C(graphite) + o2(g) → co2(g)δh o rxn = −393.5 kj/mol h2(g) + 1 2 o2(g) → h2o(l)δh o rxn = −285.8 kj/mol 2c2h6(g) + 7o2(g) → 4co2(g) + 6h2o(l)δh o rxn = −3119.6 kj/mol calculate the enthalpy change for the reaction below: 2 c(graphite) + 3h2(g) → c2h6(g)

Answer : The enthalpy change for the reaction,

Solution : Given,

(1)        

(2)      

(3)    

Now we have to calculate the enthalpy change for this reaction by doing:

Therefore, the enthalpy change for the reaction,

a. energy is absorbed in the reaction.

if in a given reaction the energy of products is greater than the energy of reactants it means that reaction is endothermic and during the reaction, the energy is absorbed by reactants. in an endothermic reaction, the products have more stored chemical energy than the reactants. and in this reaction the reactants are more stable than products before the compounds with less energy are always more stable.

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.