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N-butane has an enthalpy of formation of $\pu{-125.6 kJ/mol}$, whereas it is $\pu{-134.2kJ/mol}$ for 2-methylpropane. When comparing the heat set free at the combustion process, why is it higher for n-butane? Isn't the enthalpy of formation the negative of enthalpy of combustion and therefore 2-methylpropane should be higher?

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Take 4 moles of carbon and 5 moles of H2. Synthesize a mole of n-butane and measure the heat evolved (125.6 kJ). Then repeat the process, but make 2-methyl propane (you'll measure 134.2 kJ). Then, with sufficient oxygen, prepare 4 moles of CO2 and 5 moles of H2O from each of the previous two procedures and measure the heat of combustion (more for n-butane).

The total heat evolved to form the final oxidation products from carbon and H2 will be the same, but since the paths are different, the heats evolved in the two steps of each procedure will be different. n-Butane evolves less heat upon formation, therefore can (must) evolve more heat in the next step when it is oxidized to final products.

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Isn't the enthalpy of formation the negative of enthalpy of combustion

No! That certainly isn't true.

Consider this equation: $\ce{CH4 + O2 -> CO2 + H2O}$. If you reverse it, you'll get: $\ce{CO2 + H2O -> CH4 + O2}$ which is clearly not an enthalpy of formation because:

  1. the left hand side of the reaction does not have elements instead compounds
  2. the right hand side of the reaction should have had a single product, but instead has two

I cannot comment on the second part of your question because I could not find standard data for heats of combustion of both the compounds.

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