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?

1 Answer
May 20, 2017

All of the gas A and some of the gas B will react to produce A_3B_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 kPa.

Explanation:

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

The reaction will be:

3A + 5B -> A_3B_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_3B_5 and gas A.

The reaction in the equation above will occur 12/5 = 2.4 times, so there will be 2.4 mol of A_3B_5. 2.4xx3=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:

PV=nRT

Rearranging:

P=(nRT)/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 \prop n_1T_1 = 40xx150=6000

P_2 \prop n_2T_2 = 23.2xx160=3712

The pressure will change by 2288 kPa.