Asked by BBaa

Consider the following reaction.

2NO2(g)⇌N2O4(g)

When the system is at equilibrium, it contains NO2 at a pressure of 0.724 atm , and N2O4 at a pressure of 0.0524 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?

𝑃NO2 =
atm
𝑃N2O4=
atm

Answers

Answered by DrBob222
..................2NO2(g) ⇌ N2O4(g)
E............0.724 atm........0.0524 atm
Kp = p(N2O4)/p^2(NO2) = 0.0524/(0.724)^2 = 0.1 (rounded to 0.1)
Reducing the volume to 1/2 means pN2O4 becomes 0.0524 x 2 = 0.105 atm and pNO2 = 0.724 x 2 = 1.45 atm. Then, Le Chatelier's Principles tells us that when pressure is increased in a system at equilibrium that the equilibrium will shift to the side with fewer moles. So it will shift to the right producing more N2O4 at the expense of NO2.
.........................2NO2(g) ⇌ N2O4(g)
I.........................1.45...............0.105
C.......................-2x.....................+x
E.....................1.45-2x................0.105+x
Plug the E line into the Kp expression and solve for x, then evaluate 1.45 - 2x and 0.105 + x. Note that I have rounded here and there so you should start from the beginning and use your numbers and not mine.
Post your work if you get stuck.
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions