1. Chapter2. Some Thermodynamics Aspects of Intermolecular Forces Chapter2. Some Thermodynamics Aspects of Intermolecular Forces 한국과학기술원 화학과 계면화학 제 1 조 김동진 최윤정 조우경

2. 2.1 Interaction Energies of Molecules in Free Space and in Interaction potential w(r) between two molecules a Medium.  The pair potential : interaction potential between two particles  The potential of mean force : interaction in a solvent medium.  F = - dw(r)/dr The work can be done by F  w(r) : the free energy or available energy. In considering the forces between two molecules in liquids, Solvent effects several effects should be involved that do not arise when the interaction occurs in free space.

3. 1. Two solute in solvent 1) Pair potential w(r) = solute-solute interaction energy + change in solute-solvent interaction energy + change in solvent-solvent interaction energy 2) The net force depends on the attraction(solutes – solvents) 3) Solutes behavior In free space : attraction In a medium : repel each other

4. 2. The ‘structure’ of solvents molecules is perturbed. If the free energy varies with r (solute-solute) Solvation or Structural force 3. Solute – solvent interaction can change the properties of solute. e.g. dipole moment, charge, etc So, the properties of dissolved molecules may be different in different media.

5. 4. Cavity formation by solvent Gas Solute Condensed medium : consider the cavity energy expended by the medium Cohesive energy(self-energy) :  i : The energy of an individual molecule in a medium (gas or liquid) = The sum of the interaction with all the surrounding molecules. Now, how are  i and w(r) related? In the gas phase, w(r) is written like this W(r) = C/r n for r > σ ( where n > 3) = ∞ for r < σ σ : hard sphere diameter of the molecules

6.  i gas = Example 1. Molecules in a liquid or solid contact with 12 other molecules(close packing)  When a molecule is introduced in a own liquid Net energy change :  i liq  6w(σ) The molar cohesive energy : U = N o  i liq  6 N o w(σ) And  = 1/(molecular volume) = 1/[(4  /3)(  /2 ) 3 ] ∴ Cf. For n=6,  i liq  4w(  ) or U  - 4N o w(  )  ∞ W(r)  4  r 2 dr = -4  C  /(n-3)  n-3  i liq = 1 2 W(r)  4  r 2 dr = -12C (n – 3)  n  - 12 (n – 3) w(  )  ∞

7. Example 2. Solute molecule is dissolved in a solvent medium.  Solute molecule is surrounded by 12 solvent molecules  Solute(s) size  Solvent(m) size  i liq  -[6 w mm (  ) - 12 w sm (  )] 2.2. The Boltzmann Distribution The effective pair potential between two dissolved solute molecules in a medium is just the change in the sum of their free energies  i as they approach each other.  When the  i 1 is not the same  i 2 in two regions of a system. X 1 = X 2 exp[-(  i 1 -  i 2 )/kT] by Boltzmann distribution For dilute system  i 1 + kTln X 1 =  i 2 + kTln X 2 X n : equilibrium concentrations In many different regions of states in a system  i n + kTln X n =  in equilibrium for all states n = 1,2,3, …

8. 2.3. The Distribution of Molecules and Particles in System at Equilibrium Case 1.  i z + kTlnρ z =  i o + kTlnρ o ρ z = ρ o exp[-(  i z -  i o ) / kT] since (  i z -  i o ) = mgz z: altitude, ρ z = ρ o exp[-mgz / kT] : gravitational distribution law Case 2. For charged molecules or ions ρ 2 = ρ 1 exp[-e( ψ 2 - ψ 1 ) / kT] ψ 1, ψ 2 : electric potentials From case 1 and 2, the interaction energy did not arise from local intermolecular interactions, but from interactions with an externally applied gravitational or electric field. Case 3. Two phase system If one of the phaes is a pure solid or liquid,  i 1 =  i 2 + kTlnX 2 X 2 = X 1 exp [-(  i 2 -  i 1 ) / kT] = X 1 exp( -Δ  i / kT) Generally, X 2 = X 1 exp [-Δ  i + mgΔz + eΔψ / kT]

9. 2.4. The Van der Waals Equation of State.  i gas = - 4  C  /(n-3)  n-3 = -A  For molecules of finite sizes, X gas = 1/(v – B) =  /(1 - B  ) V = : the gaseous volume occupied per molecule B = 4  3 /3 : the excluded volume ∴ Chemical potential  of the gas (∂μ/∂P) T = v = 1/  or (∂P/∂  ) T =  (∂μ/∂P) T Now pressure P is related to , So, P = o   ∂∂ ∂∂ T dd = o  - A  + (1 -B  ) kT dd = - A  2 - ln(1 - B  ) 1 2 kT B  =  i gas + kTlnX gas = -A  + kT ln [  / (1 - B  )] 1 

10. For B  < 1, ln(1 - B  ) = -B  - (B  ) 2 + …  - B  (1 + B  )  -B  / ( 1 - B  )  - B / ( v - B) 1 2 1 2 1 2 1 2 ∴ P + a v2v2 ( v – b ) = kT : Van der Waals equation a = A = 2  C/(n-3)  n-3 and b = B = 2  3 / 3 1 2 1 2 Therefore, conceptually, the constants a and b can be thought of as accounting for the attractive and repulsive forces between the molecules. b depends on only on the molecular size  and on stabilizing repulsive contribution to the total pair potential.

11. Van der Waals Coefficients Gas a (L 2 atm/mole 2 ) b (L/mole) Ar1.3630.0322 CH 4 2.2830.0428 C6H6C6H6 18.240.1154 CO 2 3.6400.0427 H2H2 0.2480.0266 H2OH2O5.5360.0305 He0.03460.0237 N2N2 1.4080.0391 O2O2 1.3780.0318 SO 2 6.8030.0564 Xe4.2500.0511 a coefficients correlates with the degree of polarity of substance. The most polar of these molecules have the highest a coefficients. The pressure of these gases is most significantly affected by intermolcular attractions b coefficients increase with the size of the atom or molecle

12. 2.5. The Criterion of the thermal energy kT for gauging the strength of an interaction  How strong the intermolecular attraction must be if it is to condense molecules into a liquid at a particular temperature and pressure.  i gas + kT logX gas =  i liq + kT logX liq ( in equilibrium)  i gas –  i liq  –  i liq = -kT log(X liq /X gas ) ( ∵  i liq >>  i gas )  -kT log(22400/20) (at STP condition, V gas ~22400cm -3,V liq ~20cm -3 ) –  i liq  7 kT B or -N 0  i liq / T B  7N 0 k = 7R (T B =boiling temp.) : the T B of liq. Is simply proportional to the energy needed to take a molecule from liq. into vap.

13.  For one mole of molecules, U vap =-N 0  i liq, L vap = H vap = U vap + PV  U vap + RT B L vap / T B  ( U vap / T B ) + R  7R + R = 8R  70JK -1 mol -1 L vap / T B  80JK -1 mol -1 (cohesive energy  i liq  9kT)  Trouton’s rule

14.  Trouton’s rule (at the Normal Boiling Point, G° vap = 0.) the latent heat of vaporization divided by the boiling point (in kelvin) is approximately constant for a number of liquids. This is because the standard entropy of vaporization is itself roughly constant, being dominated by the large entropy of the gas. Boiling point of a substances provides a reasonably accurate indication of the strength of the cohesive forces or energies holding molecules together in condensed phases.

16.  The molecules will condense once their cohesive energy  i liq with all the other molecules in the condensed phase exceeds about 9kT. When,  i liq  6w(  ) ∴ pair interaction energy of two molecules or particles in contact exceeds about 3/2kT, then it is strong enough to condense them into a liquid or solid.  standard reference for gauging the cohesive strength of an interaction potential

17. X(  2 ) = X(  1 )exp kT Orientational distribution w(r,  2 ) –w(r,  1 )

18. 2.6. Classification of Forces Intermolecular forces  Purely electrostatic (from Coulomb force b.t. charges)  interaction b.t. charges, permanent dipoles, quadruples …  Polarization forces(from dipole moments induced)  Quantum mechanical  covalent or chemical bonding, repulsive steric or exchange interaction

22. the chemical potential is = Four forces

23. Hellman.Feynman theorem: "Once the spatial distribution of the electron clouds has been determined by solving the Schrödinger equation, the intermolecular forces may be calculated on the basis of straightforward classical electrostatics."