# Enthalpy of Different Chemical Processes

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#### Enthalpy of Formation (ΔfH° or ΔformationH°):

• The change in enthalpy of a chemical reaction at a given temperature and pressure, when one mole of the substance is formed from its constituent elements in their standard states is called as the heat of formation.

#### Explanation:

• Consider following reaction

C(s)  + O2(g)  → CO2(g)   ,  ΔformationH°  =  -395.39 kJ mol-1

• Since one mole of carbon dioxide gas is formed we can say that the heat of formation of carbon dioxide gas is -395.39 kJ.

#### Notes:

• Standard state of the element is that stable state of the element in which it exists at 1 atm. Pressure and 298 K
• Enthalpies of elements in their standard states are arbitrarily taken as zero.
• Enthalpy of a compound is equal to its heat of formation.
• When solving problems on heat of formation make sure that the product side of the thermochemical equation has one mole of the substance whose heat of formation is to be calculated.

#### Enthalpy of Dissociation (ΔdissociationH°):

• Change in enthalpy of a chemical reaction at a given temperature and pressure, when one mole of a substance is dissociated into its constituent elements is called heat of dissociation.

#### Explanation:

• Consider following reaction

H2(g) →  2H(g),    ΔdissociationH° = + 435.136 kJ mol-1

• Since one mole of hydrogen gas is dissociated the heat of dissociation of hydrogen gas is + 435.136 kJ

#### Enthalpy of Combustion (ΔcH° or ΔcombustionH°):

• Change in the enthalpy of a chemical reaction at a given temperature and pressure when one mole of a substance is combusted (burn)  completely in excess of oxygen is called heat of combustion.

#### Explanation:

• Consider following reaction

C(s)  + O2(g)  → CO2(g)   , ΔH  =  -395.39 kJ mol-1

• Since one mole of carbon is combusted completely the heat of combustion of carbon is – 395.39 kJ.

#### Enthalpy of Neutralization (ΔneutralizationH°):

• Change in the enthalpy of a chemical reaction at a given temperature and pressure when one gram equivalent weight of an acid is completely neutralized by one gram equivalent weight of the base is called heat of neutralization.

#### Explanation:

• Consider following reaction

HCl(aq) + NaOH(aq)  →   NaCl  +    H2O     ΔneutralizationH°  =  -56.9 kJ

• The heat of neutralization of HCI by NaOH is -56.9 KJ.
• For all strong acids and bases, the heat of neutralization is same because, in their neutralization reaction, there is a combination of H+ ions of an acid with OH- ions of the base to produce un-dissociated water.

### Change of Phase:

• Change of phase involves the change in the physical state of matter. During the phase change, the chemical properties of the substance do not change but physical properties change. Following are the types of phase changes.

#### Fusion:

• This is the process in which the matter changes from solid state to liquid state.
• It is endothermic process
• e.g. melting of ice H2O(s) → H2O(l)

#### Vapourization:

• This is the process in which the matter changes from liquid state to gaseous state.
• It is endothermic process
• e.g. boiling of water H2O(l) → H2O(g)

#### Sublimation:

• This is the process in which the matter changes from the solid state into gaseous state directly.
• It is endothermic process
• e.g. heating of camphor Camphor(s) → Camphor(g)

#### Characteristics of Change of Phase:

• The phase change always takes place at constant pressure and temperature.
• During phase transition, there is an equilibrium between the two phases. Thus both the phases exist simultaneously.
• The Change in temperature takes place only when completion of phase transition.

#### Enthalpy of Fusion (ΔfusH°):

• The enthalpy-change that accompanies the fusion of one mole of a solid without the change in a temperature at constant pressure is called its enthalpy of fusion.

#### Explanation:

• Consider following reaction

H2O(s) → H2O(l),               ΔfusionH  = +6.01  kJ mol-1

• Thus, the equation indicates that when one mole of ice melts at 0° C at 1 atm pressure, the enthalpy-change is 6.01 kJ. i.e. 6.01 kJ of energy is absorbed.

#### Enthalpy of Freezing (ΔfreezeH°):

• The enthalpychange that accompanies the freezing of one mole of a liquid without change in a temperature at constant pressure is called its enthalpy of freezing.

#### Explanation:

• Consider following reaction

H2O(l) → H2O(s),               ΔfreezeH  = +6.01  kJ mol-1

• Thus, the equation indicates that when one mole of water freezes at 0° C at 1 atm pressure, the change in enthalpy is -6.01 kJ. i.e. 6.01 kJ of energy is released.

#### Enthalpy of Vaporization  (ΔvaporizationH°):

• The enthalpy change that accompanies the vaporization of one mole of a liquid without a change in a temperature at constant pressure is called its enthalpy of vaporization.

#### Explanation:

• Consider following reaction

H2O(l) → H2O(g),   ΔvapourizationH  = + 40.7  kJ mol-1

• Thus, the equation indicates that when one mole of water vaporizes at 100° C at 1 atm pressure, the change in enthalpy is + 40.7 kJ. i.e. 40.7 kJ of energy is absorbed.

#### Enthalpy of Condensation (ΔcondensationH°):

• The enthalpy change that accompanies the condensation of one mole of a liquid without a change in a temperature at constant pressure is called its enthalpy of condensation.

#### Explanation:

• Consider following reaction

H2O(l) → H2O(g),   ΔcondensationH  = – 40.7  kJ mol-1

• Thus, the equation indicates that when one mole of water vapours condense at 100° C at 1 atm pressure, the enthalpy-change is – 40.7 kJ. i.e. 40.7 kJ of energy is released.

#### Enthalpy of Sublimation  (ΔsublimationH°):

• The direct conversion of solid to vapour without going through the liquid state is called sublimation.The enthalpy-change that accompanies the condensation of one mole of a solid directly into vapours at a constant temperature at constant pressure is called its enthalpy of sublimation.

#### Explanation:

• Consider following reaction

H2O(s) → H2O(g),       ΔsublimationH  = +51.08  kJ mol-1

• Thus, the equation indicates that when one mole of ice sublimes at O° C at 1 atm pressure, the enthalpychange  is + 51.08 kJ. i.e. 51.8 kJ of energy is absorbed.
• It is to be noted that ΔsublimationH = ΔfusionH + ΔvapourizationH
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