At 25°c, the following heats of reaction are known: 2clf(g) + o2(g) → cl2o(g) + f2o(g) δh°rxn = 167.4 kj/mol 2clf3(g) + 2o2(g) → cl2o(g) + 3f2o(g) δh°rxn = 341.4 kj/mol 2f2(g) + o2(g) → 2f2o(g) δh°rxn = –43.4 kj/mol at the same temperature, use hess's law to calculate δh°rxn for the reaction: clf(g) + f2(g) → clf3(g)

-108.7 KJ/mol

Explanation:

Hess's law state that in a multiple stage reaction, the enthalpy change is the sum of all the enthalpy changes.

2CIF (g) + O₂ (g) → Cl₂O(g) + F₂O (g)        ΔH°rxn =167.4  KJ/mol

2ClF₃(g) + 2O₂(g) → Cl₂O(g) + 3F₂O(g) ΔH°rxn = 341.4 kJ/mol

flip the above reaction

Cl₂O(g) + 3F₂O(g) → 2ClF₃(g) + 2O₂(g) ΔH°rxn = -341.4 kJ/mol

2CIF (g) + O₂ (g) → Cl₂O(g) + F₂O (g)        ΔH°rxn = 167.4  KJ/mol

2F₂(g) + O₂(g) → 2F₂O(g) ΔH°rxn = –43.4 kJ/mol

add all the enthalpy change = -341.4 kJ/mol + 167.4  KJ/mol – 43.4 kJ/mol = -217.4

2CIF (g) + 2F₂(g)  → 2ClF₃(g)          ΔH°rxn =  -217.4 KJ/mol

ClF(g) + F2(g) → ClF3(g) =         (ΔH°rxn =  -217.4 KJ/mol) / 2 = -108.7 KJ/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.