# Question c3b18

May 28, 2017

["CH"_2"O"] = "0.0200 M"

#### Explanation:

All you have to do here is to write the expression for the equilibrium constant that describes this equilibrium.

The equilibrium constant tells you the ratio that exists between the product of the concentrations of the products raised to the power of their respective stoichiometric coefficients and product of the concentrations of the reactants raised to the power of their respective stoichiometric coefficients.

So

${K}_{c} = \text{product of concentrations for products"/"product of concentrations for reactants}$

In your case, you have a single product and $2$ reactants, so the equilibrium constant will be equal to

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

Now, you know that

${K}_{c} = 1$

This tells you that, at equilibrium, you have

$\left[{\text{CH"_2"O"] = ["CO"] * ["H}}_{2}\right]$

Plug in your values to find

["CH"_2"O"] = "0.200 M" * "0.100 M"

["CH"_2"O"] = color(darkgreen)(ul(color(black)("0.0200 M")))#