HW #63

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Sarah Kang 2A
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Joined: Fri Sep 26, 2014 2:02 pm

HW #63

Postby Sarah Kang 2A » Sat Jan 17, 2015 11:10 pm

For homework problem #63 in chapter 7, why are the standard reaction enthalpies in the solutions manual listed next to their respective combustion equations? Aren't those reaction enthalpies for the compounds in the original equation?

Also, why do we have to use several combustion reactions in order to get to our final equation? Can't we just calculate the standard reaction enthalpy by subtracting the reaction enthalpies of the reactants from the reaction enthalpy of the product?

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

Re: HW #63

Postby Niharika Reddy 1D » Sun Jan 18, 2015 11:54 am

The data given in the problem is the standard enthalpy of combustion of each of the products and reactants of the final reaction. The standard enthalpy of combustion is the standard reaction enthalpy for the combustion of one mole of substance. This is why those reaction enthalpies are given next to their respective combustion reactions, since standard enthalpies of combustion are given, not standard enthalpies of formation.

As for the second part of your question, I think you're confusing enthalpy of reaction with standard enthalpies of formation. You can calculate the reaction enthalpy by subtracting standard enthalpies of formation of reactants from the standard enthalpies of formation of products but the question specifically asks to use the data given (ie standard enthalpies of combustion of each of the products and reactants) to determine the reaction enthalpy for the hydrogenation of ethyne to ethane. Since it specifies that we should use that information, we need to write out the individual combustion reactions and manipulate them, using Hess's Law to get the final reaction and reaction enthalpy.

Calvin Nguyen 1H
Posts: 15
Joined: Fri Sep 26, 2014 2:02 pm

Re: HW #63

Postby Calvin Nguyen 1H » Sat Jan 24, 2015 9:26 pm

Actually, the enthalpy is calculated by subtracting the standard enthalpy of the products by the stand enthalpy of the reactants.

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

Re: HW #63

Postby Niharika Reddy 1D » Sun Jan 25, 2015 12:30 am

Yes, that's another way of saying it, but it's the same as what was stated above. Subtracting standard enthalpies of formation of reactants from the standard enthalpies of formation of products is the same thing as subtracting the standard enthalpy of the products by the standard enthalpy of the reactants.


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