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Vaporization

Enthalpy of Formation
The enthalpy of formation of a given compound is defined as the enthalpy change when 1 mole of a given compound is formed, starting from the elements in their stable states of aggregation.

A few examples are
H2(g) + ½ O2(g) H2O(1) Δ (H2O,1)=-285.77 kJ
C(s)+ O2(g) CO2(g) Δ (CO2,g)=-393.5 kJ

Problem
From the following thermochemical equations, calculate the enthalpy of formation of cane sugar (C12H22O11).

Solution
  1. C12H22O11(s) + 12O2(g) 12CO2(g) + 11H2O(1) Δ H=-5.644 MJ
  2. C(s)+O2(g) CO2(g)ΔH=-0.393 MJ
  3. H2 (g) + ½ O2(g) H2O(1) ΔH=-0.286 MJ
Multiplying Eq. (b) by 12 and Eq. (c) by 11 and adding them, we get
12 [C(s) + O2(g) CO2(g)] ΔH=-12 0.393 MJ
11 [H2(g) + ½ O2(g) H2O(1)] Δ H=-11 0.286 MJ
12C(s) + 11 H2(g) +
Δ H=-7.862 MJ

Subtracting Eq. (a) from the above resulting equation, we get
12C(s) + 11H2(g) +ΔH=-7.862 MJ
-[C12H22O11(s)+12O2(g) 12CO2(g) + 11H2O(1)]

Δ H=+5.644 MJ
12C(s)+11H2(g)+
ΔH=-2.218 MJ

Enthalpy of Combustion
Enthalpy of combustion of a given substance is defined as the enthalpy change when one mole of this substance combines with the requisite amount of oxygen to give product in their stable states of aggregation. A few examples are
CH4(g)+2O2(g) CO2(g)+2H2O(l) Δ Hcomb=-891.85 kJ mol-1
C (graphite) + O2(g) CO2(g) Δ Hcomb=-393.51 kJ mol-1

Enthalpy of Fusion
Enthalpy of fusion of a given substance is defined as the enthalpy change when one mole of the given substance in the solid form is converted into liquid form at the said temperature. A few examples are
H2O(s)
MgCl2(s)

Enthalpy of Vaporization
Enthalpy of vaporization of given substance is defined as the enthalpy change when one mole of the given substance in liquid form is converted into vapour form at a constant temperature. A few examples are
H2O(1)  
C6H6(1)




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