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?

Answer 1

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.

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_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.

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

1. 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.

2. 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.