Chemistry: A Molecular Approach (3rd Edition)

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

Chapter 7 - Sections 7.1-7.6 - Exercises - Problems by Topic - Page 330: 45b

Answer

239 kJ/mol

Work Step by Step

Energy of one mole of photons = $hν\times\frac{6.022\times10^{23}}{1 mole} = \frac{hc}{λ}\times\frac{6.022\times10^{23}}{1 mole}$ h = $6.626\times10^{-34}Js$ c = $2.997925\times10^{8}ms^{-1}$ λ = 500 nm = $500\times10^{-9}m$ E = $\frac{6.626\times10^{-34}Js\times2.997925\times10^{8}ms^{-1}}{500\times10^{-9}m}\times\frac{6.022\times10^{23}}{1 mole}$ = $2.39\times10^{5} Jmol^{-1}$ = 239 kJ/mol.
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