Match the entries listed in Column I with appropriate entries listed in Column II.{:(,"Column I",,"Column II"),((A),"Hess' law",(p),2.303 log.(P_(2))/(P_(1))=(Delta_("vap")H)/(R)((T_(2)-T_(1))/(T_(1)T_(2)))),((B),"Combustion reaction",(q),Delta_("vap")H=88 JK^(-1) mol^(-1)xx"Boiling point in Kelvin"),((C),"Trouton's law",(r),"Exothermic"),((D),"Clausius-Cal-peyron equation",(s),DeltaH" remains the same irrespective steps"):}

Match the entries listed in Column I with appropriate entries listed i

1.	Match the entries listed in Column I with appropriate entries listed in Column II.{:(,"Column I",,"Column II"),((A),"Hess' law",(p),2.303 log.(P_(2))/(P_(1))=(Delta_("vap")H)/(R)((T_(2)-T_(1))/(T_(1)T_(2)))),((B),"Combustion reaction",(q),Delta_("vap")H=88 JK^(-1) mol^(-1)xx"Boiling point in Kelvin"),((C),"Trouton's law",(r),"Exothermic"),((D),"Clausius-Cal-peyron equation",(s),DeltaH" remains the same irrespective steps"):}
Answer» Solution :(A)HESS' law STATES that enthalpy change in a reaction remains the same whether the reaction takes place in one step or in several steps. (B) Combustion reactions are exothermic. (C ) `(Delta_(vap)H)/("Boiling POINT in K")=88 J K^(-) mol^(-1)` (D)`2.303 log.(P_(2))/(P_(1))=(Delta_(vap)H)/(T)((T_(2)-T_(1))/(T_(1)T_(2)))` It is clausius-clapeyron equation.