# 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, ${K}_{c} = 1.6 \times {10}^{-} 5$ for the reaction $2 N O C l \left(g\right) \setminus \leftrightarrow 2 N O \left(g\right) + C {l}_{2} \left(g\right)$

Jul 1, 2018

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

Work is incomplete; can someone check my process?

#### Explanation:

ICE table:
$2 N O C l \left(g\right) \setminus \leftrightarrow 2 N O \left(g\right) + C {l}_{2} \left(g\right)$
$\text{ " " 0 " " " " " " 2.0 " " " " " " " } 1.0$
$\text{ " "+2x" " " " "-2x" " " " " " } - x$
$\text{ " " } 2 x$$\text{ " } 2.0 - 2 x$ $\text{ " " " " } 1.0 - x$

${K}_{c} = 1.6 \times {10}^{-} 5 = \frac{\left(2.0 - 2 x\right) \left(1.0 - x\right)}{2 x}$
As ${K}_{c}$ is small, $x$ is negligible during subtraction.
$\setminus \therefore 2 x \setminus \cong \frac{2.0 \setminus \cdot 1.0}{1.6 \times {10}^{-} 5}$

As a result, the equilibrium concentrations are
$\left[N O C l\right] = 2 x \ldots$
$\left[N O\right] = 2.0 - 2 x \ldots$
$\left[C {l}_{2}\right] = 1.0 - x \ldots$

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" $\text{NOCl}$ from zero concentration).

Since ${K}_{c}$ is small for the forward direction,

$\left[\text{products}\right]$ $\text{<<}$ $\left[\text{reactants}\right]$

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 $\text{1.0 L}$ flask, the mols are equal to the concentration in molarity, which is what really goes into the ICE table:

${\text{2NOCl"(g) rightleftharpoons 2"NO"(g) + "Cl}}_{2} \left(g\right)$

$\text{I"" "0" "" "" "" "" "2.0" "" "" } 1.0$
$\text{C"" "+2x" "" "" "-2x" "" } - x$
$\text{E"" "2x" "" "" "" "2.0-2x" } 1.0 - x$

Given ${K}_{c} = 1.6 \times {10}^{- 5}$, and remembering that coefficients go into the exponents and change in concentration,

$1.6 \times {10}^{- 5} = \frac{{\left(2.0 - 2 x\right)}^{2} \left(1.0 - x\right)}{2 x} ^ 2$

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:

$\frac{1}{1.6 \times {10}^{- 5}} = 62500 = {\left(2 x\right)}^{2} / \left({\left(2.0 - 2 x\right)}^{2} \left(1.0 - x\right)\right)$

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

$62500 \left(4.0 - 8 x + 4 {x}^{2}\right) \left(1.0 - x\right) = 4 {x}^{2}$

$= 62500 \left(4.0 - 4 x - 8 x + 8 {x}^{2} + 4 {x}^{2} - 4 {x}^{3}\right)$

$\implies - 4 {x}^{3} + 12 {x}^{2} - 12 x + 4.0 = \frac{1}{62500} \cdot 4 {x}^{2}$

$\implies - {x}^{3} + 3 {x}^{2} - 3 x + 1.0 = \frac{1}{62500} \cdot {x}^{2}$

$\implies {x}^{3} - 3 {x}^{2} + 3 x - 1.0 \approx 0$

This cubic has one real solution, which is $\text{0.9752 M}$.

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 ${K}_{c}$:

$\frac{1}{62500} = 1.6 \times {10}^{- 5} = \frac{{\left(2.0 - 2 x\right)}^{2} \left(1.0 - x\right)}{2 x} ^ 2$

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

$= 1.60386 \times {10}^{- 5} \approx 1.6 \times {10}^{- 5}$ color(blue)(sqrt"")