# What volume flask should be used to get 0.37 moles I present for every mole I_2 at equilibrium?

## For the dissociation of ${I}_{2} \text{(g)}$ at 1200°C, ${K}_{c} = 0.011$. What volume flask should we use if we want 0.37 moles of $I$ to be present for every mole of ${I}_{2}$ present at equilibrium? ${I}_{2} \text{(g)"\rightleftharpoons2I"(g)}$ My work: (can't see? it sets equilibrium pressures as stated below:) ${P}_{I} = \setminus \frac{0.37 \cdot 0.0821 \left(1200 + 273\right)}{V} = \setminus \frac{0.37 \cdot 121}{V} \text{ atm}$ ${P}_{{I}_{2}} = \setminus \frac{1.00 \cdot 0.0821 \left(1200 + 273\right)}{V} = \setminus \frac{1.00 \cdot 121}{V} = \frac{121}{V} \text{ atm}$ after using those pressures for ${K}_{p}$ in the formula $\setminus \textcolor{red}{{K}_{c} = {K}_{p} {\left(R T\right)}^{\setminus \Delta n}}$, I got $V = \frac{0.0307}{0.011} \setminus \approx 0.279 \text{ L}$

Jul 19, 2018

Are we missing information? We need to be able to figure out the exact number of mols of $\text{I}$ or ${\text{I}}_{2}$... no ratios, an actual number.

We begin with the ICE table:

$\text{I"_2(g) " "rightleftharpoons" " 2"I} \left(g\right)$

$\text{I"" "["I"_2]_i" "" "" "" } 0$
$\text{C"" "-x" "" "" } + 2 x$
$\text{E"" "["I"_2]_i-x" "" } 2 x$

So we write the mass action expression to get:

${K}_{c} = 0.011 = \left(\left[{\text{I"]^2)/(["I}}_{2}\right]\right)$

$= {\left(2 x\right)}^{2} / \left({\left[{\text{I}}_{2}\right]}_{i} - x\right)$

What we apparently want is

(2x)/(["I"_2]_i - x) = 0.37 = (["I"])/(["I"_2]) = ("mols I"//cancel"L")/("mols I"_2//cancel"L")

We now have a system of equations:

${\left(2 x\right)}^{2} / \left({\left[{\text{I}}_{2}\right]}_{i} - x\right) = 0.011$ $\text{ "" } \boldsymbol{\left(1\right)}$
(Note that $0.011$ is in implied units of $\text{M}$.)

$\frac{2 x}{{\left[{\text{I}}_{2}\right]}_{i} - x} = 0.37$ $\text{ "" } \boldsymbol{\left(2\right)}$
(Note that $0.37$ is unitless.)

By inspection, we have:

${\left(2 x\right)}^{2} / \left({\left[{\text{I}}_{2}\right]}_{i} - x\right) = 0.011 = 0.37 \left(2 x\right) = 0.74 x$

So,

x = ("0.011 M")/0.74 = ul"0.0149 M"

As a result, take $\left(2\right)$ to get

$\left(2 \cdot \text{0.0149 M")/(["I"_2]_i - "0.0149 M}\right) = 0.37$

2 cdot "0.0149 M" = 0.37(["I"_2]_i - "0.0149 M")

$\text{0.0297 M" = 0.37["I"_2]_i - "0.0055 M}$

Therefore, the initial concentration of ${\text{I}}_{2}$ is:

color(blue)(["I"_2]_i) = ("0.0352 M")/(0.37) = color(blue)("0.0952 M")

and the fraction of dissociation is

$\alpha = \frac{x}{{\left[{\text{I}}_{2}\right]}_{i}} = 0.1565$

Now we just need to know the ACTUAL mols of either $\text{I}$ or ${\text{I}}_{2}$... the ratio desired can be ANY NUMBER of combinations of quantities...

The fraction of dissociation $\alpha$ and the concentration lost $x$ also vary with concentration ${\left[{\text{I}}_{2}\right]}_{i}$, so we cannot use those to determine the volume.