# Question ef994

Apr 20, 2017

Here's what I got.

#### Explanation:

Start by calculating the initial concentrations of the two reactants.

Since you're working with a $\text{1.00-L}$ container, you can treat the number of moles and the concentration interchangeably.

You will thus have

["O"_ 2]_ 0 = "0.0560 M"

["N"_ 2"O"]_ 0 = "0.020 M"

Now, you know that you have

$2 {\text{N"_ 2"O"_ ((g)) + 3"O"_ (2(g)) rightleftharpoons 4"NO}}_{2 \left(g\right)}$

By definition, the equilibrium constant for this equilibrium reaction is equal to

${K}_{c} = \left({\left[{\text{NO"_2]^4)/(["N"_2"O"]^2 * ["O}}_{2}\right]}^{3}\right)$

At equilibrium, you know that

["NO"_2] = "0.020 M"

According to the balanced chemical reaction, every $2$ moles of nitrous oxide that take part in the reaction will consume $3$ moles of oxygen gas and produce $4$ moles of nitrogen dioxide.

We're still working with a $\text{1.00-L}$ container, which means that the reaction produced $0.020$ moles of nitrogen dioxide. You can thus say that it must have consumed

0.020 color(red)(cancel(color(black)("moles NO"_2))) * ("2 moles N"_2"O")/(4color(red)(cancel(color(black)("moles NO"_2)))) = "0.010 moles N"_2"O"

and

0.020 color(red)(cancel(color(black)("moles NO"_2))) * "3 moles O"_2/(4color(red)(cancel(color(black)("moles NO"_2)))) = "0.015 moles O"_2

You can thus say that when the equilibrium was established, the reaction consumed

["N"_ 2"O"]_ "consumed" = "0.010 M"

["O"_ 2]_ "consumed" = "0.015 M"

Therefore, the equilibrium concentrations of the two reactants are

["N"_ 2"O"] = ["N"_ 2"O"]_ 0 - ["N"_ 2"O"]_ "consumed"

["N"_ 2"O"] = "0.020 M" - "0.010 M"

["N"_ 2"O"] = "0.010 M"

and

["O"_ 2] = ["O"_ 2]_ 0 - ["O"_ 2]_ "consumed"

["O"_ 2] = "0.0560 M" - "0.015 M"

["O"_ 2] = "0.041 M"#

You are now ready to calculate the equilibrium constant -- I'll skip the units for simplicity

${K}_{c} = {0.020}^{4} / \left({0.010}^{2} \cdot {0.041}^{3}\right) = 23$

The answer is rounded to two sig figs, the number of sig figs you have for the equilibrium concentration of nitrogen dioxide.