I was wondering what I did wrong in the question below:
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I divided the bond dissociation energy by two as it produces two Chlorine atoms, and I took the negative of Chlorine's electron affinity to find the sum needed for -443 KJ/mol by solving for x. Any clue what I did wrong?
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mjc123ChemistSr. Member
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Posts: 1955Mole Snacks: +284/-12
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CyrustorzRegular MemberPosts: 29Mole Snacks: +0/-2
Actually it turns out my issue was just: As I said in my original post, I divided the bond dissociation energy by two even though the dissociation energy that they provided was already divided by two (the dissociation energy for 1/2 Cl2).I appreciate the help but I don't know why everyone on this board has to answer so cryptically. We all have to take responsibility for our own learning after class, so obviously I realize that if I just take the answer you give me I won't learn anything and I will fail on the test. You guys don't have to be the ones trying to get me to think critically... Though admittedly you don't have to be doing any of this at all. So again, the help is appreciated.

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Ah! I didn't spot that either. But the point I was making is that the lattice energy is the energy for the processCsCl(s)
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Cs+(g) + Cl-(g)and is a positive quantity. What you were calculating, correctly or incorrectly, was -U.
...I appreciate the help but I don't know why everyone on this board has to answer so cryptically. We all have to take responsibility for our own learning after class, so obviously I realize that if I just take the answer you give me I won't learn anything and I will fail on the test. You guys don't have to be the ones trying to get me to think critically... Though admittedly you don't have to be doing any of this at all. So again, the help is appreciated.

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I would suggest that you do not digress with statements like this in a regular thread.We have a forum where you can discuss such things.SeeComments for Staff and Comments from StafforGeneric DiscussionYou have done this kind of digression on other posts and it leads to a disjointed thread.Some time soon I will probably remove the distracting part of the post on this thread and possibly the others.