Chemistry: Principles and Practice (3rd Edition)

Published by Cengage Learning
ISBN 10: 0534420125
ISBN 13: 978-0-53442-012-3

Chapter 13 - Chemical Kinetics - Questions and Exercises - Exercises - Page 564: 13.27


Rate of disappearance of $NO_{2}$= 0.011 M/s. Rate of the reaction= 0.0055 M/s.

Work Step by Step

Rate of the reaction= $\frac{\Delta [N_{2}O_{5}]}{\Delta t}=-\frac{1}{2}\times\frac{\Delta [NO_{2}]}{\Delta t}= 0.0055 M/s$ $\implies \frac{\Delta [NO_{2}]}{\Delta t}= -2\times\frac{\Delta [N_{2}O_{5}]}{\Delta t}$ $=-2\times0.0055M/s= -0.011 M/s$. Negative sign indicates the disappearance and the rate of disappearance of $NO_{2}$ is 0.011 M/s.
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