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 583b: 17


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Work Step by Step

$n_{COBr_2,Reacted}=x, [COBr_2]_{equil}=(0.0500-x)/2.00$ $[CO]_{Equil}=x/2.00, [Br_2]_{Equil}=x/2.00$ Equilibrium constant: $K=[CO][Br_2]/[COBr_2]$ $0.190=x/2\times x/2\div (0.0500-x)/2$ $0.190\times (0.0500-x)/2\times4=x^2$ $0.019-0.38x=x^2$ $x^2+0.38x-0.019=0$ $x=0.0447\ mol$ At equilibrium: $[CO]=0.0224\ M=[Br_2], [COBr_2]=2.63\times10^{-3}\ M$ Fraction decomposed of COBr2 $f=x/0.05=0.954=95.4\%$
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