Question

A `3 L` container holds `16` mol and `24` mol of gasses A and B, respectively. Groups of five of molecules of gas B bind to three molecules of gas A and the reaction changes the temperature from `340^oK` to `480^oK`. How much does the pressure change?

Answer 1

The pressure after the reaction will be approximately 0.226 times the pressure before the reaction.

Explanation:

The overall strategy will be to calculate the number of moles of gas before and after the reaction and then use the ideal gas law to calculate the change in pressure. We will assume

1) The container is both rigid and closed so that its volume does not change.

2) The reaction is irreversible and goes to completion.

Our reaction is

3A + 5B `rarr` C

Let

`n_A^o=` the number of moles of A before the reaction = 16.

`n_B^o=` the number of moles of B before the reaction = 24.

`n_C^o=` the number of moles of C before the reaction = 0.

`n_o=` the total number of moles of gas before the reaction.

Note that

`n_o=n_A^(o)+n_B^(o)+n_C^o=16+24+0=40`.

Also let

`n_A=` the number of moles of A after the reaction.

`n_B=` the number of moles of B after the reaction.

`n_C=` the number of moles of C before the reaction.

`n=` the total number of moles of gas after the reaction.

Note that

`n=n_A+n_B+n_C`.

First determine the limiting reagent. Let's assume that B is the limiting reagent. 24 moles of B would require

24 moles B * (5 moles B/3 moles A) = 14.4 moles of A.

Since we have more than 14.4 moles of A (we have 16), ALL of the B will react with A, so this means that B is the limiting reagent.

At the end of the reaction, we will have

`n_A=n_A^(o)-14.4=16-14.4=1.6` moles of A left,

`n_B=0` moles of B left, and

`n_C=` 24 moles B * (1 mole C)/(5 moles B) = 4.8 moles C.

So `n=n_A+n_B+n_C=1.6+0+4.8=6.4` total moles.

Now we use the ideal gas law to determine the pressure change. If `P` = the pressure after the reaction, and `P_o` = the pressure before the reaction, then

`P/P_o=(n/n_o)(T/T_o)=(6.4/40)(480/340)~~0.226`