Question

A `4 L` container holds `19` mol and `20` mol of gasses A and B, respectively. Groups of four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from `210^oK` to `120^oK`. How much does the pressure change?

Answer 1

The pressure drops from 168 atm to 22.2 atm.

Explanation:

I'll answer a question for a fellow Wisconsinite!

The trickiest part of this problem is determining the change in the number of moles from the reaction.

Let

`n_0^A` = the initial number of moles of gas A = 19.

`n_0^B` = the initial number of moles of gas B = 20.

`n_0` = the total number of moles of gas present before the reaction = 19 + 20 = 39.

We are given all the information we need to calculate the initial pressure if we make the assumption that gasses A and B are ideal. (In reality this is not often the case at 120K!)

`P^0=(n^0RT^0)/V^0=(39(0.082057)(210))/4~~168` atm

When the two types of gasses react, they form a new gas C. The problem states that the reaction goes as follows:

`3A+4BrarrC`

We know that the limiting reagent here is gas B since we require 33% more B than A for the reaction yet we only have about 5% more B than A before the reaction. We will assume that this reaction goes to completion so that there is no gas B after the reaction.

Let

`x` = the number of moles of gas B that reacts = 20.

`n^A` = the number of moles of A left after the reaction.

`n^B` = the number of moles of B left after the reaction = 0.

`n^C` = the number of moles of C formed by the reaction.

Because of the reaction stoichiometry

`n^A=n_0^A-(3/4)x=19-(3/4)20=4` moles of gas A.

`n^C=x/4=20/4=5` moles of gas C

The total number of moles of gas after the reaction, `n`, is

`n=n^A+n^B+n^C=4+0+5=9` moles of gas.

Again assuming the ideal gas law holds we can calculate the pressure after the reaction.

`P=(nRT)/V=(9(0.082057)(120))/4~~22.2` atm