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: 16


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Work Step by Step

$n_{N_2O_4,Reacted}=x, [N_2O_4]_{Equil}=(15.6\ g/92.01\ g/mol-x)/5.000\ L$ $n_{NO_2,Formed}=2x, [NO_2]_{Equil}=2x/5.000$ Equilibrium constant: $K=[NO_2]^2/[N_2O_4]$ $170=(2x/5.000)^2/[(0.170-x)/5.000]$ $170\times(0.170-x)/5\times(5/2)^2=x^2$ $x^2=36.13-212.5x$ $x^2+212.5x-36.13=0$ $x=0.1694\ mol$ At equilibrium: $n_{NO_2}=0.339\ mol$ $f=(x)\div(15.6/92.01)=0.9992=99.92\%$
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