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?

1 Answer
Feb 10, 2017

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%