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A branched alkane is more stable because it forms a crystal lattice structure in the solid phase. So it packs better and this maximizes attractive forces. So it is more stable. Lower heat of combustion = more stable.
i.e. why do branched isomers give off more heat when combusted than non-branched.
Great question. It's not commonly answered in organic chemistry textbooks, nor is it generally tested for, because it goes into hyperconjugation and molecular orbital theory. Move on. -------- *only if you MUST know* Essentially, in linear alkenes you're going to have a greater number of stabilizing hyperconjugative interactions where a C-C sigma bond is donating into a C-C sigma* orbital, and these are more stabilizing than interactions between C-H and C-C bonds due to better orbital overlap. E.g. in 2,2-dimethylpropane you can't have any C-C(sigma) - C-C sigma* bonds, whereas in n-pentane there are 4 potential interactions. Therefore n-pentane is going to be more stable.
These are very subtle effects and go way beyond the scope of what you're expected to know. For instance the difference in heat of combustion between 2,2,4-trimethylpentane and n-octane is only 30 kJ/mol. (5461 kJ/mol vs. 5430 kJ/mol). we're talking less than 1% difference here.
Hyperconjugation also explains why substituted alkenes are more stable, but that's another story.