A #3 L# container holds #9 # mol and #12 # 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 #160# #K# to #360# #K#. How much does the pressure change?

Question

A #3 L# container holds #9 # mol and #12 # 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 #160# #K# to #360# #K#. How much does the pressure change?

Nora Arzate · Accepted Answer

The pressure will be reduced by the chemical reaction but increased by the temperature change. The reaction prevails, so the final pressure is approximately #0.54# times the initial pressure.

Since we are not informed, we must first assume that A and B are monatomic. The equation that is balanced will be: #2A + 3B to A_2B_3# Gas B will be the limiting reagent, since the reaction requires #3# #mol# of B for every #2# #mol# of A, and 12 is less than #3/2# times 9. That means all #12# #mol# of B will react, with #8# #mol# of A, producing #4# #mol# of #A_2B_3#. There will still be #1# #mol# of unreacted A, for a total of #5# #mol# of gas. Since the volume remains constant, the shift in pressure is taken into consideration in two steps: We have gone from #21# #mol# of gas to #5# #mol# of gas in the same volume, to the pressure will be #5/21# of its initial value. The temperature has changed from #160# #K# to #360# #K#, which means the pressure will be #360/160# of the initial value. Combining these two things, the final pressure will be #5/21xx360/160=1800/3360~~0.54# times as great.

Adrian Ayala · Answer

undefined The pressure increases by a factor of 4.