A 5L container holds 12 mol and 9 mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from 360oK to 420oK. By how much does the pressure change?

1 Answer
Jul 6, 2018

the pressure change from 126 Bar to 12 Bar

Explanation:

the reaction is
3B+4A=B3A4+Q
so if at the beginning you have (12+9) = 21 mol of gases, after the reaction you have 3 mol of gas( we don't know if we have still gas?)
at the beginning with the gases'law, we have:
P=nRTV=21mol×8,31Jmol×K×360K0,005m3=126×105Jm3=126×105Nm2=126×105Pascal=126Bar
after the reaction:
P=nRTV=3mol×8,31Jmol×K×420K0,005m3=12×105Jm3=12×105Nm2=12×105Pascal=12Bar