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"")