Question

A `6 L` container holds `5` mol and `6` 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 `480^oK` to `420^oK`. How much does the pressure change?

Answer 1

Depending on how you interpret the question, the answer is that the pressure drops to either 34.5% or to 21.2% of its starting value.
See the explanation below.

Explanation:

If I am understanding the question properly, it says that 3 molecules of A and 2 molecules of B bind together to form a single molecule. What I'm not clear on is whether that new molecule is one of gas, or not. Assuming it is, here's what must be done.

First, we have a limiting reagent problem to determine how many molecules of B remain once all of A has reacted.

Because it takes 3 molecules of A to react with only 2 of B, the 5 moles of A present will react with `5" mol of A"xx(2" mol of B")/(3" mol of A") = 3 1/3` mol of B will be used up and `2 2/3` mol of B will remain, unreacted.

So, the reaction will result in the container now holding `2 2/3` mol of B. Plus, there will be `5/3` of a mol of the new product present, because all of A is used, at a rate of three molecules per one molecule of product formed.

Note: If the problem meant that this new product was not a gas, then change my answer by leaving these `5/3` of a mole out.

Next, we apply the ideal gas law:

`(P_1V_1)/(n_1T_1)=(P_2V_2)/(n_2T_2)`

In this problem, `V_1=V_2=6L` and we leave this out.

`n_1=11` mol

`n_2=4 1/3` mol (or `2 2/3` mol of you leave out the product.)

`T_1=480K`

`T_2=420K`

We will determine the value of `P_2/P_1` (the fractional change in P)

From the ideal gas law:

`P_2/P_1=(n_2T_2)/(n_1T_1) = (4 1/3 * 420)/(11*480)=0.345`

The pressure drops to 34.5 % of its initial value.

If you leave out the product (assume it to be solid), the answer is

`P_2/P_1=(n_2T_2)/(n_1T_1) = (2 2/3 * 420)/(11*480)=0.212`

The pressure drops to 21.2% of its original value in this case.