Question

A `5 L` container holds `9` mol and `12` 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 `320^oK` to `210 ^oK`. By how much does the pressure change?

Answer 1

The Pressure within the container decreases by
`Delta P=9.43*10^6color(white)(l)"Pa"`

Explanation:

Number of moles of gaseous particles before the reaction:
`n_1=9+12=21color(white)(l)"mol"`

Gas A is in excess.
It takes `9*3/2=13.5color(white)(l)"mol">12 color(white)(l)"mol"` of gas B to consume all gas A and `12*2/3=8 color(white)(l)"mol"<9 color(white)(l)"mol"` vice versa.
`9-8=1color(white)(l)"mol"` of gas A would be in excess.

Assuming that every two molecules of A and three molecules of B combine to yield a single gaseous product molecule, the number of moles of gas particles present in the container after the reaction would equal to
`color(darkblue)(n_2)=12*color(darkblue)(1)/3+1=color(darkblue)(5)color(white)(l)"mol"`

The volume of the container in the appropriate SI unit would be
`V=5 color(white)(l)"dm"^3=5*10^(-3) color(white)(l)m^3`

The temperature drops from `T_1=320 color(white)(l)"K"` to `T_2=210 color(white)(l)"K"` during the reaction. Apply the ideal gas law with

`R=8.314color(white)(l)"m" ^3* "Pa"* "mol" ^(-1)* "K" ^(-1)`

gives

`P_1=(n_1*R*T_1)/(V)=1.12*10^7 color(white)(l)"Pa"`

`color(darkblue)(P_2)=(color(darkblue)(n_2)*R*T_1)/(V)=color(darkblue)(1.75*10^6) color(white)(l)"Pa"`

`color(darkblue)(Delta P)=color(darkblue)(P_2)-P_1=color(darkblue)(-9.43*10^6)color(white)(l)"Pa"`

Hence the pressure decrease by `color(darkblue)(9.43*10^6)color(white)(l)"Pa"`.

Note that the figures in dark blue are dependent on the assumption that every two moles of gas A and three moles of gas B combine to form one mole of product, which is also a gas. See if you can find the right `Delta P` based on further information given in the question.

Reference
[1] The Ideal Gas Law, http://www.science.uwaterloo.ca/~cchieh/cact/c120/idealgas.html