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?

1 Answer
May 7, 2018

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