Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 15 - Questions and Problems - Page 709: 15.45

Answer

$ P_{ COCl_2 } = 0.703 \space atm$ $ P_{ CO }=0.0570 \space atm$ $ P_{ Cl_2 }= 0.0570 \space atm$

Work Step by Step

1. Write the equilibrium constant expression: - The exponent of each concentration is equal to its balance coefficient. $$K_c = \frac{[Products]}{[Reactants]} = \frac{P_{ CO }P_{ Cl_2 }}{P_{ COCl_2 }}$$ 2. At equilibrium, these are the concentrations of each compound: $ P_{ COCl_2 } = 0.760 \space M - x$ $ P_{ CO } = 0 \space M + x$ $ P_{ Cl_2 } = 0 \space M + x$ $$4.63 \times 10^{-3} = \frac{(x)(x)}{(0.760 - x)}$$ x = 0.0570 $ P_{ COCl_2 } = 0.760 \space M - 0.0570 M = 0.703 atm$ $ P_{ CO } = 0 \space M + 0.0570 M = 0.0570 atm$ $ P_{ Cl_2 } = 0 \space M + 0.0570 M = 0.0570 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.