We saw in Section 6.7 that the reaction of solid sodium with giseous chlorine to yield solid sodium chloride ( Na + CI − ) is favorable by 411 kJ/mol. Calculate the energy change for the alternative reaction that yields chlorine sodide ( Cl + Na − ) , and then explain why sodium chloride formation is preferred. 2Na ( s ) + Cl 2 ( g ) → 2Cl + Na − ( s ) Assume that the lattice energy for Cl + Na − is the same as that for Na + Cl − . The following data are needed in addition to that found in Section 6.7: E ea for Na ( g ) = − 52 .9 kJ/mol E i1 for Cl ( g ) = + 1251 kJ/mol
We saw in Section 6.7 that the reaction of solid sodium with giseous chlorine to yield solid sodium chloride ( Na + CI − ) is favorable by 411 kJ/mol. Calculate the energy change for the alternative reaction that yields chlorine sodide ( Cl + Na − ) , and then explain why sodium chloride formation is preferred. 2Na ( s ) + Cl 2 ( g ) → 2Cl + Na − ( s ) Assume that the lattice energy for Cl + Na − is the same as that for Na + Cl − . The following data are needed in addition to that found in Section 6.7: E ea for Na ( g ) = − 52 .9 kJ/mol E i1 for Cl ( g ) = + 1251 kJ/mol
Solution Summary: The author explains the Born-Haber cycle of the formation of ionic solids from its constituent elements.
We saw in Section 6.7 that the reaction of solid sodium with giseous chlorine to yield solid sodium chloride
(
Na
+
CI
−
)
is favorable by 411 kJ/mol. Calculate the energy change for the alternative reaction that yields chlorine sodide
(
Cl
+
Na
−
)
, and then explain why sodium chloride formation is preferred.
2Na
(
s
)
+
Cl
2
(
g
)
→
2Cl
+
Na
−
(
s
)
Assume that the lattice energy for
Cl
+
Na
−
is the same as that for
Na
+
Cl
−
. The following data are needed in addition to that found in Section 6.7:
E
ea
for Na
(
g
)
=
−
52
.9 kJ/mol
E
i1
for Cl
(
g
)
=
+
1251
kJ/mol
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