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### Topic: Hess's Law explanation please need help  (Read 35964 times)

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#### sam12103

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• Mole Snacks: +0/-0 ##### Hess's Law explanation please need help
« on: October 16, 2009, 09:04:31 PM »
I have been trying to do the homework for Chem, but unfortunately because my professor never explains anything, all the students in the class are always left confused. Can someone please explain Hess's law, I would prefer if you can give me an example problem, and solve it step by step. I am very confused with this concept. Because this is my first year taking Chemistry I am very new to this material, if you find it easy can you please explain this problem that I am going to give the easiest way, so I can grasp the concept.

Thank you very much

44.

The enthalpy changes of the following reactions can be measured:

C2H4(g) + 3O2(g) -> 2CO2 (g) +2H20 (l)     Change of rH degree= -1411.1 kJ/mol rxn
C2H5OH(l) + 3O2(g) -> 2CO2 (g) + 3H2O (l)  Change of rH degree= -1367.5 kJ/mol rxn

Use the values and Hess's law to determine the enthalpy change for the reaction

C2H4(g) +H2O (l) -> C2H5OH (l)

#### UG

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« Reply #1 on: October 16, 2009, 09:15:41 PM »
Hess's Law is really just an implication of the Law of Conservation of Energy and it states that: the enthalpy change in converting reactants to products is the same regardless of which route occurs and providing the initial conditions of the reactants and final conditions of the products are the same.
For example, a practical example of Hess's Law is the calculation of the standard enthalpy of formation of carbon monoxide, because it is impossible to measure  :delta:fHo (CO,g) direstly since some carbon dioxide is always formed. By measuring the enthalpies of combustion of both carbon monoxide and its component elements and linking the values obtained in an energy cycle, it is possible to determine the :delta:fHo (CO,g).
So...we want to find :delta:fHo (CO,g)

C(s) + 0.5O2 (g) CO (g)     :delta:H1 we do not know

So we can use information from two other reactions:
The formation of CO2 from C, which is really just the combustion of C
C(s) + O2 (g) CO2 (g)  :delta:H2 = -393 kJ per mole

And the formation of CO2 from CO, which is really just the combustion of CO

CO(g) + 0.5O2 (g) CO2 (g)  :delta:H3 = -283 kJ per mole

BY applying Hess's law, the overal enthalpy change for the converstion of C(s) to CO2 is independent of the route taken, therefore,  :delta:H2 =  :delta:H1 +  :delta:H3

Are you following, I sorry, my explanation sounds a bit confusing but the main thing is that energy is conserved so no matter if it only takes one step or a hundred, the enthalpy change is the same.
You can just substitute the numbers in:

-393 = -282 + :delta:fHo (CO,g)
So the :delta:fHo (CO,g) is -110 kJ per mole

« Last Edit: October 16, 2009, 09:27:19 PM by UG »

#### UG

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« Reply #2 on: October 16, 2009, 09:39:55 PM »
The enthalpy changes of the following reactions can be measured:

C2H4(g) + 3O2(g) -> 2CO2 (g) +2H20 (l)     Change of rH degree= -1411.1 kJ/mol rxn
C2H5OH(l) + 3O2(g) -> 2CO2 (g) + 3H2O (l)  Change of rH degree= -1367.5 kJ/mol rxn

Use the values and Hess's law to determine the enthalpy change for the reaction

C2H4(g) +H2O (l) -> C2H5OH (l)

So the equation you want is: C2H4(g) +H2O (l) -> C2H5OH (l)

You have to rearrange the other two:

C2H4(g) + 3O2(g) -> 2CO2 (g) +2H20 (l)     Change of rH degree= -1411.1 kJ/mol rxn
C2H5OH(l) + 3O2(g) -> 2CO2 (g) + 3H2O (l)  Change of rH degree= -1367.5 kJ/mol rxn

First thing you do is you must switch the second one around so you get C2H5OH(l) on the right hand side

2CO2 (g) + 3H2O (l) C2H5OH(l) + 3O2(g)  Change of rH degree= +1367.5 kJ/mol rxn

C2H4(g) + 3O2(g) -> 2CO2 (g) +2H20 (l)  Change of rH degree= -1411.1 kJ/mol rxn

From here on, the molecules on each side will cancel:

2CO2 (g) + 3H2O (l) C2H5OH(l) + 3O2(g)  Change of rH degree= +1367.5 kJ/mol rxn

C2H4(g) + 3O2(g) -> 2CO2 (g) +2H20 (l)  Change of rH degree= -1411.1 kJ/mol rxn

You'll see that what's left is the one you want:

C2H4(g) +H2O (l) -> C2H5OH (l)

So the enthalpy change is simply: 1367.5 + (-1411.1)
Sometimes it will not be simple and you might need to switch around multiple equations and/or multiply the coefficient.