A 3 L3L container holds 9 9 mol and 12 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 160160 KK to 360360 KK. How much does the pressure change?

1 Answer
Apr 4, 2016

The pressure will be reduced by the chemical reaction but increased by the temperature change. The reaction prevails, so the final pressure is approximately 0.540.54 times the initial pressure.

Explanation:

We need to assume that A and B are monatomic initially, because we are not told.

The balanced equation will be:

2A + 3B to A_2B_32A+3BA2B3

Gas B will be the limiting reagent, since the reaction requires 33 molmol of B for every 22 molmol of A, and 12 is less than 3/232 times 9.

That means all 1212 molmol of B will react, with 88 molmol of A, producing 44 molmol of A_2B_3A2B3. There will still be 11 molmol of unreacted A, for a total of 55 molmol of gas.

The volume does not change, so there are two steps in considering the change in pressure:

  1. We have gone from 2121 molmol of gas to 55 molmol of gas in the same volume, to the pressure will be 5/21521 of its initial value.

  2. The temperature has changed from 160160 KK to 360360 KK, which means the pressure will be 360/160360160 of the initial value.

Combining these two things, the final pressure will be 5/21xx360/160=1800/3360~~0.54521×360160=180033600.54 times as great.