# Question c3e2b

Oct 16, 2016

Here's what I got.

#### Explanation:

The key here is the number of moles of gas that remains after the reaction is complete.

• four moles of carbon monoxide, $4 \times \text{CO}$
• six moles of hydrogen gas, $6 \times {\text{H}}_{2}$

After the reaction is complete, the container holds

• one mole of carbon monoixide, $1 \times \text{CO}$
• three moles of methanol, $3 \times \text{CH"_3"OH}$

As you can see, the number of moles of gas decreases, which means that you should expect the pressure to decrease as well.

Now, when temperature and volume are kept constant, the pressure of a gas is directly proportional to the number of moles of gas present.

You can thus say that

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {P}_{1} / {n}_{1} = {P}_{2} / {n}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

${n}_{1}$, ${P}_{1}$ - the initial number of moles of gas and the volume they occupy
${n}_{2}$, ${P}_{2}$ - the final number of moles of gas and the volume they occupy

In your case, the number of moles of gas goes from

${n}_{1} = \text{10 moles }$ to $\text{ "n_2 = "4 moles}$

This means that the final pressure of the gas will be

${P}_{2} = {n}_{2} / {n}_{1} \cdot {P}_{1}$

${P}_{2} = \left(4 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(10color(red)(cancel(color(black)("moles}}}}\right) \cdot {P}_{1}$

$\textcolor{p u r p \le}{{P}_{2} = \frac{4}{10} \cdot {P}_{1}}$

Now, the partial pressure of carbon monoxide before the reaction takes place is equal to the mole fraction of the gas multiplied by the total pressure inside the container, as given by Dalton's Law of Partial Pressures.

The mole fraction of carbon monoxide is simply the number of moles of carbon monoxide divided by the total number of moles of gas present in the container

chi_"CO" = (4 color(red)(cancel(color(black)("moles"))))/((4 + 6)color(red)(cancel(color(black)("moles")))) = 4/10#

Since we've said that the pressure inside the container before the reaction takes place is equal to ${P}_{1}$, the partial pressure of carbon monoxide is

$\textcolor{p u r p \le}{{P}_{\text{CO}} = \frac{4}{10} \cdot {P}_{1}}$

Therefore, the total pressure inside the container after the reaction takes place is equal to the partial pressure of carbon monoxide before the reaction takes place.