Question #9ad66
1 Answer
Explanation:
For the sake of simplicity, I will use the following notation
"CCl"_3"CH"("OH")_2 -> "reactant"CCl3CH(OH)2→reactant "CCl"_3"CHO" -> "product"CCl3CHO→product
The balanced chemical equation that describes this equilibrium will be--keep in mind that you have a
"reactant"_ ((sol)) rightleftharpoons "product"_ ((sol)) + "H"_ 2"O"_ ((sol))reactant(sol)⇌product(sol)+H2O(sol) I used
("sol")(sol) to denote the fact that the compounds are dissolved in a solution that does not have water as the solvent.
Now, you know that for every
The initial concentration of chloral hydrate is
["reactant"]_ 0 = "0.010 moles"/"1 L" = "0.010 M"[reactant]0=0.010 moles1 L=0.010 M
The problem tells you that the equilibrium concentration of water is equal to
So the equilibrium concentration of chloral will be
["product"] = ["H"_ 2"O"] = "0.0020 M"[product]=[H2O]=0.0020 M
and the equilibrium concentration of chloral hydrate will be
["reactant"] = ["reactant"]_ 0 - "0.0020 M"[reactant]=[reactant]0−0.0020 M
["reactant"] = "0.00080 M"[reactant]=0.00080 M
By definition, the equilibrium constant will be equal to
K_c = (["product"] * ["H"_2"O"])/(["reactant"])Kc=[product]⋅[H2O][reactant] Keep in mind that water is a solute here, so its concentration must be included in the expression of the equilibrium constant.
Plug in your values to find--I'll leave the answer without added units.
color(darkgreen)(ul(color(black)(K_c))) = (0.0020 * 0.0020)/(0.0080) = color(darkgreen)(ul(color(black)(5.0 * 10^(-4))))
The answer is rounded to two sig figs.
