Given #K_c = 0.36# at #2000^@ "C"# for #"N"_2"O"_4(g) rightleftharpoons "NO"_2(g)#, if the initial concentration of #"NO"_2# is #"1 M"#, what are the equilibrium concentrations of #"N"_2"O"_4(g)# and #"NO"_2(g)#?

1 Answer
Aug 23, 2017

Answer:

#"N"_2"O"_4: 0.33# #M#

#"NO"_2: 0.34# #M#

Explanation:

We're asked to find the equilibrium concentrations of #"N"_2"O"_4# and #"NO"_2#, given the initial #"NO"_2# concentration.

The equilibrium constant expression is given by

#K_c = (["NO"_2]^2)/(["N"_2"O"_4]) = ul(0.36)" "# #(2000color(white)(l)""^"o""C")#

We'll do our I.C.E. chart in the form of bullet points, for fun. Then, our initial concentrations are

INITIAL

  • #"N"_2"O"_4:# #0#

  • #"NO"_2:# #1# #M#

According to the coefficients of the reaction, the amount by which #"NO"_2# decreases is two times as much as the amount by which #"N"_2"O"_4# increases:

CHANGE

  • #"N"_2"O"_4:# #+x#

  • #"NO"_2:# #-2x#

And so the final concentrations are

FINAL

  • #"N"_2"O"_4:# #x#

  • #"NO"_2:# #1# #M# #- 2x#

Plugging these into the equilibrium constant expression gives us

#K_c = ((1-2x)^2)/(x) = ul(0.36)" "# #(2000color(white)(l)""^"o""C"#, excluding units#)#

Now we solve for #x#:

#(4x^2 - 4x + 1)/x = 0.36#

#4x^2 - 4x + 1 = 0.36x#

#4x^2 - 4.36x + 1 = 0#

Use the quadratic equation:

#x = (4.36+-sqrt((4.36)^2 - 4(4)(1)))/(8) = 0.328color(white)(l)"or"color(white)(l)0.762#

If we plug the larger solution in for #x# in the final #"NO"_2# concentration #(1-2x)#, we would obtain a worthless negative value. Therefore, we use the smaller solution to calculate the final concentrations:

#color(red)("final N"_2"O"_4) = x = color(red)(ulbar(|stackrel(" ")(" "0.33color(white)(l)M" ")|)#

#color(blue)("final NO"_2) = 1-2x = 1-2(0.328) = color(blue)(ulbar(|stackrel(" ")(" "0.34color(white)(l)M" ")|)#

each rounded to #2# significant figures.