Question

Consider the following reaction.


`2 NO_2 (g) \rightleftharpoons N_2O_4 (g)`
When the system is at equilibrium, it contains NO₂ at a pressure of 0.882 atm , and N₂O₄ at a pressure of 0.0778 atm . The volume of the container is then reduced to half its original volume. What is the pressure of each gas after equilibrium is reestablished?
P NO₂ = atm
P N₂O₄ = atm

Answer 1

NO2= 1.764 atm

N2O4= 0.1556 atm

Step-by-step explanation:

This values of pressure has doubled, i.e multiplied by 2, because as the volume is halved, the pressure doubles, this is in accordance with Boyle's law