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?

1 Answer
Apr 28, 2018

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