# How many mols of "CO"_2 were added at constant temperature and volume to a "1-L" container containing "0.400 mols" of "CO"_2 and of "H"_2, and "0.200 mols" of "CO" and of "H"_2"O"(g) (K_c = 4) to cause "CO"_2 to drop to "0.300 mols"?

##### 1 Answer
Jun 23, 2017

I got ${\text{0.900 mols CO}}_{2}$ added. At constant temperature and volume, the mols added only affect the partial pressures of each gas and thus their concentrations.

This is just asking you to:

1. Determine the equilibrium constant for an initial state.
2. Use this previous equilibrium configuration as your new initial state, where the amount of ${\text{CO}}_{2}$ you now start with is plus an unknown amount, $y$
3. The reaction will go backwards to reach its second equilibrium configuration, and thus the change in mols, $x$, will have the reversed sign compared to the forward reaction.

Since the volume of the container is shared, we can use just the mols of everything instead of the molarity. The first equilibrium constant will stay the same, and thus the starting equilibrium configuration has

${K}_{c} = \left(\left[\text{CO"_2]["H"_2])/(["CO"]["H"_2"O}\right]\right) = \frac{{0.400}^{2}}{{0.200}^{2}} = 4$

We want to add ${\text{CO}}_{2}$, so we must have that the equilibrium shifts left by Le Chatelier's Principle, right, because adding product consumes the product and pushes the reaction backwards to form more reactants.

Your initial ICE table should look like this (after reaching the first equilibrium configuration already, and adding the ${\text{CO}}_{2}$ to disturb it, with $x$'s presumed unknown):

${\text{CO"(g) " "+" " "H"_2"O"(g) " "rightleftharpoons" " "CO"_2(g) + "H}}_{2} \left(g\right)$

$\text{I"" "0.200" "" "" "0.200" "" "" "" "0.400 + y" "" } 0.400$
$\text{C"" "+x" "" "" "+x" "" "" "" "" "" "-x" "" } - x$
$\text{E"" "0.200 + x" "0.200 + x" "0.400 + y - x" } 0.400 - x$

Now, we want the final mols of $\text{CO}$ to be $0.300$ AFTER this second equilibrium configuration is established, so we know that:

${n}_{C O} = 0.200 + x = 0.300$

Thus, $x = 0.100$, $0.400 + y - x = 0.300 + y$, etc, and we assume that we get the same equilibrium constant...

$\frac{\left(0.300 + y\right) \left(0.300\right)}{\left(0.300\right) \left(0.300\right)} = 4$

Solve for the amount of ${\text{CO}}_{2}$ added, $y$:

$\frac{4 \left({0.300}^{2}\right)}{0.300} = 0.300 + y$

$\implies \textcolor{b l u e}{y} = \frac{4 \left({0.300}^{2}\right)}{0.300} - 0.300 = \textcolor{b l u e}{\text{0.900 mols CO"_2 " added}}$