Is it really appropriate to measure entropy change using enthalpy?
Yes.  So to get the change in entropy, you need to identify a reversible path between thermodynamic equilibrium state A and thermodynamic equilibrium state B.  If it is a phase change, you start out with liquid  at temperature T and equilibrium vapor pressure P(T) (state A), and end up with vapor at the same temperature and pressure (state B).  This can be done reversibly by putting one mole of a liquid in a cylinder with a massless frictionless piston and holding the external pressure on the piston constant at P while adding heat very gradually (reversibly) until all the liquid changes to vapor.  During this process, the volume of material in the cylinder increases from that of one mole of liquid to that of one mole of vapor.  Since the pressure is held constant throughout, the work done is $P(V_v-V_l)$.  From the first law, the change in internal energy is equal to the heat added minus the work done:  $(U_v-U_l)=Q-P(V_v-V_l)$.  Rearranging this equation gives:$$(U_v+PV_v)-(U_l+PV_v)=H_v-H_l=\Delta H_{lv}=Q$$So the heat that you add reversibly is equal to the change in enthalpy in going from the liquid at T and equilibrium vapor pressure P(T) to the vapor at T and equilibrium vapor pressure P(T).  Since this is a reversible process at constant temperature, $\Delta S=\frac{Q}{T}=\frac{\Delta H_{lv}}{T}$.  This completes what I wanted to say about the phase change situation.  I will move on to the chemical reaction situation after you have had a chance to look this over and ask questions.