The compounds n-butane, ch3(ch2)2ch3, and trimethylamine, n(ch3)3, have very similar molecular weights. however, their melting points are appreciably different. select the compound with the lowest melting point, and the correct reason why.

            The lowest boiling point is of n-Butane because it only experiences London Dispersion Forces between molecules.

                   Lets take start with the melting point of both compounds.

                                      n-Butane  =  - 140 °C

                                      Trimethylamine  =  - 117 °C

Intermolecular Forces in n-Butane:

                                                      As we know n-Butane is made up of Carbon and Hydrogen atoms only bonded via single covalent bonds. The electronegativity difference between C and C atoms is zero while, that between C and H atoms is 0.35 which is less than 0.4. Hence, the bonds in n-Butane are purely non polar in nature. Therefore, only London Dispersion Forces are found in n-Butane which are considered as the weakest intermolecular interactions.

Intermolecular Forces in Trimethylamine:

                                                             Trimethylamine (a tertiary amine) is made up of Nitrogen, Carbon and Hydrogen atoms bonded via single covalent bonds. The electronegativity difference between N and C atoms is 0.49  which is greater than 0.4. Hence, the C-N bond is polar in nature. Therefore, Dipole-Dipole interactions will be formed along with London Dispersion Forces which are stronger than Dispersion Forces. Therefore, due to Dipole-Dipole interactions Trimethylamine will have greater melting point than n-Butane.

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.

the answer your looking for is c).