Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 6 - Electronic Structure of Atoms - Exercises: 6.29d

Answer

The average energy of these bonds is $368.1kJ/mol$.

Work Step by Step

1) In the question, they wrote that the 325-nm radiation provides exactly the energy to break an average chemical bond in the skin. What they mean is that an average chemical bond has an energy equal to the energy of a single photon of the 325-nm radiation, or $E_{bond}=E_{photon}\approx6.112\times10^{-19}J$. 2) Now the question asks for the average energy of these bonds in $kJ/mol$. In other words, it asks for the total energy of 1 mol of these average bonds (or about $6.022\times10^{23}$ bonds). Therefore, the average energy of these bonds is: $E=(6.022\times10^{23})\times E_{bond}=(6.022\times10^{23})\times(6.112\times10^{-19})\approx3.681\times10^5J/mol\approx368.1kJ/mol$
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