Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 15 Principles of Chemical Reactivity: Equilibria - Study Questions - Page 583e: 50


See the answer below.

Work Step by Step

$P=p(NO_2)+p(N_2O_4)\rightarrow, p(NO_2)=x, p(N_2O_4)=0.15-x$ $K_p=p(N_2O_4)/p(NO_2)^2$ $7.1=0.15-x/x^2$ $7.1x^2+x-0.15=0$ $x=0.091$ $p(NO_2)=0.091\ atm, p(N_2O_4)=0.059\ atm$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.