Thermochemistry And Finding Enthalpy Of Formation...

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Offline OmniBishounen

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Thermochemistry and finding enthalpy of formation...
« on: July 23, 2008, 08:49:37 PM » I've been racking my brains and trying many different solutions to this problem literally all day. I finally broke down and joined this site. :(Nitroglycerine, C3H5(NO3)3(l) is an explosive. When ignited, it produces water, nitrogen, carbon dioxide, and oxygen. Detonation of one mole of trinitroglycerine liberate 5725kJ of heat. What is the standard enthalpy of formation for C3H5(NO3)3(l)? Give your answer in kiloJoules.The first thing I did was balance the equation.4 C3H5(NO3)3(l) -> 10 H2O + 6 N2 + 12 CO2 + O2It is given that the enthalpy of water is -241.8 kJ per mole and for carbon dioxide it is -393.5 kJ per mole. The other two gases have none as they are standalone. For this side of the equation, I got a total of -7140. I then multiplied 5725 by 4 since 4 moles of trinitroglycerine liberates 22900 kJ of heat. From here, my frustration multiplies. Since the 22900 is a loss of heat, that would make the enthalpy of the trinitroglycerine 30040 kJ, right? I've had answers of 30040 kJ, wrong, I divided that answer by 4 thinking "per mole", still wrong. I even used the liquid equivalent of water for some reason and that was still wrong. I swear I've gone through this problem forwards and backwards trying to figure this out. On a side note, my professor told us to put the answer in as a positive although the answer is technically negative. Thanks in advance. As much as I hate to post my own problems, I've been at this one problem for hours and it's a sticking point for me. I'm sure the solution is something totally obvious that I'm simply missing. Any help would be appreciated. Logged

Offline Yggdrasil

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Re: Thermochemistry and finding enthalpy of formation...
« Reply #1 on: July 23, 2008, 09:02:59 PM » Perhaps it may be clearer if you write out the chemical reactions for the formation of water and the formation of nitroglycerin, then figure out how you add the equations together to get the correct values. Logged

Offline enahs

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Re: Thermochemistry and finding enthalpy of formation...
« Reply #2 on: July 23, 2008, 09:31:38 PM » Your problem is that the enthalpy value for the reaction of one mole of trinitroglycerine is actually the value for, coincidentally, 4 moles of trinitroglycerine. Logged

Offline OmniBishounen

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Re: Thermochemistry and finding enthalpy of formation...
« Reply #3 on: July 23, 2008, 10:52:28 PM » For the products: 10 H2O(g) + 6 N2(g) + 12 CO2(g) + O2(g)10 moles H2O x -241.8 = -2418 kJ 6 N2 = 012 moles CO2 x -393.5 = -4722 kJI add these up and get -7140 kJHrxn = -5725 x 4 = -22900-22900 = -7140 - x I even tried x = 15760 as well as x = 15760 / 4 = 3940This is in addition to my other attempts and I'm still getting it wrong. Logged

Offline Yggdrasil

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Re: Thermochemistry and finding enthalpy of formation...
« Reply #4 on: July 23, 2008, 11:18:02 PM » I get 3940 kJ/mol as well. Logged

Offline OmniBishounen

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Re: Thermochemistry and finding enthalpy of formation...
« Reply #5 on: July 23, 2008, 11:35:53 PM » Wow. I haven't the foggiest idea then. I put that in as an answer and I was counted as being wrong. I'm officially out of ideas and about ready to wave the white flag. Could be a mistake on their end. Logged

Offline macman104

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Re: Thermochemistry and finding enthalpy of formation...
« Reply #6 on: July 24, 2008, 12:03:15 AM » Not the first time online homework answers have been wrong. Check with your professor. Logged

Offline OmniBishounen

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Re: Thermochemistry and finding enthalpy of formation...
« Reply #7 on: July 24, 2008, 12:39:26 AM » I got it. I was using the wrong enthalpy for water. It was supposed to be the liquid form after all.Thanks for your help everyone. Logged

Offline enahs

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Re: Thermochemistry and finding enthalpy of formation...
« Reply #8 on: July 24, 2008, 08:10:18 AM » It does not matter what the computer system say. The correct answer is ~400 kJ.What you are getting for the mathematical correct answer for the problem is fine, I am just point out that the information in the question is wrong. Logged
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