# Question a4aeb

Nov 30, 2015

Here's what I got.

#### Explanation:

So, you're dealing with an equilibrium reaction between nitrogen tetroxide, ${\text{N"_2"O}}_{4}$, and nitrogen dioxide, ${\text{NO}}_{2}$

${\text{N"_2"O"_text(4(g]) rightleftharpoons color(red)(2)"NO}}_{\textrm{2 \left(g\right]}}$

Looking at the equilibrium reaction, it's important to realize what the $1 : \textcolor{red}{2}$ mole ratio that exists between the two gases means.

For every mole of nitrogen tetroxide that dissociates, the reaction produces $\textcolor{red}{2}$ moles of nitrogen dioxide. This means that regardless of how many moles of nitrogen tetroxide dissociate, the reaction will always produce two times more moles of nitrogen dioxide.

So, you know that you're starting with $1$ mole of nitrogen tetroxide and that 20% dissociates to form nitrogen dioxide. So, what does that tell you?

For every $100$ moles of nitrogen tetroxide, $20$ moles will dissociate and $80$ moles will remain undissociated, i.e. as nitrogen tetroxide.

This means that out of $1$ mole, you will have

1 color(red)(cancel(color(black)("mole N"_2"O"_4))) * "20 moles dissociate"/(100color(red)(cancel(color(black)("moles N"_2"O"_4)))) = "0.2 moles N"_2"O"_4

that will dissociate to form nitrogen dioxide. Using the aforementioned mole ratio, this will result in

0.20color(red)(cancel(color(black)("moles N"_2"O"_4))) * (color(red)(2)" moles NO"_2)/(1color(red)(cancel(color(black)("mole N"_2"O"_4)))) = "0.40 moles NO"_2

produced by the equilibrium reaction. So, at equilibrium, the reaction vessel will contain

• $0.80$ moles of ${\text{N"_2"O}}_{4} \to$ left undissociated
• $0.40$ moles of ${\text{NO}}_{2} \to$ formed by the reaction

Now, the total pressure of the mixture is said to be equal to $\text{1 atm}$. As you know, each component of a gaseous mixture will contribute to the total pressure proportionally to its mole fraction - this is known as Dalton's Law of Partial Pressures.

In your case, the mole fractions of the two gases will be

${\chi}_{{N}_{2} {O}_{4}} = \left(0.80 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/( (0.80 + 0.40) color(red)(cancel(color(black)("moles}}}}\right) = \frac{2}{3}$

${\chi}_{N {O}_{2}} = \left(0.40 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/((0.80 + 0.40) color(red)(cancel(color(black)("moles}}}}\right) = \frac{1}{3}$

This means that you have

${P}_{{N}_{2} {O}_{4}} = {\chi}_{{N}_{2} {O}_{4}} \times {P}_{\text{total}}$

P_(N_2O_4) = 2/3 * "1 atm" = color(green)(2/3 "atm")

and

P(NO_2) = 1/3 * "1 atm" = color(green)(1/3"atm")

By definition, ${K}_{p}$ will be equal to

K_p = ( ("NO"_2)^color(red)(2))/(("N"_2"O"_4)) = (1/3)^color(red)(2) "atm"^2* 3/2 "atm"^(-1) = 1/9 * 3/2"atm" = color(green)(1/6" atm")#