# Question #70d16

Oct 16, 2016

Here's what I got.

#### Explanation:

For starters, the reaction you're looking at here is

$3 {\text{H"_ (2(g)) + "N"_ (2(g)) -> 2"NH}}_{3 \left(g\right)}$

Notice that the reactants' side contains

• three moles of hydrogen gas, $3 \times {\text{H}}_{2}$
• one mole of nitrogen gas, $1 \times {\text{N}}_{2}$

The product's side contains

• two moles of ammonia, $2 \times {\text{NH}}_{3}$

Now, let's take each cube to be equivalent to one unit of volume. You can thus say that you have

• $\text{4 moles of gas in 4 cubes } \to$ for the reactants' side

• $\text{2 moles of gas in 2 cubes } \to$ for the product's side

Notice that the ratio of moles of gas to number of cubes remains unchanged. You have

• $\text{4 moles gas"/"4 cubes" = "1 mole/cube } \to$ for the reactants' side

• $\text{2 mole gas"/"2 cubes" = "1 mole/cube } \to$ for the product's side

This essentially means that the pressure remains unchanged because you get the same moles of gas to volume ratio on both sides of the reaction.

$\textcolor{w h i t e}{a}$
PROVE THIS USING BOYLE'S LAW

To prove this, let's assume that the volume remains constant, i.e. that you end up with $2$ moles of gas in $4$ cubes on the product's side.

Since pressure is directly proportional to the number of moles of gas when temperature and volume are kept constant, decreasing the number of moles of gas by a factor of $2$ would cause the pressure to decrease by the same factor.

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 this case, you'd have

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

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

This is the pressure on the reactant's side if you have $2$ moles of gas in $4$ cubes.

Now focus on the product's side alone. You keep the number of moles of gas and the temperature constant and decrease the volume from $4$ cubes to $2$ cubes.

You know from Boyle's Law that pressure and volume have an inverse relationship when the number of moles of gas and the temperature are kept constant

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {P}_{\text{1 product" * V_"1 product" = P_"2 product" * V_"2 product}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In this case, you'd have

${P}_{\text{2 product" = V_"1 product"/V_"2 product" * P_"1 product}}$

${P}_{\text{2 product" = (4 color(red)(cancel(color(black)("cubes"))))/(2color(red)(cancel(color(black)("cubes")))) * P_"1 product}}$

${P}_{\text{2 product" = 2 * P_"1 product}}$

But you know that ${P}_{\text{1 product}}$, which is the pressure you get after the reaction is complete, is equal to

${P}_{\text{1 product}} = \frac{1}{2} \cdot {P}_{1}$

Here ${P}_{1}$ is the pressure you start with on the reactants' side.

Therefore,

${P}_{\text{2 product}} = {P}_{2} = 2 \cdot \frac{1}{2} \cdot {P}_{1}$

${P}_{\text{2 product}} = {P}_{1}$

Once again, this goes to show that the pressure remains unchanged because you decreased the number of moles of as by a factor of $2$ and decreased the volume by a factor of $2$.