\({ Mg }_{ (s) }+{ Br }_{ 2(I) }\longrightarrow { MgBr }_{ 2(s) }\)     \({ \Delta H }_{ f }^{ o }=-524\quad KJ{ mol }^{ -1 }\)
Sublimation         : \({ Mg }_{ (s) }\longrightarrow { Mg }_{ (s) }\quad { \Delta H }_{ 1 }^{ o }=+148\quad KJ{ mol }^{ -1 }\)
Ionisation            :  \({ Mg }_{ (g) }\longrightarrow { { Mg }^{ 2+ } }_{ (g) }\quad { \Delta H }_{ 2 }^{ o }=+2187\quad KJ{ mol }^{ -1 }\)
Vapourisation      : \({ Br }_{ 2(I) }\longrightarrow { Br }_{ 2(g) }\quad { \Delta H }_{ 3 }^{ o }=+31\quad KJ{ mol }^{ -1 }\)
Dissociation        : \({ Br }_{ 2(g) }\longrightarrow { 2Br }_{ (g) }\quad { \Delta H }_{ 4 }^{ o }=+193\quad KJ{ mol }^{ -1 }\)
Electron affinity   : \({ 2Br }_{ (g) }+{ 2e }^{ - }\longrightarrow { 2{ Br }^{ - } }_{ (g) }\quad { \Delta H }_{ 5 }^{ o }=-662\quad KJ{ mol }^{ -1 }\)
Lattice enthalpy: \({ Mg }_{ (g) }+2{ Br }_{ (g) }^{ - }\longrightarrow { MgBr }_{ 2(s) }\quad { \Delta H }_{ 6 }^{ o }=?\)
\({ \Delta H }_{ f }^{ o }={ \Delta H }_{ 1 }^{ o }+{ \Delta H }_{ 2 }^{ o }+{ \Delta H }_{ 3 }^{ o }+{ \Delta H }_{ 4 }^{ o }+{ \Delta H }_{ 5 }^{ o }+{ \Delta H }_{ 6 }^{ o }\)
-524 = 148 + 2187 + 31 + 193 - 662 + \({ \Delta H }_{ 6 }^{ o }\)
\({ \Delta H }_{ 6 }^{ o }\) = -2421 KJ mol^-1
\({ Mg }_{ (g) }^{ 2+ }+{ 2Br }_{ (g) }^{ - }\longrightarrow { MgBr }_{ 2(s) }{ \Delta H }_{ 6 }^{ o }=-2421\quad KJ\quad { mol }^{ -1 }\)
\({ MgBr }_{ 2(s) }\longrightarrow { Mg }_{ (g) }^{ - }+{ 2Br }_{ (g) }^{ - }={ \Delta H }_{ 6 }^{ 0 }=-2421\quad KJ\quad { mol }^{ -1 }\)

\({ H }_{ 2(g) }+\frac { 1 }{ 2 } { O }_{ 2 }\longrightarrow { H }_{ 2 }{ O }_{ (1) }\quad \quad { \Delta H }^{ o }=-285.8\quad KJ\quad (1)\)
\({ C }_{ 9graphite) }+{ O }_{ 2 }\longrightarrow { CO }_{ 2 }\quad ​​​​{ \Delta H }^{ o }=-393.5\quad KJ\quad (2)\)
\({ CH }_{ 4(g) }+{ 2O }_{ 2 }\longrightarrow { CO }_{ 2(g) }+{ 2H }_{ 2 }{ O }_{ (1) }\quad { \Delta H }^{ o }=-890\quad KJ\quad (3)\)
equation (1) X₂ + (2) - (3)
\({ C }_{ (graphite) }+{ 2H }_{ 2(g) }\longrightarrow { CH }_{ 4(g) }\quad { \Delta H }_{ f }^{ o }=-74.7\quad KJ\)
(2 x -285.8) + (-393.5) - (-890.4)
= -571.6 - 393.5 + 890.4
= -965.1 + 890.4
= -74.7 KJ

Heat change at constant pressure (at atmospheric pressure) can be measured using a coffee cup calorimeter. A schematic representation of a coffee cup calorimeter is given in Figure. Instead of bomb, a styrofoam cup is used in this calorimeter. It acts as good adiabatic wall and doesn't allow transfer of heat produced during the reaction to its surrounding. This entire heat energy is absorbed by the water inside the cup. This method can be used for the reactions where there is no appreciable change in volume. The change in the temperature of water is measured and used to calculate the amount of heat that has been absorbed or evolved in the reaction using the following expression.
q =m_w C_wΔT

where m_w is the molar mass of water and C_w is the molar heat capacity of water (4184 kJ K^-1 mol^-1)·

A thermo chemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change (ΔH)
The following conventions necessarily adopted in thermo chemical equations:
(i) The coefficients in a balanced thermo chemical equation refer to number of moles of reactants and products involved in the reaction.
(ii) The enthalpy change of the reaction ΔH_r has unit KJ.
(iii) When the chemical reaction is reversed the value of ΔH is reversed in sign with the same magnitude.
(iv) Physical states of all species must be specified in a thermo chemical reaction.
(v) If the thermo chemical equation is multiplied throughout by a number, the enthalpy change is also be multiplied by the same number.
(vi) The negative sign of ΔH_r indicates an is exothermic and the positive sign of ΔH_r  indicates an endothermic type of reaction.
For example, consider the following reaction,
2H_2(g) + O_{2(g) ⟶} 2H₂O_(g) ΔH_r^o=-967.4 kJ
2H₂O_(g) ⟶2H_2(g) + O_2(g) ⟶ ΔH_r^o= + 967.4 kJ

Mathematical statement of the first law of thermodynamics is
ΔU = q+w
Case 1: For a cyclic process involving isothermal expansion of an ideal gas
ΔU= 0; ∴q = -w
In other words, during a cyclic process, the amount of heat absorbed by the system is equal to work done by the system.
Case 2: For an isochoric process (no change in volume) there is no work of expansion.
ΔV= 0, w =0, ΔU=q_v
In other words, during isochoric process, the amount of heat supplied to the system is converted to its internal energy.
Case 3: For an adiabatic process there is no change in heat. i.e. q = O. Hence
q =0;ΔU=w
In other words, in an adiabatic process, the decrease in internal energy is exactly equal to the work done by the system on its surroundings.
Case 4: For an isobaric process. There is no change in the pressure. P remains constant. Hence
ΔU=q+w
ΔU = q-PΔV
In other words, in an isobaric process a part of heat absorbed by the system is used for PV expansion work and the remaining is added to the internal energy of the system.