Question #ef994

1 Answer
Apr 20, 2017

Answer:

Here's what I got.

Explanation:

Start by calculating the initial concentrations of the two reactants.

Since you're working with a #"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"N"_ 2"O"_ ((g)) + 3"O"_ (2(g)) rightleftharpoons 4"NO"_ (2(g))#

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

#K_c = (["NO"_2]^4)/(["N"_2"O"]^2 * ["O"_2]^3)#

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 #"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/( 0.010^2 * 0.041^3) = 23#

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