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
Answer:
Explanation:
We're asked to find the equilibrium concentrations of
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
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 = ((12x)^2)/(x) = ul(0.36)" "# #(2000color(white)(l)""^"o""C"# , excluding units#)#
Now we solve for
#(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
#color(red)("final N"_2"O"_4) = x = color(red)(ulbar(stackrel(" ")(" "0.33color(white)(l)M" "))#
#color(blue)("final NO"_2) = 12x = 12(0.328) = color(blue)(ulbar(stackrel(" ")(" "0.34color(white)(l)M" "))# each rounded to
#2# significant figures.