Hess's Law Problem

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TexasFluoride

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Even after reading the explanation, I still don't understand this problem.

Could somebody further explain.

Why do we solve it with (delta H reactant - delta H product). I solved it with the opposite since I thought Hess's law used (delta H product -delta H reactant).

The answer is D

Thank you

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So you have to figure out the Hf of C6H5OH.

The equation is given and the Hf is also given for that equation (the top one)
Now you do Products - Reactants

the molecule you are trying to figure out is one of the reactant in that equation.

(6(CO2) + 3(H20))-((1(C6H5OH)+ 7(O2)) = 729.8

plug everything in and consider C6H5OH x and O2 has zero heat of formation because it is in its standard form and then you'll get the answer.
 
So you have to figure out the Hf of C6H5OH.

The equation is given and the Hf is also given for that equation (the top one)
Now you do Products - Reactants

the molecule you are trying to figure out is one of the reactant in that equation.

(6(CO2) + 3(H20))-((1(C6H5OH)+ 7(O2)) = 729.8

plug everything in and consider C6H5OH x and O2 has zero heat of formation because it is in its standard form and then you'll get the answer.

I was under the impression that since this is a Heat of "formation" problem and since we are solving for the heat of formation for C6H5Oh. We will be forming C6H5OH, therefore C6H5OH will be the product not a reactant.
Could somebody please explain why my reasoning is flawed? I'm still confused.
Thanks
 
i cant remember how to tell if Hf is 0. i.e. how do you know when a substance is in standard form?
 
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hesslawproblem.jpg


Even after reading the explanation, I still don't understand this problem.

Could somebody further explain.

Why do we solve it with (delta H reactant - delta H product). I solved it with the opposite since I thought Hess's law used (delta H product -delta H reactant).

The answer is D

Thank you




I don't even want to look at that terrible explanation and confuse myself lol. What is that Kaplan? ---Just terrible.


First you need to write the eq to form phenol:

6C + 3H2 + 1/2O2 -----> C6H5OH


Now you need to alter the 3 given equations in order for them to give you your "desired" equation for the formation of C6H5OH.

First, flip rxn A.
Then, multiply Rxn B by 6
and then rxn C by 1.5

Whatever you do to the rxn you do to the delta H.
Thus A becomes -A (-729 in this case), B becomes (6 x 94.4) etc...


doing so should give you the right answer...
iits like -729 + (6x94) + (1.5 x 140) = approx positive 40


Its not product minus reactants - thats a somewhat similar concept, but a different kind of problem
this is application of hess' law.
 
the explanation given in the example is dead on right! this problem asks you to calculate the standard heat of formation of one of the reactants (C6H5OH) in the equation given Hf overall rxn, CO2, H20, and O2.
there is a typo in second to last line (should be 3x 68 not 3 x 136).

The Hf for the overall REACTION = sum Hf Products - sum Hf reactants. the Hfs of products and reactants are experimentally determined standard heats of formation that come from a table. here they are given to you (all but O2 which is O and Phenol which you have to solve for). so you are given all of the Hfs of reactants and products except for phenol which you can figure out algebraically.
 
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the explanation given in the example is dead on right! this problem asks you to calculate the standard heat of formation of one of the reactants (C6H5OH) in the equation given Hf overall rxn, CO2, H20, and O2.
there is a typo in second to last line (should be 3x 68 not 3 x 136).

The Hf for the overall REACTION = sum Hf Products - sum Hf reactants. the Hfs of products and reactants are experimentally determined standard heats of formation that come from a table. here they are given to you (all but O2 which is O and Phenol which you have to solve for). so you are given all of the Hfs of reactants and products except for phenol which you can figure out algebraically.


explanation still sucksssss
 
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