A 7 L container holds 28  mol and 12  mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to five molecules of gas A and the reaction changes the temperature from 150^oK to 160^oK. How much does the pressure change?

May 20, 2017

All of the gas A and some of the gas B will react to produce ${A}_{3} {B}_{5}$. The final pressure will change due to the change in the temperature and the change in the number of moles of gas present.

The pressure will change by $2288$ $k P a$.

Explanation:

(we need to assume that the gases are monatomic before the reaction, since we are not told otherwise)

The reaction will be:

$3 A + 5 B \to {A}_{3} {B}_{5}$

That is, 8 mol of reactant gases combine to produce each 1 mol of product gas.

The gases are not present in this ratio, so one will be 'in excess' of the other and be left over after the reaction.

There is much more of gas A than gas B, but we need more of gas B for each reaction, so all the gas B will be used up and some gas A will be left over. After the reaction, the container will include product gas ${A}_{3} {B}_{5}$ and gas A.

The reaction in the equation above will occur $\frac{12}{5} = 2.4$ times, so there will be $2.4$ mol of ${A}_{3} {B}_{5}$. $2.4 \times 3 = 7.2$ mol of gas A will react, so $28 - 7.2 = 20.8$ mol of gas A will remain.

That means there will be a total of $20.8 + 2.4 = 23.2$ mol of gas in the container in total after the reaction. There were $28 + 12 = 40$ mol of gas in the container before the reaction.

We can use the universal gas equation:

$P V = n R T$

Rearranging:

$P = \frac{n R T}{V}$

In this case R, the gas constant, and V, the volume of the container, are constant, so we can remove them from the expression:

${P}_{1} \setminus \propto {n}_{1} {T}_{1} = 40 \times 150 = 6000$

${P}_{2} \setminus \propto {n}_{2} {T}_{2} = 23.2 \times 160 = 3712$

The pressure will change by $2288$ $k P a$.