What is the change in enthalpy of the first reaction below, given the enthalpies of the other two reactions?

∆H0 = -222kJ/mol

Explanation:

Using Hess's law, we can find the ΔH of a reaction from the sum of another related reactions as follows:

Using the reactions:

(1) C(s) + O2(g) → CO2(g) ∆H0= -394 KJ/mol

(2) CO(s) + 1/2 O2(g) → CO2(g) ∆H0= -283 KJ/mol

Twice (1):

2C(s) + 2O2(g) → 2CO2(g) ∆H0= 2*-394 KJ/mol = -788kJ/mol

The inverse reaction of (2):

-(2) CO2(g) → CO(g) + 1/2 O2(g) ∆H0= 283 KJ/mol

Twice this reaction:

2*-(2) 2CO2(g) → 2CO(s) + O2(g) ∆H0= 2*283 KJ/mol= 566kJ/mol

Now, the sum of 2*(1) - 2*(2) produce:

2C(s) + 2O2(g) + 2CO2(g)→ 2CO2(g) + 2CO(g) + O2(g) ∆H0= -788kJ/mol +  566kJ/mol

Subtracting the molecules that ar in both sides of the reaction:

2C(s) + O2(g) → 2CO(g) ∆H0 = -222kJ/mol

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.