Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Johnny 2375 on September 11, 2019, 12:57:00 PM
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Can't seem to figure out this one could you help me please thanks.
Calculate the average C-C bond enthalpy in benzene (C6H6) given the following data:
C(s) → C(g)
enthalpy change = +715 kJmol-1
H2(g) → 2 H(g)
enthalpy change = +436 kJmol-1
C6H6(l) → C6H6(g)
enthalpy change = +31 kJmol-1
enthalpy of formation (C6H6(l)) = +49 kJmol-1
E (C-H) = +413 kJmol-1
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It is the forum rule that you have to show your work to receive help.
What have you tried so far?
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I got the right answer which is 507 KJ mol-1 but my working out was wrong and accidental
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formulated this equation: 6C + 3H2 → C6H6
Then I tried using hess's law
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"working out was wrong and accidental" is rather vague, particularly since you say you got the right answer. Also, your formulated equation doesn't include any states of matter, which are obviously important to setting up the problem.
Here's a hint: start with the definition of the enthalpy of formation in terms of some of the chemical species noted in the information given to you. Then you can work backwards.
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I don't understand why there are 3 equations its confusing
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Ok, look: One isolated molecule of benzene contains 6 C-H bonds and 6 C-C bonds. You could theoretically form these bonds from 6 lone carbon atoms and 6 lone hydrogen atoms, all in the gaseous state. Whatever energy is stored in a benzene molecule from a starting state of 6 isolated carbon and 6 isolated hydrogen molecules goes into forming those bonds. You have to use gaseous states here because condensed phases have intermolecular energies to worry about. (The problem uses moles of molecules, but the idea here is the same.)
Represented by a chemical equation, this is 6C(g) + 6H(g) :rarrow: 1 C6H6
You have to assemble all those other pieces of information to find out the enthalpy change associated with this process. Your starting point, which is the tricky part to see, is the enthalpy of formation of liquid benzene. The enthalpy of formation has a very specific definition. I suggest you look up the definition of the enthalpy of formation, and write a chemical equation as it relates to benzene. You can be pretty sure it will involves some of the species in the equations you are given, and I suggest you keep close track of the phases of matter, because that's important. Then you can use a standard Hess law approach to determine the delta H of the above reaction.
There's one more logic step after that, but see if you can get to that point and the rest isn't too hard.
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Why isn't it 3H2 instead of 6H
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((6(413)+31+6x)) - ((6(715)+6(436)) = 49
6x - 4397 = 49
6x = 4446
x= 741
I don't know what i'm doing wrong
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I don't know either, because that's just a bunch of numbers. In chemistry you have be careful about writing out exactly what you're doing, and few problems won't benefit from explicitly writing out a chemical reaction formula. This is especially the case if you're trying to get someone to help you.
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What is the working out?
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For starters, your right hand side of the equation is wrong,it should be 49+31,you would have to add both the enthalpy of formation of liquid benzene and its subsequent conversion to gaseous