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: 47


See the answer below.

Work Step by Step

a) $p=p(NO_2)+p(N_2O_4)\rightarrow p(N_2O_4)=1.50\ atm-x, p(NO_2)=x$ $K_p=p(NO_2)^2/p(N_2O_4)$ $0.148=x^2/(1.50-x)$ $x^2+0.148x-0.222=0$ $x=0.403$ $f=2x/(1.50-x)\times100\%=73.5\%$ b) It would increase, because, since more moles of the product are formed than moles of reactants are consumed, decreasing pressure shifts the equilibrium to the right.
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.