Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 13 - Sections 13.1-13.7 - Exercises - Problems by Topic - Page 638: 25a

Answer

$Rate=-\frac{1}{2}\frac{\Delta[HBr]}{\Delta t}=\frac{\Delta[H_{2}]}{\Delta t}=\frac{\Delta[Br_{2}]}{\Delta t}$

Work Step by Step

$\Delta$ represents change so ${\Delta t}$ means change in time. Rate is equal to the change in concentration over the change in time so for $H_{2}$ and $Br_{2}$ this is simply $\frac{\Delta[H_{2}]}{\Delta t}$ and $\frac{\Delta[Br_{2}]}{\Delta t}$. For $HI$ its concentration decreases as time passes so we add a negative sign. Also, two moles of it decompose to make just one mole of each product making its concentration decrease twice as fast as the products increase. This means we need to multiply it by $\frac{1}{2}$ to get the correct rate.
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