + I2
I.....1.60.....0....0
C.....-2x......x....x
E....1.60-2x...x....x
Substitutethe E line into the Keq expression and solve for x.