Question

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)`

Answer 1

`[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...`

Answer 2

`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"")`