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January 26, 2015

January 26, 2015

Posted by **Hey!** on Sunday, August 6, 2006 at 6:57pm.

5C(s) + 6H2 >>> C5H12 Hf= -173.5kJ

C+O2>>> CO2 Hf= -393.5kJ

H2 + 1/2O2 >>> H2O Hf= -241.8kJ

H2O >>> H2O Hf= +44.0kJ

THEN I am given this:

mass of pentane= 2.15g

volume of water equivalent to calorimeter= 1.24L

initial temp. of calorimeter & contents= 18.4 degrees celcius

final temp. of calorimeter & contents= 37.6 degrees celcius

The one thing I dont get is why the give me the equations with the Hf. I dont get it. As well, what am I supposed to do with the rest of the information? Can you please help me get started. Thanks*

The equations, I think, are there to allow you to calculate, from the values given, the heat of combustion that you SHOULD get from experimental data. Reverse equation 1, then add equation 2 (multiplied by 5) and add equation 3 (multiplied by 6). You left something off the last equation but I assume it is H2O liquid ==> H2O steam (or it may be the other way around). You will need to add or subtract that 44 kJ (remember that is per mole and you have 6 mols H2O ) as appropriate. Also, remember that if you multiply an equation by 5 or 6 to also multiply the kJ and if you reverse the equation you must reverse the sign given. Finally, the last part is an experimental part.

Delta H = mass x specific heat water x (Tf - Ti) where Tf is the final temperature and Ti is the initial temperature. I would assume the density of water to be 1.00 g/L here so the mass of H2O would be 1,240 g. The specific heat of water is 4.18 J/g*C. This should give you the experimental value for the combustion of pentane

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