An 8L container holds 6 mol and 5 mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from 320oK to 450oK. How much does the pressure change by?

1 Answer
Mar 31, 2018

The final pressure will be 0.767 times the initial pressure.

Explanation:

The overall strategy for this problem is to determine the change in the number of moles present and then use the ideal gas law to calculate the pressure change.

We have a reaction where

A+ 2B -> C

Let nA represent the number of moles of A after the reaction and n0A=6 represent the number of moles of A before the reaction.

Let nB represent the number of moles of B after the reaction and n0B=5 represent the number of moles of B before the reaction.

Let nC represent the number of moles of C after the reaction and n0C=0 represent the number of moles of C before the reaction.

If all 5 moles of B react, then

nA=n0A52=652=72

nB=n0B5=55=0

nC=n0C+52=0+52=52

The total number of moles, n, after the reaction will be

n=nA+nB+nC=72+0+52=6 moles.

Initially there were n0=6+5+0=11 total moles.

Now we can use the Ideal Gas Law to calculate the ratio of the final pressure, P to the initial pressure, P0. Let T0=320K be the temperature before the reaction and let T=450K be the temperature after the reaction. The Ideal Gas Law tells us that

PP0=nn0TT0=611450320=1351760.767.