Andrew Polyak Posted February 22, 2013 Posted February 22, 2013 The heats of combustion (to CO2 and H2O) of some C5 compounds are given below. Z-2-pentene –3369 kJ/mol E-2-pentene –3365 kJ/mol 1-pentene –3376 kJ/mol pentane–3536 kJ/molThe heat of formation of CO2 is –393 kJ/mol and that of H2O is –285 kJ/mol. What is the heat of hydrogenation of 1-pentene (in kJ/mol)? There are three questions on my OCHEM practice exam like this, and I can't understand any of them. I understand everything else, but I want the extra 50 points. I know how to calculate delta h using BDE, but that's where my knowledge stops.
Enthalpy Posted February 23, 2013 Posted February 23, 2013 1) What do you obtain by hydrogenating 1-pentene? 2) Can you find a series of transformations, each having a known heat, which is equivalent to this hydrogenation: same beginning, same end? Can you then combine the individual heats? Don't forget the additional H2O. Keep in mind: there is a gross and a net heat of combustion (translation to other languages may not be direct). One makes liquid water at rooms temperature, the other gaseous water at room temperature. The difference is big and is a very frequent source of mistakes. Always check the consistency of the heats of combustion. -285 kJ/mol is liquid water. Added warning: heats of formation are often measured by burning the compounds. It's universal and apparatus exists for that purpose BUT the combustion heat of hydrocarbons exceeds vastly their heat of formation, so that a slightly imprecise heat of combustion reults in a botched heat of formation or hydrogenation.
John Cuthber Posted February 23, 2013 Posted February 23, 2013 You also need to be aware of this http://en.wikipedia.org/wiki/Hess's_law
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