8.73 part C

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ClaireHW
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8.73 part C

Postby ClaireHW » Tue Jan 23, 2018 10:43 pm

Use bond enthalpies in Table 8.6 and 8.7 to estimate the reaction enthalpy for

CH4 + CCl4 -> CHCl3 + CH3Cl

I was confused by the method shown in the Solutions Manuel. Could someone explain step by step how this is done?

Thanks!

(Claire Woolson Dis 1K)

Rachel Lu_dis1H
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Re: 8.73 part C

Postby Rachel Lu_dis1H » Tue Jan 23, 2018 11:13 pm

the reaction enthalpy of the whole equation is the bond enthalpy of reactants - bond enthalpy of products. This is because breaking bonds is endothermic (+) and requires energy while forming bonds releases energy and is exothermic (-). For me, I draw out the lewis structures and count how many different bonds there are. So for CH4 there are 4 C-H bonds. Then you find how much energy it takes to break a C-H bond and multiply it by 4. You do this for each reactant and product and then plug them into the equation I mentioned in the first sentence.

mhuang 1E
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Re: 8.73 part C

Postby mhuang 1E » Tue Jan 23, 2018 11:56 pm

I was also confused on how the enthalpy of the rxn would be 0 from the solutions manual. How does 1 C-C bond, 4 C-H bonds, and 1 H-H bond equal to 1 C-C bond, and 6 C-H bonds from the products?

Jared Smith 1E
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Re: 8.73 part C

Postby Jared Smith 1E » Wed Jan 24, 2018 12:04 am

The enthalpy of the reaction is 0 because the bonds being broken and formed have the exact same energy as one another. For example 4 C-H bonds are broken and 4 Cl-C bonds are broken. Then between the two reactant molecules 4 C-H bonds are formed and 4 Cl-C bonds are formed. The amount of bonds broken and formed are the same, they atoms just bond to the the carbons in different proportion than the reactants.


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