If 2.0 mol #NO# and 1.0 mol #Cl_2# are placed in a 1.0-L flask, calculate the equilibrium concentrations of all species?

(I am aware this likely isn't the proper constant. But I'd like to know if I did everything else correctly...)

At #35°C#, #K_c=1.6xx10^-5# for the reaction
#2NOCl(g)\leftrightarrow2NO(g)+Cl_2(g)#

2 Answers
Jul 1, 2018

#[NOCl]=?#
#[NO]=?#
#[Cl_2]=?#

Work is incomplete; can someone check my process?

Explanation:

ICE table:
#2NOCl(g)\leftrightarrow2NO(g)+Cl_2(g)#
#" " " 0 " " " " " " 2.0 " " " " " " " "1.0#
#" " "+2x" " " " "-2x" " " " " " "-x#
#" " " "2x##" " " 2.0-2x# #" " " " " "1.0-x#

#K_c=1.6xx10^-5=((2.0-2x)(1.0-x))/(2x)#
As #K_c# is small, #x# is negligible during subtraction.
#\therefore2x\cong(2.0\cdot1.0)/(1.6xx10^-5)#

As a result, the equilibrium concentrations are
#[NOCl]=2x...#
#[NO]=2.0-2x...#
#[Cl_2]=1.0-x...#

Jul 1, 2018

#color(blue)(["NO"]) = 2.0 - 2("0.9752 M") = color(blue)("0.0496 M")#
#color(blue)(["Cl"_2]) = 1.0 - 0.9752 = color(blue)("0.0248 M")#
#color(blue)(["NOCl"]) = 2(0.9752) = color(blue)("1.9504 M")#

It is important to realize that #K_c# as-given is for the forward reaction, and the reasonable direction to proceed is the reverse reaction (you cannot lose "reactant" #"NOCl"# from zero concentration).

Since #K_c# is small for the forward direction,

#["products"]# #"<<"# #["reactants"]#

is expected.

Hence, what we should have done is the following.


Gas-phase equilibrium constants aren't as well-known, so it's fine to use the one you are given. (#K_a# and #K_b# have been determined for many acids and bases already.)

Since these gases are in a #"1.0 L"# flask, the mols are equal to the concentration in molarity, which is what really goes into the ICE table:

#"2NOCl"(g) rightleftharpoons 2"NO"(g) + "Cl"_2(g)#

#"I"" "0" "" "" "" "" "2.0" "" "" "1.0#
#"C"" "+2x" "" "" "-2x" "" "-x#
#"E"" "2x" "" "" "" "2.0-2x" "1.0-x#

Given #K_c = 1.6 xx 10^(-5)#, and remembering that coefficients go into the exponents and change in concentration,

#1.6 xx 10^(-5) = ((2.0-2x)^2(1.0-x))/(2x)^2#

However, it is unreasonable to use the small #x# approximation here, because #K_c# for the forward direction is small, NOT the reverse direction.

The way this reaction is written, it is clearer to flip everything to get:

#1/(1.6xx10^(-5)) = 62500 = (2x)^2/((2.0-2x)^2(1.0-x))#

And so, instead, we have to say that #x# is NOT small. This would have to be expanded in full.

#62500(4.0 - 8x + 4x^2)(1.0 - x) = 4x^2#

#= 62500(4.0 - 4x - 8x + 8x^2 + 4x^2 - 4x^3)#

#=> -4x^3 + 12x^2 - 12x + 4.0 = 1/62500 cdot 4x^2#

#=> -x^3 + 3x^2 - 3x + 1.0 = 1/62500 cdot x^2#

#=> x^3 - 3x^2 + 3x - 1.0 ~~ 0#

This cubic has one real solution, which is #"0.9752 M"#.

#color(blue)(["NO"]) = 2.0 - 2("0.9752 M") = color(blue)("0.0496 M")#
#color(blue)(["Cl"_2]) = 1.0 - 0.9752 = color(blue)("0.0248 M")#
#color(blue)(["NOCl"]) = 2(0.9752) = color(blue)("1.9504 M")#

Verify #K_c#:

#1/62500 = 1.6 xx 10^(-5) = ((2.0-2x)^2(1.0-x))/(2x)^2#

#stackrel(?" ")(=) ((0.0496^2)(0.0248))/(1.9504^2)#

#= 1.60386 xx 10^(-5) ~~ 1.6 xx 10^(-5)# #color(blue)(sqrt"")#