Most of the chemistry happens in open containers. Open containers have constant pressure. At constant pressure work is done to carry out the reaction and also to "push" the air up (called as expansion work). This expansion work is $${\Delta PV}$$. Thus some part of the heat leaks as expansion work without being converted into Internal Energy.
So we define a new term $${\Delta H ~(Enthalpy) = \Delta U + \Delta PV}$$. It accounts for expansion work and internal energy. As the heat is used only for internal energy and expansion work, Enthalpy is equal to heat supplied. There are various types of enthalpies depending upon Internal Energy is used to do what kind of work.
Phase is something that is uniform in composition and physical state. It is similar to state of matter (solid, liquid and gas), but these is only one state of matter (solid carbon) but there can be many phases (graphite, diamond)
Enthalpy of phase transition
It is the heat supplied at constant 1 atm pressure to bring about a change in phase. It can be $$\ce{H2O (liquid) <=> H2O (gas) -> +44KJ}$$ or $$\ce{C (graphite) <=> C (diamond) -> +4KJ}$$ Conversion of graphite into diamond is thermodynamically possible but kinetics does not allow it, as there are very strong bond to break and rearrange. This is true for every thermodynamical reaction.
Enthalpy of vapourisation and Enthalpy of Condensation
Enthalpy of vapourisation is the heat required to vaporise 1 mole of liquid into gas. (a phase change) Energy needs to be supplied and hence enthalpy is positive.
As Enthalpy is a state function, the reaction can proceed from the final state (gas) to initial state (liquid) by absorbing the same amount of energy. This heat absorbed at constant pressure is called as enthalpy of condensation and is numerically equal to negative of enthalpy of vapourisation.
Enthalpy of Fusion and Enthalpy of freezing
The heat required to melt 1 gm of solid into liquid at a constant pressure of 1 atm is enthalpy of fusion. 273
The enthalpy required for the reverse reaction is (liquid to solid) is called enthalpy of freezing. It follows the same lines as of above two enthalpies.
Enthalpy of sublimation and Enthalpy of vapour deposition
Some substances directly vaporise into gas, without forming liquid and liquid is not stable at any temperature and pressure. The enthalpy required to do so is enthalpy of sublimation. As enthalpy is a state function, the enthalpy required to deposit the gas into a solid is called enthalpy of vapour deposition.
Enthalpy of atomic and molecular changes
Enthalpy of ionisation and Enthalpy of electron capture
The energy required to remove the first electron of one mole of gas at constant pressure is called enthalpy of ionisation. The energy required to remove the second electron is called Second enthalpy of ionisation. The enthalpy of ionisation of 1 mole of gas at 273.15 Kelvin and 1 atm pressure is called "Ionisation Energy".
The enthalpy of capturing an electron by a neutral atom (neutral, not positive ion) is called enthalpy of electron capture. At 273.15 K and 1 atm, for 1 mole this energy is also called "Standard electron gain Enthalpy"
Enthalpy of disassociation and Enthalpy of bond formation
The heat required to disassociate a molecule into 2 entities is called enthalpy of bond dissociation Eg $$\ce{H2O -> H2 + O ~~~~ \Delta H = +927 KJ}$$
the opposite of it is enthalpy of bond formation Eg. $$\ce{H2 + O2 -> H2O}$$ Here the enthalpy of (H-O-H) is not equal to 2 times enthalpy of (O-H)
for this particular reaction, $$\ce{H2O(gas) -> H + HO^. ~~~~\Delta H = +428KJ }$$ and $$\ce{OH^. -> H + O^. ~~~~ \Delta H = +428 KJ}$$. Therefore total enthalpy is +927KJ. As Enthalpy of disassociation is dependent upon reaction itself $$\ce{\Delta H(O-H)~is~the~average~value~of~\Delta H(O-H)~of~H-OH~,~\Delta H(O-H)~of~CH3OH~etc.}$$ 979