Answer:
0.0184
Explanation:
There is some info missing. I think this is the original question.
Gaseous hydrogen iodide is placed in a closed container at 425 °C, where it partially decomposes to hydrogen and iodine: 2 HI(g) ⇄ H₂(g) + I₂(g). At equilibrium, it is found that HI = 3.59 × 10⁻³ M, H₂= 4.87 × 10⁻⁴ M, and I₂= 4.87 × 10⁻⁴ M.
What is the value of Kc at this temperature?
Express the equilibrium constant to three significant digits.
Let's consider the following reaction at equilibrium.
2 HI(g) ⇄ H₂(g) + I₂(g)
The concentration equilibrium constant (Kc) is the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.
Kc = [H₂] × [I₂]/[HI]²
Kc = 4.87 × 10⁻⁴ × 4.87 × 10⁻⁴/ (3.59 × 10⁻³)²
Kc = 0.0184
The value of Kc at this temperature is 0.0184.
The following equation should be used at equilibrium
2 HI(g) ⇄ H₂(g) + I₂(g)
Here the Kc i.e. concentration equilibrium represents the multiplication of the concentration of the product that should be raised to the coefficient.
So,
Kc = [H₂] × [I₂]/[HI]²
Kc = 4.87 × 10⁻⁴ × 4.87 × 10⁻⁴/ (3.59 × 10⁻³)²
Kc = 0.0184
hence, The value of Kc at this temperature is 0.0184.
Learn more about temperature here: https://brainly.com/question/9636950
