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Mathematics 16 Online
OpenStudy (anonymous):

Prove that Hess Law works. Show this by displaying which two thermochemical equations add up to the other. Also show how the corresponding two engergy values add up the other.

OpenStudy (anonymous):

The equations are NaOH(s)->NaOH(aq)+42q/mol NaOH(s) +HCl(aq)->H20+NaCl(aq)+91.1q/mol NaOH(aq)+HCl(aq)->H2O + NaCl(aq) +.59q/mol

OpenStudy (anonymous):

Chemistry question in math?

OpenStudy (anonymous):

are you sure you have the right value for the enthalpy change in the third equation?

OpenStudy (anonymous):

Yes.

OpenStudy (anonymous):

can't be.

OpenStudy (anonymous):

the source is probably wrong.

OpenStudy (anonymous):

because?

OpenStudy (anonymous):

the energy change in the second equation should be the sum of the energy changes in the first and third equations

OpenStudy (anonymous):

The equations are NaOH(s)->NaOH(aq)+42q/mol NaOH(s) +HCl(aq)->H20+NaCl(aq)+91.1q/mol NaOH(aq)+HCl(aq)->H2O + NaCl(aq) +.59q/mol add the first and third equations, then cancel the NaOH(aq) from both sides

OpenStudy (anonymous):

you end up with 2 copies of the second equation with different energy outputs and that shouldn't happen

OpenStudy (anonymous):

oh.. Ok. this is starting to make some sense now. I didnt know how to find the equations to add. It does say in the next question why are they not exactly the same as the Hess Law suggest.

OpenStudy (anonymous):

it definitely shouldn't be THAT far off.

OpenStudy (anonymous):

I got the entropies off a lab. So too many variables that could have happened.

OpenStudy (anonymous):

where are you that you use q/mol? all the labs I've seen used kJ

OpenStudy (anonymous):

California. Oh yah. Im looking at it right now. it should be kj/mol.

OpenStudy (anonymous):

Also if I had used half the grams of NaOH would it affect the heat absorbed by water?

OpenStudy (anonymous):

so then say experimental error for what makes it not exact.

OpenStudy (anonymous):

if you used half the amount of NaOH you would produce about half the heat?

OpenStudy (anonymous):

and your uncertainty would be higher if you have to worry about that

OpenStudy (anonymous):

Sorry I dont understand what you just said

OpenStudy (anonymous):

then you don't have to worry about it :)

OpenStudy (anonymous):

So it wouldnt matter about the grams?

OpenStudy (anonymous):

@Peter14

OpenStudy (anonymous):

no, if you halved the grams the reaction would produce half the energy so the water would absorb half the energy.

OpenStudy (anonymous):

thanks. My chem teacher loves giving us labs that she never taught about. @Peter14

OpenStudy (anonymous):

good luck

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