# Question 9ad66

Jan 1, 2018

${K}_{c} = 5.0 \cdot {10}^{- 4}$

#### Explanation:

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

• $\text{CCl"_3"CH"("OH")_2 -> "reactant}$
• $\text{CCl"_3"CHO" -> "product}$

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!

${\text{reactant"_ ((sol)) rightleftharpoons "product"_ ((sol)) + "H"_ 2"O}}_{\left(s o l\right)}$

I used $\left(\text{sol}\right)$ 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 $\text{0.0020 M}$. This means that in order for the reaction to produce $\text{0.0020 M}$ of water, it must also produce $\text{0.0020 M}$ of chloral and consume $\text{0.0020 M}$ 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

${K}_{c} = \left(\left[\text{product"] * ["H"_2"O"])/(["reactant}\right]\right)$

Keep in mind that water is a solute here, so its concentration must be included in the expression of the equilibrium constant.

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{K}_{c}}}} = \frac{0.0020 \cdot 0.0020}{0.0080} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{5.0 \cdot {10}^{- 4}}}}$