A #3 L# container holds #5 # mol and #5 # mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from #340^oK# to #320^oK#. How much does the pressure change?

1 Answer
Dec 26, 2016

We have to take two things into account:
the change in moles and the change in temperature.

Explanation:

(1) change caused by change in number of molecules:
Reaction equation: #2A+3B->A_2B_3#
So #5# moles of B will react with #2/3xx5=3 1/3# moles of A.
#1 2/3# moles of A will be left.

Since every 3 moles of B will form 1 mole of #A_2B_3#:
#5/3=1 2/3# mole of #A_2B_3# will be formed

Total moles before reaction: #5+5=10# moles
Total moles after reaction: #1 2/3 + 1 2/3=3 1/3# moles
(mixture of #A_2B_3# and left-over #A#)

Pressure change because of amount of matter:
#3 1/3div5=10/3div5=xx2/3#

(2) Change caused by temperature change:
Temperature goes down from #34oK# to #320K#

Pressure change: #320/340=xx16/17#

Total change:
Pressure is decreased by a factor of #2/3xx16/17=32/51~~xx0.63#

Or: pressure decreases to #63%# of original (or by #37%#)

Note:
Of course you could have used the general gas-formula

#p=(nRT)/V# to calculate both pressures.