How do I calculate [N2O4] at equilibrium when [NO2]=2.40×10−2mol/L.?
For the reaction N2O4(g)⇌2NO2(g), the value of K at 25∘C is 7.19×10−3.
help me solve this. Thanks.
More Questions From This User See All
For the reaction N2O4(g)⇌2NO2(g), the value of K at 25∘C is 7.19×10−3.
help me solve this. Thanks.
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
[N2O4]o = 0,012 M
[N2O4]eq = 0,012 - x
Kc = 0,00719 = 0,024^2 / (0,012 - x)
x = - 0,068
[N2O4]eq = 0,080 M