A 3L container holds 5 mol and 5 mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from 340oK to 320oK. How much does the pressure change?

1 Answer
Dec 26, 2016

We have to take two things into account:
the change in moles and the change in temperature.

Explanation:

(1) change caused by change in number of molecules:
Reaction equation: 2A+3BA2B3
So 5 moles of B will react with 23×5=313 moles of A.
123 moles of A will be left.

Since every 3 moles of B will form 1 mole of A2B3:
53=123 mole of A2B3 will be formed

Total moles before reaction: 5+5=10 moles
Total moles after reaction: 123+123=313 moles
(mixture of A2B3 and left-over A)

Pressure change because of amount of matter:
313÷5=103÷5=×23

(2) Change caused by temperature change:
Temperature goes down from 34oK to 320K

Pressure change: 320340=×1617

Total change:
Pressure is decreased by a factor of 23×1617=3251×0.63

Or: pressure decreases to 63% of original (or by 37%)

Note:
Of course you could have used the general gas-formula

p=nRTV to calculate both pressures.