Question

A `5 L` container holds `5` mol and `6` mol of gasses A and B, respectively. Every three of molecules of gas B bind to two molecule of gas A and the reaction changes the temperature from `360^oK` to `210 ^oK`. By how much does the pressure change?

Answer 1

The pressure decreases by either 54.65 atm or 61.543 atm, depending on whether the product is gaseous or solid.

Explanation:

We are changing three things here: Pressure, Temperature, and number of moles of gas. In terms of the Ideal Gas Law, that is P, T, and n. V and R remain constant. To keep all this straight, let's make a table of initial and final states:

                  Initial state:                                  Final State

`V" "5.0 l" "5.0 l`
`R" "0.082054 (l*(atm)/(mol*K))" "0.082054 (l*(atm)/(mol*K))`
`n" "11 mol" "3 mol or 1 mol`(see below)
`T" "360 K" "210 K`
`P" "TBD" "TBD`

A couple of points.
1) The question does not specify whether the product of the reaction is gaseous or not. If it is gaseous, then you will be left with three moles (you will lose 4 A and 6 B, leaving one mole of A and two moles of `A_2B_3`). If it is not gaseous, then you are simply left with one mole of A.
2) Just a note on proper notation: the temperature is in kelvins, not "degrees kelvin". No degree symbol is used with K.

We can easily calculate the initial pressure:
`P = nRT/V`
`P = 11mol xx 0.082057(l*(atm)/(mol*K)) xx (360K) / (5l) = 64.989 atm`

Now for the final pressure (we will calculate both answers since the question didn't specify whether the final product was a gas (unlikely, given the temperature) or a solid:

`P_3 = 3mol xx 0.082057(l*(atm)/(mol*K)) xx (210K) / (5l) = 10.339 atm`

or

`P_1 = 1mol xx 0.082057(l*(atm)/(mol*K)) xx (210K) / (5l) = 3.446 atm`

in the `P_3` case (product remains gaseous) the pressure decreases by 54.65 atm, or `(54.65)/(64.989) xx 100% = 84%`

in the `P_1` case (product is solid) the pressure decreases by 61.543 atm, or `(61.543)/(64.989) xx 100% = 95%`