# 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?

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 $0.54$ 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:

$2 A + 3 B \to {A}_{2} {B}_{3}$

Gas B will be the limiting reagent, since the reaction requires $3$ $m o l$ of B for every $2$ $m o l$ of A, and 12 is less than $\frac{3}{2}$ times 9.

That means all $12$ $m o l$ of B will react, with $8$ $m o l$ of A, producing $4$ $m o l$ of ${A}_{2} {B}_{3}$. There will still be $1$ $m o l$ of unreacted A, for a total of $5$ $m o l$ of gas.

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

1. We have gone from $21$ $m o l$ of gas to $5$ $m o l$ of gas in the same volume, to the pressure will be $\frac{5}{21}$ of its initial value.

2. The temperature has changed from $160$ $K$ to $360$ $K$, which means the pressure will be $\frac{360}{160}$ of the initial value.

Combining these two things, the final pressure will be $\frac{5}{21} \times \frac{360}{160} = \frac{1800}{3360} \approx 0.54$ times as great.