# Question #7a92c

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that you need to use the initial equilibrium partial pressures to find the *equilibrium constant*,

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

#"I"_text(2(g]) rightleftharpoons color(red)(2)"I"_text((g])#

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

#P_"total" = P_(I_2) + P_I#

#P_"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 * V_1 = n * R * T#

for the final state of the system, you have

#P_2 * V_2 = n * R * T#

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

#(P_1V_1)/(P_2V_2) = (color(red)(cancel(color(black)(n * R * T))))/(color(red)(cancel(color(black)(n * R * T)))) implies P_1V_1 = P_2V_2#

But you also know that

#P_1 * color(red)(cancel(color(black)(V_1))) = P_2 * color(red)(cancel(color(black)(V_1)))/2 implies P_2 = 2P_1#

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

#P_2 = 2 * "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 *increase* **relative** to that of

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 * P_"total" " "# , where

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

#K_p = (chi_(I)^2 * P_2^color(red)(cancel(color(black)(2))))/(chi_(I_2) * color(red)(cancel(color(black)(P_2)))) = chi_(I)^2/chi_(I_2) * 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/(1-chi_(I)) * 0.52#

#chi_(I)^2 = 0.256/0.52 * (1-chi_(I))#

#chi_(I)^2 + 0.45714 * chi_(I) - 0.45714 = 0#

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

#chi_(I) = 0.485#

The *mole fraction* of

#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")#