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Topic: help with a simple yet tricky problem  (Read 4325 times)

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Offline brycebb

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help with a simple yet tricky problem
« on: August 06, 2012, 11:33:10 PM »
The decomposition of nitroglycerin produces carbon dioxide, nitrogen, water and oxygen gases.
a. write a balanced equation
2C3H5N3O9---> 12CO2 + 6N2+ 10H20 + O2
I got that part right, not that difficult
b. If the decomposition of nitroglycerin releases 6.23kj/g of energy in the form of heat, what is the standard molar enthalpy of formation for nitroglycerin?

now for part b, my understanding of standard molar enthalpy of formation is the amount of energy evolved or absorbed in order to create one mol of a stable and pure substance from its component compounds.  I have no idea how to figure this out though,  I've tried doing this
1g(1mol/227.0872g)(1mol-rxn/1mol)(x)= -6.23kj
227.0872 is the MM of nitroglycerin
that method did not work, so I tried (227.0872g/1mol)(6.23kj/1g)=-1414.75kj/mol
that is the wrong answer as well.  Can anybody point me in the right direction, the correct answer is -3.7Xto^2kj/mol. 

Offline brycebb

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Re: help with a simple yet tricky problem
« Reply #1 on: August 07, 2012, 01:04:17 AM »
i forgot to mention that in part b you are given 1g of nitro

Offline sjb

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Re: help with a simple yet tricky problem
« Reply #2 on: August 07, 2012, 02:22:04 AM »
The decomposition of nitroglycerin produces carbon dioxide, nitrogen, water and oxygen gases.
a. write a balanced equation
2C3H5N3O9---> 12CO2 + 6N2+ 10H20 + O2
I got that part right, not that difficult
b. If the decomposition of nitroglycerin releases 6.23kj/g of energy in the form of heat, what is the standard molar enthalpy of formation for nitroglycerin?

now for part b, my understanding of standard molar enthalpy of formation is the amount of energy evolved or absorbed in order to create one mol of a stable and pure substance from its component compounds.  I have no idea how to figure this out though,  I've tried doing this
1g(1mol/227.0872g)(1mol-rxn/1mol)(x)= -6.23kj
227.0872 is the MM of nitroglycerin
that method did not work, so I tried (227.0872g/1mol)(6.23kj/1g)=-1414.75kj/mol
that is the wrong answer as well.  Can anybody point me in the right direction, the correct answer is -3.7Xto^2kj/mol. 

For b, it's from the elements in their standard states. So you also need to take into account your decomposition forms carbon dioxide and water.

Offline brycebb

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Re: help with a simple yet tricky problem
« Reply #3 on: August 07, 2012, 02:30:09 AM »
The decomposition of nitroglycerin produces carbon dioxide, nitrogen, water and oxygen gases.
a. write a balanced equation
2C3H5N3O9---> 12CO2 + 6N2+ 10H20 + O2
I got that part right, not that difficult
b. If the decomposition of nitroglycerin releases 6.23kj/g of energy in the form of heat, what is the standard molar enthalpy of formation for nitroglycerin?

now for part b, my understanding of standard molar enthalpy of formation is the amount of energy evolved or absorbed in order to create one mol of a stable and pure substance from its component compounds.  I have no idea how to figure this out though,  I've tried doing this
1g(1mol/227.0872g)(1mol-rxn/1mol)(x)= -6.23kj
227.0872 is the MM of nitroglycerin
that method did not work, so I tried (227.0872g/1mol)(6.23kj/1g)=-1414.75kj/mol
that is the wrong answer as well.  Can anybody point me in the right direction, the correct answer is -3.7Xto^2kj/mol. 

For b, it's from the elements in their standard states. So you also need to take into account your decomposition forms carbon dioxide and water.

so would I approach it like a Hess's Law problem, where I take known chemical reactions and their change in enthalpies and rearrange it to get the answer?

Offline AWK

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Re: help with a simple yet tricky problem
« Reply #4 on: August 07, 2012, 02:45:47 AM »
Quote
2C3H5N3O9---> 12CO2 + 6N2+ 10H2O + O2
wrong balancing
AWK

Offline brycebb

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Re: help with a simple yet tricky problem
« Reply #5 on: August 07, 2012, 03:05:30 AM »
Quote
2C3H5N3O9---> 12CO2 + 6N2+ 10H2O + O2
wrong balancing
ahhh I typed it wrong, I had it balanced two different ways, one way was
4C3H5N3O9---> 12CO2 + 6N2 +10H2O +O2
and the other was
2C3H5N3O9--->6CO2 + 3N2 + 5H20 + 1/2O2
I apologize for that, but any idea on where to go in this problem?

Offline Dan

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Re: help with a simple yet tricky problem
« Reply #6 on: August 07, 2012, 03:16:58 AM »
so would I approach it like a Hess's Law problem, where I take known chemical reactions and their change in enthalpies and rearrange it to get the answer?

Yes
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Offline sjb

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Re: help with a simple yet tricky problem
« Reply #7 on: August 07, 2012, 03:57:36 AM »
so would I approach it like a Hess's Law problem, where I take known chemical reactions and their change in enthalpies and rearrange it to get the answer?

Well, not quite like a Hess's law problem, it is a Hess's law problem to my mind... ;)

Offline brycebb

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Re: help with a simple yet tricky problem
« Reply #8 on: August 07, 2012, 06:50:08 PM »
so would I approach it like a Hess's Law problem, where I take known chemical reactions and their change in enthalpies and rearrange it to get the answer?

Well, not quite like a Hess's law problem, it is a Hess's law problem to my mind... ;)
haha that's what I meant, but thank you for the help, I eventually figured it out.  I greatly appreciate it.

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