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Topic: Formation reactions and hess's law  (Read 28368 times)

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Formation reactions and hess's law
« on: February 20, 2006, 06:22:13 PM »
Hi everyone, im just wondering if anyone can confirm whether i have used the right method to solve this question, im a bit lost because the reaction in the question looks like part of the solvay process which is used to produce sodium carbonate, which if it is, I have used the wrong reactions in the hess's law calculation (below), but the reason why i did use those reactions is because the question states to write out all of the formation reactions adding up to...

so in other words all of the reactants must apear in their standard state form dont they? for e.g O3 is wrong, it should be O2 + 1/2O2

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.

rxn 4 (reversed): 2NaHCO3(s)---->2Na(s) + H2(g) + 2C(s) + O2(g) + 1/2O2(g)

so: Delta H rxn 4 (reversed) = -(Delta Hrxn (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 out on this q it would be greatly appreciated.

cheers,

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Re:Formation reactions and hess's law
« Reply #1 on: February 20, 2006, 11:58:41 PM »
Have I made sense with what ive written above? any ideas?
The only stupid question is a question not asked.

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Re:Formation reactions and hess's law
« Reply #2 on: February 21, 2006, 03:29:42 AM »
It should be:
(-1810.11KJ)+(1901.62KJ)

Thermal decomposition of NaHCO3 is slightly endothermic reaction
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Re:Formation reactions and hess's law
« Reply #3 on: February 21, 2006, 04:46:34 AM »
It should be:
(-1810.11KJ)+(1901.62KJ)

Thermal decomposition of NaHCO3 is slightly endothermic reaction

OK this is good, means i didnt have too much wrong with my method just a miscalculation at the end,

BUT, from what youve said in your post AWK about it should be

(-1810.11KJ)+(1901.62KJ)

even though ive had to change the sign of dH for rxn 4 to positive, by doing what your suggesting breaks , what I thought to be the core rule to calculating dH (or dG, for that matter) as:

dH rxn = sigma dHf (products) - sigma dHf (reactants)

I know this is somthing fairly simple but i just cant get my head around it???

for example say the dH value for the reactants was -1901.62 KJ then

dH rxn = (-1810.11KJ)-(-1901.62KJ)

which also = (-1810.11KJ)+(1901.62)

what im asking is, why did you change it from (products - reactants) to (products + reactants), and how were you able to do so???

AAAAAAAAhhhhhhhh!! my head hurts (such a numbnut lol)

:stupid:

« Last Edit: February 21, 2006, 04:51:38 AM by madscientist »
The only stupid question is a question not asked.

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Re:Formation reactions and hess's law
« Reply #4 on: February 21, 2006, 07:14:07 AM »
dH rxn = sigma dHf (products) - sigma dHf (reactants)

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

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Re:Formation reactions and hess's law
« Reply #5 on: February 21, 2006, 06:16:26 PM »
Allright heres how i did it breifly:

rxn 1: C(s) + O2(g) ------> CO2(g)        Delta H1 = -393.51 KJ

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

rxn 3: 2Na(s) + C(s) + O2(g) + 1/2O2(g) ----->Na2CO3(s)
Delta H3 = -1130.77KJ

rxn 4: 2Na(s) + H2(g) + 2C(s) + O2(g) + 1/2O2(g)----> 2NaHCO3(s)                                                  Delta H4 = -1901.62 KJ

Delta H'4 = 1901.62 KJ

adding these equations and cancelling common species gives the following net rxn:

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

Delta Hnet = Delta H1 + Delta H2 + Delta H3 + Delta H'4

= (-393.51KJ)+(-285.83KJ)+(-1130.77KJ)+1901.62
= 91.51 KJ

I was trying to calculate it directly from rxn4 (reversed) but although the reactant is the same, the products are not the same in the rxn which im looking for an enthalpy value for.

I cant beleive it, its actually sunk in!!