Question #ef994
1 Answer
Here's what I got.
Explanation:
Start by calculating the initial concentrations of the two reactants.
Since you're working with a
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
We're still working with a
#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.