Calculating delta H
posted by Anne
I honestly don't know where I went wrong with my calculations.
I've been adding the products separately. Then I add the reactants separately.
Finally, I take the product and subtract from the initial.
Could you do this so I can double check my work?
HCl + NaOH = Na + Cl + H2O
HCl= 1.672x10^2 = 167.2
NaOH= 4.258x10^2 = 425.8
Na = 2.40x10^2 = 240
H2O= 2.858x10^2 = 285.8
Cl= 1.6710x10^2 = 167.1
I took the HCl and NaOH and added them.
I took the sum of Na, H20, and Cl .
I took the products and subtracted and got 99.9= 9.99x10^1
I have to put my answer in scientific notation with 3 sig figs. this is what I got and it's apparently wrong. So my question is, where did my calculations go wrong? Can you double check and show the answer?
What's tricky is the negatives.

DrBob222
dHo rxn = (n*dHo products)  (n*dHo reactants)
It appears to me that you have left off the  sign. It isn't 99.9 but 99.9. Show your work if you're still stuck. 
Anne
Ok. Could you double check these two for me as well?
1.) NaOH = Na + OH
NaOH= 4.26x10^2
Na= 2.403x10^2
OH= 2.300x10^2
(426)(240.3+230)=44.3? 4.43x10^1
or should this be negative too?
2.) HCl + OH = H2O + Cl
HCl= 1.67 x10^2
OH= 2.3002 x10^2
H2O= 2.858 x10^2
Cl= 1.6710 x10^2
Work: (167+230.02)  (285.8+167.1)= 55.88 or 5.59 x10^1
Is this negative too? I've been doing the same steps. . I first changed the notation into standard form. This is the only work I've done. 
Anne
the answers are right with the negatives. My question is:
Could you show me on the calculator, typing in the parenthesis, of how you plugged in the numbers to get negative answers? I'm just confused on the negative. 
DrBob222
Perhaps we should straighten what you are trying to calculate. I have assume it is dHrxn and that is products  reactants. Are you calculating dH reaction or bond energies.
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