Posted by Kathryn on Tuesday, April 7, 2009 at 10:27pm.
I am beyond desperate. I do not understand how to calculate heat change at all and I have been looking online and reading my book for the last three hours. Below is posted a problem. It would be amazing if someone could work through it and explain each step they are doing and why they are doing it.
Calculate the heat change for the formation of lead(IV) chloride by the reaction of lead(II) chloride with chlorine
PbCl2(s) + Cl2(g) > PbCl4(l) H=?
___
Use the following thermochemical equations:
Pb(s) + 2Cl2(g) > PbCl4(l) h= 329.2kj
Pb(s) + Cl2(g) > PbCl2(s) h= 359.4kj

Chemistry  Kathryn, Tuesday, April 7, 2009 at 10:35pm
I flipped the second equation around to go with Hess's law which gives me a h value of 359.4 for the second one but I am lost after that.

Chemistry  Anon, Tuesday, April 7, 2009 at 10:58pm
ask your teacher.

Chemistry  DrBob222, Tuesday, April 7, 2009 at 11:03pm
I think you're trying to make it too hard.
Use equation 1 as is, reverse equation 2 (and change the sign of delta H for equation 2), add 1 to 2 and add delta H1 to the new delta H 2.
Pb(s) + 2Cl2(g) ==> PbCl4(l)
PbCl2(s) ==>Pb(s) + Cl2(g)

Pb(s)+2Cl2(g)+PbCl2(s)==>PbCl4(l)+Pb(s)+Cl2(g)
Note that Pb(s) cancels.
Note that 2Cl2 on the left and 1 Cl2 on the right leaves 1 Cl2 on the left to leave you with
Pb(s) + Cl2(g) ==>PbCl4(l)
329.2 + 359.4 = ??

Chemistry  GleekOut, Wednesday, February 6, 2013 at 8:07pm
to Drbob222 you actually are left with PbCl2(s) + Cl2(g) ==> PbCl4 because theres 2Cl2 on the first equation and the reverse equation only cancels out one of those Cl2's
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