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### 8.57 Hess's Law

Posted: **Fri Jan 26, 2018 1:08 am**

by **Hellen Truong 2J**

How do you find deltaH given enthalpies of combustion? I checked with the solution manual and I was confused about its setup of writing out all the combustion equations and then rearranging them to get the final equation so I keep getting positive 312kj/mol instead of negative.

### Re: 8.57 Hess's Law

Posted: **Fri Jan 26, 2018 10:32 am**

by **Shannon Wasley 2J**

The first step in finding the change in enthalpy of combustion reactions is creating each combustion reaction and balancing each equation. The problem gives you the compounds in which you have to make a combustion reaction for. For each compound you add 02 as a reactant and the products would be CO2 and H2O, but you have to remember to balance the equations.

The next step is adding the equations together to get the reaction of hydrogenation of ethyne to ethane. First find one of your three combustion reactions that has a reactant that is also a reactant in the hydrogenation reaction. For example, you could choose the combustion of C2H2, which has an enthalpy of (-1300).

Then you find one of your combustion reactions that has a product that is a product in the hydrogenation reaction. Since C2H6 is the only product in the hydrogenation reaction, you must flip your combustion reaction of C2H6, in order to make it a product in the reaction. This would flip the sign of the change in enthalpy of the combustion reaction of C2H6. It would change from -1560 to +1560.

You then add the previous to equations together (the normal combustion reaction of C2H2 and the reversed reaction of C2H6). This would add the enthalpies together, making the total +260 (-1300 + 1560)

Then you need to add the combustion reaction of H2, in order to cancel out all the other reactants and products that are not in the hydrogenation reaction. However, in the hydrogenation reaction, there is a 2H2, which means that you have to multiply the entire combustion reaction of H2 by 2, which also multiplies the change in enthalpy by 2 (-286 x 2 = -572). The reactants and products will cancel and the final answer of the change in enthalpy will be (-1300 + 1560 + -572) -312 kJ/mol

### Re: 8.57 Hess's Law

Posted: **Sun Jan 28, 2018 10:54 pm**

by **Anna Li 2E**

For the equation in 8.57 all of the compounds combusts into CO2+H2O, so the equation becomes (delta H)= (Hc C2H2) +2(Hc H2) -(Hc C2H6). Here we use Hessâ€™s Law to determine the entire reaction enthalpy so it would be (-1300 kJ/mol)+2(-286 kJ/mol)-(-1560 kJ/mol) which results in the answer -312 kJ/mol

### Re: 8.57 Hess's Law [ENDORSED]

Posted: **Mon Jan 29, 2018 1:23 am**

by **Kyra LeRoy 1E**

In a simple form, you write out the reactions for combustion for those given enthalpies (add + O2 to all of them, reactants will be CO2 & H2O, except for with H2, then all you get as a product is H20).

Then you just proceed as normal by multiplying/adding the equations together to cancel things out & find the final reaction enthalpy of the given equation.

If you add the enthalpies of combustion in the same way you would standard enthalpies then you will get 312, but you have to go through the process of multiplying equations and adding/canceling out to get -312.