enthalpy of hydration
One possible way of analyzing the reaction involves breaking it into separate sublimation, ionization, and hydration steps:
Cu2+(aq) Cu2+(g) H° = +2289 kJThis process is endothermic because strong attractive ion-dipole forces between the copper(II) ion and the water must be overcome to vaporize the ion.
Cu2+(g) + 2 e- Cu(g) H° = -2703.4 kJThis process is the reverse of ionization; H is equal to minus the sum of the first and second ionization enthalpies for copper. Capture of electrons by a cation is always exothermic.
Cu(g) Cu(s) H° = -337.4 kJH for condensation of gaseous copper is minus the heat of formation for gaseous copper. The process is exothermic because strong metallic bonds are formed in the condensation.
Zn(s) Zn(g) H° = +130.5 kJVaporization of solid zinc is endothermic because strong metallic bonds are being broken.
Zn(g) Zn2+(g) + 2 e- H° = +2639 kJIonization of zinc atoms is endothermic because the strong attraction the valence electrons have for the nucleus must be overcome.
Zn2+(g) Zn2+(aq) H° = -2247 kJHydration of zinc ion is exothermic because strong water-ion attractions are formed.
Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq) H° = -229 kJwith a reaction enthalpy that is probably good to +/- 30 kJ, given the uncertainties in the hydration enthalpies we used.
Although the reaction doesn't actually proceed by a series of sublimation-ionization-hydration steps, the calculation does give us a valuable insight into where the energy is actually coming from in the reaction. Try the following questions to explore some reasonable deductions that can be made about the significance of ion-water interactions, metallic bonding, and attraction for valence electrons in determining the sign on H in this reaction.
Author: Fred Senese email@example.com
Copyright © 1997-2010 by Fred Senese
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Last Revised 02/15/10.URL: http://antoine.frostburg.edu/chem/senese/101/thermo/faq/print-why-are-metal-displacements-exothermic.shtml