Quiz 1 Winter 2014 #3

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Ishkhan 3O
Posts: 24
Joined: Fri Sep 26, 2014 2:02 pm

Quiz 1 Winter 2014 #3

Postby Ishkhan 3O » Sat Jan 24, 2015 8:13 pm

I got 4372 kJ when the answer is -2186 kJ. I balanced the equation and made the coefficients 2,13,8,10, but they should be 1,13/2,4,10. I thought that coefficients had to be whole numbers. Also, why is the answer negative?

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

Re: Quiz 1 Winter 2014 #3

Postby Neil DSilva 1L » Sat Jan 24, 2015 8:23 pm

The question asks for the enthalpy for the combustion of one mole of butane. The balanced reaction (with all whole number coefficients) is the combustion of two moles of butane, so you divide the enthalpy you get using the balanced reaction by two to get the answer for one mole.

Yes, when you balance a reaction, you should have all whole number coefficients, but in the case of this problem, if you used the balanced equation with the fraction coefficients, you would still get the correct answer.

The answer should be negative because it's a combustion reaction. Combustion reactions are always exothermic, so they release energy. Make sure when you calculate the enthalpy you are adding the bond enthalpies for the bonds you are breaking and subtracting the bond enthalpies for the bonds you are forming.

Maria Davila 2I
Posts: 12
Joined: Fri Sep 26, 2014 2:02 pm

Re: Quiz 1 Winter 2014 #3

Postby Maria Davila 2I » Tue Jan 27, 2015 12:04 am

Add all bond enthalpies. Remember that when you are breaking bonds you need to use energy to break the bonds, therefore bond enthalpy is positive but when you are forming the bonds , those bond enthalpies will be negative because you are releasing. Then you add the positives (from the bonds broken) and the negatives (bonds formed) and you will get your enthalpy.

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