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Topic: Enthalpy and hess's law applied  (Read 14463 times)

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TNT_

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Enthalpy and hess's law applied
« on: February 16, 2006, 05:08:58 PM »
Hey there, how do u do?  :-X
Can i ask a question please?

If one knows enthalpy for reaction of NaHC03 and HCl, as well as Na2C03 and HCl, how is it possible to find enthalpy change of thermal decomposition of NaHC03?I think i know, but i may be completely wrong...

Say i chose the following values to use:
For this reaction: 2HCl + Na2C03 > 2NaCl + C02 + H20    delta H is   -7.794 KJ/mol
For this reaction: HCl + NaHC03  > NaCl + C02 + H20  delta H is  +9.659 KJ/mol

I have drawn the cycle as shown, and attached it. Is the cycle ok? and is my calculation correct or am I on the completely wrong track?

Using Hess’s law,
 delta h1= delta h2 + (-delta h3)

 delta h1= +9.659 - -7.794 = 17.453 KJ/mol
 

« Last Edit: February 18, 2006, 11:17:17 AM by TNT_ »

Offline plu

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Re:Enthalpy and hess's law applied
« Reply #1 on: February 17, 2006, 09:16:20 AM »
I'm sorry but I don't quite understand what you mean by "thermal decomposition of NaHCO3"  :-\  Do you have a chemical equation for this?

TNT_

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Re:Enthalpy and hess's law applied
« Reply #2 on: February 17, 2006, 07:18:32 PM »
yes of course,
2NaHC03 --heat--> Na2C03 + C02 +H20

I think im in the wrong forum... am i? ;)

Offline Borek

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Re:Enthalpy and hess's law applied
« Reply #3 on: February 17, 2006, 07:26:51 PM »
Roasted Na2CO3 decomposes further.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

TNT_

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Re:Enthalpy and hess's law applied
« Reply #4 on: February 18, 2006, 11:11:28 AM »
im just asking about the cycle. Simply whether i have the cycle which allows me to work out the enthalpy change for that reacton correct. Only the intermediate reaction for the thermal decomposition...
« Last Edit: February 18, 2006, 11:19:50 AM by TNT_ »

Offline madscientist

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Re:Enthalpy and hess's law applied
« Reply #5 on: February 18, 2006, 12:54:48 PM »
Hi TNT, you wouldnt happen to be studying through the uni of New England (Australia) would you?

Ive got the exact same question that im tackling now, the reactions you chose are not formation reactions which, if you have the same question is what there asking for, in other words the reactants have to appear as they would in there standard form, e.g, 2HCl is wrong because both chlorine and hydrogen gas are diatomic molecules (Cl2, H2).

my working on this question so far, which could be completely wrong, is:

the question states:

Using Hess's law, write out all of the formation reactions that add up to and calculate delta H (25C) for the following reaction:

2NaHCO3(s) --------> Na2CO3(s) + CO2(g) + H2O(l)


formation reactions:

rxn 1: C(s) + O2(g) ------> CO2(g)        Delta H rxn= -393.51 KJ

rxn 2: H2(g) + 1/2O2(g) -----> H2O(l)   Delta H rxn= -285.83 KJ

rxn 3: 2Na(s) + C(s) + O2(g) + 1/2O2(g) ----->Na2CO3(s)   Delta H rxn= -1130.77KJ

rxn 4: 2Na(s) + H2(g) + 2C(s) + O2(g) + 1/2O2(g)----> 2NaHCO3(s)   Delta H rxn= -1901.62 KJ

so far all of the products we need are on the product side of the formation reactions, which is where we want them. all we need to do to get the sole reactant (the 2NaHCO3(s)) onto the reactant side of the formation reaction, this is done by reversing rxn 4 and changing its sign in the process.

so: Delta H rxn (4) = 1901.62 KJ

and by subtracting the sum of the enthalpy change of the products from the reactants, delta H for the final rxn can be calculated:

Delta Hrxn = [(-1130.77KJ)+(-393.51KJ)+(-285.83KJ)] - 1901.62KJ

                = (-1810.11KJ)-(1901.62KJ)
         
                = -3711.73KJ

the value i end up with seems a little high, so ive probably gone about this question the wrong way totally, if anyone can help me and TNT on this q it would be greatly appreciated.

cheers,

madscientist :albert:
The only stupid question is a question not asked.

TNT_

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Re:Enthalpy and hess's law applied
« Reply #6 on: February 18, 2006, 01:55:56 PM »
no, im from South England...
It's quite a popular problem i presume. But your question is dfferent to mine madscientist, mine was not about the formation, but about the reactions with HCl only. Urs is a bit more complicated cycle whereas mine is simple. I figuered mine is correct neway. And the proper txtbook value for the thermal decomp of Nahc03 is 91.6 kJ mol–1.

Offline madscientist

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Re:Enthalpy and hess's law applied
« Reply #7 on: February 18, 2006, 02:50:31 PM »
no worries tnt,
I have to ask aswell, what do you mean by the thermal decomposition of NaHCO3 being 91.6 KJ mol  ??? because if you are talking about enthalpy, i.e energy per mole of substance, i have looked at a couple of different sources and they all say that the enthalpy value for NaHCO3 is approx. -950 KJ mol ???


anyway can anybody help me with my problem, would be very helpfull.

cheers

madscientist :albert:
The only stupid question is a question not asked.

TNT_

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Re:Enthalpy and hess's law applied
« Reply #8 on: February 18, 2006, 05:41:21 PM »
what i mean bythermal decomposition is the break down by using heat energy. therefore i would think that it is endothermic, requiring an input of energy

Offline madscientist

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Re:Enthalpy and hess's law applied
« Reply #9 on: February 18, 2006, 06:12:57 PM »
In that case i think you should be looking at heat capacities, Cp, Cv
The only stupid question is a question not asked.

TNT_

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Re:Enthalpy and hess's law applied
« Reply #10 on: February 19, 2006, 12:17:24 PM »
are your sources reliable madscientist for the values for the thermal decomp of NaHC03?

Offline madscientist

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Re:Enthalpy and hess's law applied
« Reply #11 on: February 19, 2006, 01:06:16 PM »
If you call three major published phys chem text books reliable then yes  :Lighten:
The only stupid question is a question not asked.

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