An 8 L 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 320^oK to 450 ^oK. 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 n_A represent the number of moles of A after the reaction and n_A^0=6 represent the number of moles of A before the reaction.

Let n_B represent the number of moles of B after the reaction and n_B^0=5 represent the number of moles of B before the reaction.

Let n_C represent the number of moles of C after the reaction and n_C^0=0 represent the number of moles of C before the reaction.

If all 5 moles of B react, then

n_A=n_A^0-5/2=6-5/2=7/2

n_B=n_B^0-5=5-5=0

n_C=n_C^0+5/2=0+5/2=5/2

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

n=n_A+n_B+n_C=7/2+0+5/2=6 moles.

Initially there were n_0=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, P_0. Let T_0=320 K be the temperature before the reaction and let T=450K be the temperature after the reaction. The Ideal Gas Law tells us that

P/P_0=n/n_0*T/T_0=6/11*450/320=135/176~~0.767.