# Question 7a92c

Oct 2, 2015

${P}_{I} = \text{0.25 atm}$
${P}_{{I}_{2}} = \text{0.27 atm}$

#### Explanation:

The idea here is that you need to use the initial equilibrium partial pressures to find the equilibrium constant, ${K}_{p}$.

After that, you will need to use Boyle's Law to find the total pressure of the system after the volume of the container is halved.

So, the equilibrium reaction looks like this

${\text{I"_text(2(g]) rightleftharpoons color(red)(2)"I}}_{\textrm{\left(g\right]}}$

The total pressure of the system is equal to the sum of the partial pressure of its components

${P}_{\text{total}} = {P}_{{I}_{2}} + {P}_{I}$

${P}_{\text{total" = 0.10 + 0.16 = "0.26 atm}}$

When you decrease volume while keeping the number of moles and temperature constant, pressure will increase proportionally - this is known as Boyle's Law.

So, for the initial state of the system, you have

${P}_{1} \cdot {V}_{1} = n \cdot R \cdot T$

for the final state of the system, you have

${P}_{2} \cdot {V}_{2} = n \cdot R \cdot T$

Divide these wto equations to get the equation for Boyle's Law

$\frac{{P}_{1} {V}_{1}}{{P}_{2} {V}_{2}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{n \cdot R \cdot T}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{n \cdot R \cdot T}}}} \implies {P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

But you also know that ${V}_{2} = {V}_{1} / 2$, since the volume is halved. This means that you have

${P}_{1} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{1}}}} = {P}_{2} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{1}}}}}{2} \implies {P}_{2} = 2 {P}_{1}$

The total pressure of the system after the volume is halved will be equal to

${P}_{2} = 2 \cdot \text{0.26 atm" = "0.52 atm}$

The equilibrium constant will be

K_p = (("I")^color(red)(2))/(("I"_2)) = (0.16""^2)/0.10 = 0.256

Try to predict what will happen to the partial pressures of the two gases after the volume is halved.

Since you're dealing with an equilibrium that involves gases, increasing the overall pressure of the system will favor the side will the smaller number of moles.

So you can expect the partial pressure of ${\text{I}}_{2}$ to increase relative to that of $\text{I}$, given the fact that both partial pressures will increase as aresult of the reduction of the volume of the container.

Now, you can express the partial pressure of a gas that's part of a mixture by using its mole fraction and the total pressure of the mixture.

${P}_{i} = {\chi}_{i} \cdot {P}_{\text{total" " }}$, where

${P}_{i}$ - the partial pressure of gas $i$;
${\chi}_{i}$ - the mole fraction of gas $i$ in the mixture;
${P}_{\text{total}}$ - the total pressure of the mixture.

The equilibrium constant for the second state of the system can thus be written as

${K}_{p} = \frac{{\chi}_{I}^{2} \cdot {P}_{2}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}}{{\chi}_{{I}_{2}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{2}}}}} = {\chi}_{I}^{2} / {\chi}_{{I}_{2}} \cdot {P}_{2}$

Also, since you only have two components in the mixture, you know that

${\chi}_{I} + {\chi}_{{I}_{2}} = 1$

This means that you have

${\chi}_{{I}_{2}} = 1 - {\chi}_{I}$

and so

${K}_{p} = 0.256 = {\chi}_{I}^{2} / \left(1 - {\chi}_{I}\right) \cdot 0.52$

${\chi}_{I}^{2} = \frac{0.256}{0.52} \cdot \left(1 - {\chi}_{I}\right)$

${\chi}_{I}^{2} + 0.45714 \cdot {\chi}_{I} - 0.45714 = 0$

The solution to this quadratic is - only take the positive solution

${\chi}_{I} = 0.485$

The mole fraction of ${\text{I}}_{2}$ will thus be

${\chi}_{{I}_{2}} = 1 - 0.485 = 0.515$

The partial pressures of the two gases after the new equilibrium is established will be

P_(I) = chi_(I) * P_2 = 0.485 * "0.52 atm" = color(green)("0.25 atm")

P_(I_2) = chi_(I_2) * P_2 = 0.515 * "0.52 atm" = color(green)("0.27 atm")#