# Question #70d16

##### 1 Answer

Here's what I got.

#### Explanation:

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

#3"H"_ (2(g)) + "N"_ (2(g)) -> 2"NH"_ (3(g))#

Notice that the reactants' side contains

,three molesof hydrogen gas#3 xx "H"_2# ,one moleof nitrogen gas#1 xx "N"_2#

The product's side contains

,two molesof ammonia#2 xx "NH"_3#

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

#"4 moles of gas in 4 cubes " -># for thereactants' side

#"2 moles of gas in 2 cubes " -># for theproduct's side

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

#"4 moles gas"/"4 cubes" = "1 mole/cube " -># for thereactants' side

#"2 mole gas"/"2 cubes" = "1 mole/cube "-># for theproduct'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.

**PROVE THIS USING BOYLE'S LAW**

To prove this, let's assume that the volume remains **constant**, i.e. that you end up with **moles** of gas in **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 **decrease** by the same factor.

You can thus say that

#color(blue)(bar(ul(|color(white)(a/a)P_1/n_1 = P_2/n_2color(white)(a/a)|)))#

Here

**initial** number of moles of gas and the volume they occupy

**final** number of moles of gas and the volume they occupy

In this case, you'd have

#P_2 = n_2/n_1 * P_1#

#P_2 = (2 color(red)(cancel(color(black)("moles"))))/(4color(red)(cancel(color(black)("moles")))) * P_1 = 1/2 * P_1#

This is the pressure on the reactant's side if you have **moles** of gas in **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 **cubes** to **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

#color(blue)(bar(ul(|color(white)(a/a)P_"1 product" * V_"1 product" = P_"2 product" * V_"2 product"color(white)(a/a)|)))#

In this case, you'd have

#P_"2 product" = V_"1 product"/V_"2 product" * P_"1 product"#

#P_"2 product" = (4 color(red)(cancel(color(black)("cubes"))))/(2color(red)(cancel(color(black)("cubes")))) * P_"1 product"#

#P_"2 product" = 2 * P_"1 product"#

But you know that

#P_"1 product" = 1/2 * P_1#

Here **start with** on the reactants' side.

Therefore,

#P_"2 product" = P_2 = 2 * 1/2 * P_1#

#P_"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 **decreased** the volume by a factor of