Question #32cda

1 Answer
Jul 15, 2017

Answer:

#K_c = 1.77 xx 10^-3#

Explanation:

We're asked to calculate the #K_c# for a reaction at a certain temperature.

Let's first write the equilibrium constant expression for this reaction:

#K_c = (["NO"]^2["O"_2])/(["NO"_2]^2)#

We can create a makeshift I.C.E. chart via bullet points as follows:

Initial Concentrations:

  • #"NO"_2#: #(0.240color(white)(l)"mol")/(2.00color(white)(l)"L") = 0.120M#

  • #"NO"#: #0#

  • #"O"_2#: #0#

because there are no #"NO"# or #"O"_2# initially.

From the coefficients of the reaction equation, we expect the changes to be

Change in concentration:

  • #"NO"_2#: #-2x#

  • #"NO"#: #+2x#

  • #"O"_2#: #+x#

Knowing that the equilibrium concentration of #"NO"_2# is

#(0.179color(white)(l)"mol")/(2.00color(white)(l)"L") = 0.0895M#

We can calculate the actual change in concentration of #"NO"_2#:

#"change" = 0.120M - 0.0895M = 0.0305M#

Which means #x# is equal to

#(0.0305M)/(2) = color(red)(0.01525M#

So the equilibrium concentrations are

Equilibrium Concentrations:

  • #"NO"_2#: #0.0895M#

  • #"NO"#: #2(color(red)(0.01525M)) = 0.0305M#

  • #"O"_2#: #color(red)(0.01525M#

We now plug these into the #K_c# expression:

#K_c = ((0.0305M)^2(0.01525M))/((0.0895M)^2) = color(blue)(1.77 xx 10^-3#

Which agrees with our calculations that equilibrium lies to the left (more reactants present).