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 583e: 43


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Work Step by Step

Initial concentration: $[PCl_5]=3.120\ g\div 208.24\ g/mol\div 10.0\ L=0.0015\ M$ $[PCl_3]=3.845\ g\div 137.33\ g/mol\div 10.0\ L=0.0028\ M$ $[Cl_2]=1.787\ g\div 70.91\ g/mol\div 10.0\ L=0.0025\ M$ Added: $[Cl_2]=1.418 g\div 70.91\ g/mol\div 10.0\ L=0.0020\ M$ $K=[PCl_3][Cl_2]/[PCl_5]$ $K=0.0025\times0.0028/0.0015=0.0047$ The equilibrium will shift to the left, so concentration of reacted: x $0.0047=(0.0028-x)(0.0045-x)/(0.0015+x)$ $x^2-0.012x+5.55\times10^{-6}=0$ $x=0.00048$ $[PCl_5]=0.0020\ M$ $[PCl_5]=0.0023\ M$ $[Cl_2]=0.0040\ M$
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