Question #9ad66

1 Answer
Jan 1, 2018

Answer:

#K_c = 5.0 * 10^(-4)#

Explanation:

For the sake of simplicity, I will use the following notation

  • #"CCl"_3"CH"("OH")_2 -> "reactant"#
  • #"CCl"_3"CHO" -> "product"#

The balanced chemical equation that describes this equilibrium will be--keep in mind that you have a #1:1# mole ratio between chloral hydrate and chloral!

#"reactant"_ ((sol)) rightleftharpoons "product"_ ((sol)) + "H"_ 2"O"_ ((sol))#

I used #("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 #1# mole of chloral hydrate that dissociates, you get #1# mole of chloral and #1# mole of water.

The initial concentration of chloral hydrate is

#["reactant"]_ 0 = "0.010 moles"/"1 L" = "0.010 M"#

The problem tells you that the equilibrium concentration of water is equal to #"0.0020 M"#. This means that in order for the reaction to produce #"0.0020 M"# of water, it must also produce #"0.0020 M"# of chloral and consume #"0.0020 M"# of chloral hydrate.

So the equilibrium concentration of chloral will be

#["product"] = ["H"_ 2"O"] = "0.0020 M"#

and the equilibrium concentration of chloral hydrate will be

#["reactant"] = ["reactant"]_ 0 - "0.0020 M"#

#["reactant"] = "0.00080 M"#

By definition, the equilibrium constant will be equal to

#K_c = (["product"] * ["H"_2"O"])/(["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.