# Question #a2c21

##### 1 Answer
Mar 21, 2015

The equilibrium concentration of $C O$ will be $8.6 \cdot {10}^{- 3} \text{M}$.

Before actually doing any calculations, take a second to look at the equilibrium constant for this reaction, ${K}_{c}$. Notice that ${K}_{c}$ is 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, $C O$ and ${O}_{2}$, will be very small in comparison to the concentration of the reactant, $C {O}_{2}$.

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

$C = \frac{n}{V} = \text{2 moles"/"5 L" = "0.4 mol/L}$

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

....$\textcolor{red}{2} C {O}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s \textcolor{b l u e}{2} C {O}_{\left(g\right)} + {O}_{2 \left(g\right)}$
I.....0.4.....................0..............0
C...$\textcolor{red}{\text{-2}} x$...................$\textcolor{b l u e}{\text{+2}} x$.........+x
E...(0.4-2x).............2x...............x

Use the definition of the equilibrium constant to solve for $x$.

${K}_{c} = \frac{{\left[C O\right]}^{2} \cdot \left[{O}_{2}\right]}{{\left[C {O}_{2}\right]}^{2}} = \frac{{\left(2 x\right)}^{2} \cdot x}{0.4 - 2 x} ^ \left(2\right) = \frac{4 {x}^{3}}{0.4 - 2 x} ^ \left(2\right)$

Because ${K}_{c}$ is so small, you can neglect the $2 x$ term in the denominator and simply the calculations significantly. This will get you

${K}_{c} = \frac{4 {x}^{3}}{0.4} ^ \left(2\right) = 2.0 \cdot {10}^{- 6}$

The value of $x$ will be $x = 0.004308$, which means that the equilibrium concentration of $C O$ will be

$\left[C O\right] = 2 \cdot x = 2 \cdot 0.004308 = \text{0.008616 M}$

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

$\left[C O\right] = \textcolor{red}{8.6 \cdot {10}^{- 3} \text{M}}$

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 $8.5 \cdot {10}^{- 3} \text{M}$.