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Old Dec 6th 2011, 01:15 PM   #1
m56
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Heat absorbed per mole and other math problems

Hello,

We're working on bond energies in my chemistry course, and I'm having some trouble with bond energies. I used the Table of Bond Energies, but didn't end up even close to the answer for this and other math-related questions:

The following molecules contain only single bonds: NH3(g) + 3F2(g) -> NF3(g) +3 HF(g). Based upon the data in the Table of Bond Energies, the heat evolved or absorbed per mole of NH3 that reacts is
A. +51 kJ/mol. B. +867 kJ/mol. C. –51 kJ/mol. D. –867 kJ/mol.
It is estimated that 750 Gt of carbon resides in the atmosphere and that another 5,000 Gt is stored in the earth as fossil fuels. If 10% of the total fossil fuel carbon is combusted and released into the atmosphere, what would be the percentage increase for carbon in the atmosphere? (1 Gigaton (Gt) = 109 tons)
In the sequence of reactions:
NO2 + Br → NOBr + O
O + NO2 → NO + O2 NO+NOBr→N2O2 +Br
(I'm new, so I don't know how to go subscripts, sorry!)
A. Br is an intermediate, NOBr a product, and O a reactant. B. Br is a reactant, NOBr a catalyst, and O a product.
C. Br is a catalyst, and NOBr and O intermediates.
D. Br is a product, NOBr a reactant and O a catalyst.
I'm not too sure how to know which is which.

Thanks!
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Old Dec 7th 2011, 12:50 AM   #2
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1. Find the energy required to break down NH3 into the constituent atoms.
Find the energy required to break 3 moles of F2 into atoms of F

Find the energy released when 1 mole of NF3 is formed.
Find the energy released when 3 moles of HF are formed.

Subtract the first part (two absorbing energy to break) from the second (two releasing energy when forming) to get the answer.

2. Find 10% of the carbon stored in fossil fuels. Divide this amount by the mass of carbon in the atmosphere and multiply by 100 to get the answer in percentage.

3. Here, you need to know your terms well.
An intermediate is something which is in the middle, which here means that it is something formed in the middle of the reaction and is often no more there at the end of the reaction.
A catalyst is something which is there at the start and end of a reaction. It's purpose is that it speeds up the rate of a reaction.
A product is something which appears only at the end of a reaction.
A reactant is something which appears only at the beginning of a reaction.

Can you try those out from what I gave you and post what you get?
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Old Dec 7th 2011, 07:51 AM   #3
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Thanks Unknown008!

For #1, I can't find N-F on the table of bond energies to solve the equation (I'm using this one): http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm

However, I did get 202.8 for my reactants.

For #2, I got 66%. I found 10% of 5000 and then divided that by 750 to get 2/3.

For #3, I got C, since Br is both at the start and end of an equation, and then NoBr and O only appear in the middle of the equation, never the start or end.
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Old Dec 7th 2011, 07:57 AM   #4
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2. Good
3. Good

1. Nope, that is not the correct table that you should use. You need to pay attention to the units and as you can see, there are no units mentioned in the table. But if you read below the table, it is mentioned that the values are in kcal, which is NOT the same units as in your question. Furthermore, there is N-H in this table, in the 1st column, 13th row.

Try it again please.
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Old Dec 7th 2011, 09:37 AM   #5
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Haha, I was looking for N-F, but I have a different table now (this one: http://butane.chem.uiuc.edu/cyerkes/...denergies.html)

For the reactants, I did: 391 + (3 * 154) and for the products I did 272 + (3 * 565), but I didn't get any of the answers.
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Old Dec 7th 2011, 09:42 AM   #6
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Originally Posted by m56 View Post
Haha, I was looking for N-F, but I have a different table now (this one: http://butane.chem.uiuc.edu/cyerkes/...denergies.html)

For the reactants, I did: 391 + (3 * 154) and for the products I did 272 + (3 * 565), but I didn't get any of the answers.
Oh oops... I don't know why I read N-H...

Okay, remember that in 1 molecule of NH3, you have 3 N-H bonds
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