The standard heat of formation of CaBr2 is -675 . The first ionization energy of Ca is 590 and its second ionization energy is 1145 . The heat of sublimation of Ca is 178 . The bond energy of Br2 is 193 , the heat of vaporization of Br2 is 31 , and the electron affinity of Br is -325 .
2007-11-15 07:41:57 · 5 answers · asked by Suker 1 in Science & Mathematics ➔ Chemistry
your final lattice energy is -1969!
2007-11-17 12:19:47 · answer #1 · answered by Pooh 4 · 0⤊ 3⤋
Lattice Energy Cabr2
2016-12-16 03:53:39 · answer #2 · answered by Anonymous · 0⤊ 1⤋
I suppose these are all kJ/mole. First, gasefy solid Ca:
Ca(s) ===> Ca(g) +178
Next take the first electron off Ca:
Ca(g) ===> Ca+(g) + e- +590
Then the second electron:
Ca+(g) ===> Ca2+(g) +1145
178 + 590 + 1145 = +1913 to get to Ca2+(g) ions
Now vaporize the Br2:
Br2(l) ===> Br2(g) +31
Next atomize Br2
Br2(g) ===> 2Br(g) + 193
Next form bromide ions:
2Br(g) + 2e- ===> 2Br-(g) 2 x -325 = -650
31 + 193 - 650 = -426 to make Br
So far, the process has cost 1913 - 650 = +1263
The heat of formation is -675. So the lattice energy is -1938
2007-11-15 07:59:05 · answer #3 · answered by steve_geo1 7 · 2⤊ 4⤋
The problem can be solved using Born-Haber Cycle. I tried but couldn't get solved sorry. Perhaps it will help you- put the values in correct places and do the calculation.
http://en.wikipedia.org/wiki/Born-Haber_cycle
2007-11-15 08:07:46 · answer #4 · answered by a new world 2 · 0⤊ 0⤋
delta Hf* [CaBr2 (s)] -675 delta Hf* [Ca(g)] 179 delta Hf* [Br(g)] 112 I1 (Ca) 590 I2 (Ca) 1145 E(Br) -325 Answer: + 2163 kJ/mol Mastering Chemistry (correct answer btw. ^ sam is right)
2016-05-23 07:26:30 · answer #5 · answered by ? 3 · 0⤊ 1⤋