# Question #a2c21

##### 1 Answer

The equilibrium concentration of

Before actually doing any calculations, take a second to look at the equilibrium constant for this reaction, *much smaller than one*, which means that the equilibrium will lie far to the left, so to speak.

In other words, **at equilibrium**, the concentrations of the products, *in comparison* to the concentration of the reactant,

To actually solve this problem you'll have to use the **ICE Chart method** (more here: http://en.wikipedia.org/wiki/RICE_chart).

Calculate the initial concentration of the reactant by using the number of moles and the volume of the vessel given

Now use the ICE table - remember that each species' concentration changes *proportional* to their respective stoichiometric coefficients

....

**I**.....0.4.....................0..............0

**C**...

**E**...(0.4-2x).............2x...............x

Use the definition of the equilibrium constant to solve for

Because

The value of

Rounded to two sig figs, although it would be better if the answer was rounded to one sig fig - this is because of the number of sig figs in 2 moles and 5 liters, the answer will be

**SIDE NOTE** *If you don't neglect the 2x term in the denominator, the value of x will be 0.00425, which will make the equilibrium concentration of CO equal to*