Using the heats of fusion and vaporization for water given in Exercise 99, calculate the change in enthalpy for the sublimation of water: H 2 O ( s ) → H 2 O ( g ) Using the ∆ H value given in Exercise 112 and the number of hydrogen bonds formed with each water molecule, estimate what portion of the intermolecular forces in ice can be accounted for by hydrogen bonding.

Interpretation:

• The change in enthalpy of sublimation has to be calculated.

• The portion of intermolecular forces in ice that account for the formation of hydrogen bonding has to be estimated.

Concept Introduction:

Enthalpy is heat content of the system. The value of enthalpy does not depend on the path of a reaction but depend on state of the system. It has a unique value for each state of the system. Thus, enthalpy is a state function.

Enthalpy change, denoted by ΔH , refers to heat evolved or absorbed during a reaction. If heat is evolved in the reaction that is exothermic reaction ΔH has negative value. For an endothermic reaction, ΔH has positive value. ΔH can be represented as,

                                               ΔH  =  ΔE + PΔV

                                    where,  ΔH  =   Change in enthalpy                ΔE   =  Change in Internal energy                ΔV  =   Change in volume                   P   =   Pressure               

Enthalpy of sublimation is denoted by ΔHsub . It is the enthalpy involved in sublimation process.

Internal energy of a system is total energy present in the system. In simple words, it is the sum of kinetic and potential energy of the particles in the system. According to First law of Thermodynamics, Energy of a system is conserved. It is only transferred from one state to another that is from system to surroundings and vice versa. So ΔE  can be represented as,

                                             ΔEuniverse  =  ΔEsys +  ΔEsurroundings

Further, ΔE  is also equivalent to sum of either heat gained or lost and either work done on the system or by the system.

                                            ΔE   =  q + w

                                   where  ΔE   =  change in internal energy                q      =   quantity of heat gained or heat lost                w     =    work done 

Explanation

Calculate ΔHsub.

solid →6.0 kJ liquid →40.7 kJ vapor

Therefore, ΔHsub is,

ΔHsub  =  6.0 kJ/mol + 40.7 kJ  =  46.7 kJ/mol

Bond energy of one hydrogen bond is 21 kJ/mol . Each water molecule forms two hydrogen bonds. Thus 42 kJ/mol of energy is required to break the hydrogen bonds. But calculated value is 46.7 kJ/mol