Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 13 - Chemical Kinetics - Questions & Problems - Page 609: 13.6

Answer

$a) -\frac{1}{2}\frac{[H_{2}]}{\Delta t}= -\frac{[O_{2}]}{\Delta t}=\frac{1}{2}\frac{[\Delta H_{2}O]}{\Delta t}$ $b)-\frac{1}{4}\frac{[NH_{3}]}{\Delta t}=-\frac{1}{5}\frac{[\Delta O_{2}]}{\Delta t}=\frac{1}{4}\frac{[\Delta NO]}{\Delta t}=\frac{1}{6}\frac{[\Delta H_{2}O]}{\Delta t}$

Work Step by Step

Given an equation aA → bB, use the relation: $-\frac{1}{a}\frac{[\Delta A]}{\Delta t}=\frac{1}{b}\frac{[\Delta B]}{\Delta t}$ $a) -\frac{1}{2}\frac{[H_{2}]}{\Delta t}= -\frac{[O_{2}]}{\Delta t}=\frac{1}{2}\frac{[\Delta H_{2}O]}{\Delta t}$ $b)-\frac{1}{4}\frac{[NH_{3}]}{\Delta t}=-\frac{1}{5}\frac{[\Delta O_{2}]}{\Delta t}=\frac{1}{4}\frac{[\Delta NO]}{\Delta t}=\frac{1}{6}\frac{[\Delta H_{2}O]}{\Delta t}$
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