Thermodynamics ...

$$H_2(g) + \frac{ 1 }{ 2 }O_2(g)\longrightarrow 2H_2O(l);   \Delta H = -286  kJ$$
$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l)......... kJ(\pm?)$$

The heat released when the requisite amount of ions in the gaseous state combine to give 1 mol of crystal lattice is known as:

The enthalpy change for reaction (i) is:

Gibbs- Helmholtz equation is

The relationship between the free energy change $$(\Delta G)$$ and entropy change $$(\Delta S)$$ at constant temperature $$(T)$$ is 

What can be concluded about the values of $$\Delta H$$ and $$\Delta S$$ from this graph ?

The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elements
v : $${ \Delta H }_{ atomisation }$$ K = 90 kJ/mol
w : $${ \Delta H }_{ ionisation }$$ K = 418 kJ/mol
x : $${ \Delta H }_{ dissociation }$$ H = 436 kJ/mol
y : $${ \Delta H }_{ electron affinity }$$ H = 78 kJ/mol
z : $${ \Delta H }_{ lattice }$$ KH = 710 kJ/mol

The Born Haber cycle below represents the energy changes occurring at 298K when $$KH$$ is formed from its elements
v : $${ \Delta H }_{ atomisation }$$ $$K = 90 kJ/mol$$
w : $${ \Delta H }_{ ionisation }$$ $$K = 418 kJ/mol$$
x : $${ \Delta H }_{ dissociation }$$ $$H = 436 kJ/mol$$
y : $${ \Delta H }_{ electron affinity }$$ $$H = 78 kJ/mol$$
z : $${ \Delta H }_{ lattice }$$ $$KH = 710 kJ/mol$$

The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elements
v : $${ \Delta H }_{ atomisation }$$ K = 90 kJ/mol
w : $${ \Delta H }_{ ionisation }$$ K = 418 kJ/mol
x : $${ \Delta H }_{ dissociation }$$ H = 436 kJ/mol
y : $${ \Delta H }_{ electron affinity }$$ H = 78 kJ/mol
z : $${ \Delta H }_{ lattice }$$ KH = 710 kJ/mol

What is the free energy change $$(\Delta G)$$ when 1.0 mole of water at $$100^{\circ}\! C$$ and 1 atm pressure is converted into steam at $$100^{\circ}\! C$$ and 1 atm pressure?