Find the ΔG° for the combustion of half a mole of glucose C6H12O6 in kJ

-1435 kJ

Explanation:

The Gibbs free energy, ΔG is an indication the amount of (useful) work that can be obtained from a chemical reaction involving reacting species and it is negative for spontaneous reaction

ΔG° = ΔH° - TΔS

ΔG = ΔG° + RT㏑Q

ΔG = G (reactants) - G(products)

Given that the combustion reaction of glucose is as follows;

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

The above reaction has a ΔG° value of -2870 kJ

Therefore, the free energy for half mole of glucose = -2870 kJ/2 = -1435 kJ

14,16 18 reacts with group

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.