Question

7. The following surface tensions were measured for aqueous solutions n-pentanol at 20°C: of C[ 10-2 mol.dm-3] Yo [mN.m-1] 0 72.6 1 64.6 2 3 60.0 56.8 4 54.3 5 6 10 51.9 49.8 43.0 Calculate surface excess concentrations and the average area occupied by each adsorbed molecule for bulk concentrations of 0.01M, 0.02M, 0.04M and 0.08M. Plot a -A curve for the adsorbed n-pentanol monolayer and compare it with the corresponding curve for an ideal gaseous film.

Answer 1

When, particles concentrate at surface there becomes a "surface excess". Surface excess is defined as Γ. There are several ways to look at surface excess. One is to view it as the amount of solute adsorbed at the surface per unit surface area in excess of bulk concentration (units = moles/m2 ). This "surface excess" could be viewed as a concentration in a small volume near the surface.

From Gibbs adsorption equation :

   Γ = - c/RT * dγ/ dC

• Conc. /(mol/L)	γ*10^3 / (N/m)
  0	72.6
  0.01	64.6
  0.02	60
  0.03	56.8
  0.04	54.3
  0.05	51.9
  0.06	49.8
  0.1	43

  

  y*10^3 / (N/m) vs. Conc. /(mol/L) = y*10^3 / (N/m) – 70.7 +-510x + 2380x^2 y*10^3 7 (N/m) Oo 0.025 0.05 0.075 Conc. /(mol/L)

  we get equation :

  γ(10^3 N/m) = 70.7 - 510 ( c/mol/L) + 2380 ( c/mol/L) ²

  we get : d{γ(10^3 N/m)}/ d(( c/mol/L)) = - 510 + 4720 ( c/mol/L)

  surface excess concentration for particular conc. ,

• at c = 0.01 mol/L

  dγ/ dC = - 462.8 * 10^-3 N*m^-1 mol^-1 L = - 462.8 * 10^-6 N*m² mol^-1

  Γ = - c/RT * dγ/ dC = - 0.01 mol/L / (8.314 J/K-mol * 293 K) *- 462.8* 10^-6 N*m² mol^-1

  Γ = 1.9*10^-6 mol/m²

No. of molecules/mol : 6.023*10²³

Thus average area occupied by each adsorbed molecules : ( 1/ 1.9*10^-6 mol/m² ) / 6.023*10²³ mol-1

           =   8.74*10^-19 m2   or 0.874 nm²

• at c = 0.02 mol/L

  dγ/ dC = - 415.6 * 10^-3 N*m^-1 mol^-1 L = - 415.6 * 10^-6 N*m² mol^-1

  Γ = - c/RT * dγ/ dC = - 0.02 mol/L / (8.314 J/K-mol * 293 K) *- 415.6* 10^-6 N*m² mol^-1

  Γ = 4.18*10^-6 mol/m²

  No. of molecules/mol : 6.023*10²³

  Thus average area occupied by each adsorbed molecules : ( 1/ 4.18*10^-6 mol/m² ) / 6.023*10²³ mol-1

             =   3.97*10^-19 m2   or 0.397 nm²

• at c = 0.04 mol/L

  dγ/ dC = - 321.2 * 10^-3 N*m^-1 mol^-1 L = - 321.2 * 10^-6 N*m² mol^-1

  Γ = - c/RT * dγ/ dC = - 0.04 mol/L / (8.314 J/K-mol * 293 K) *- 321.2* 10^-6 N*m² mol^-1

  Γ = 5.27*10^-6 mol/m²

No. of molecules/mol : 6.023*10²³

Thus average area occupied by each adsorbed molecules : ( 1/ 5.27*10^-6 mol/m² ) / 6.023*10²³ mol-1

           =   3.13*10^-19 m2   or 0.315 nm²

• at c = 0.08 mol/L

  dγ/ dC = - 132.4 * 10^-3 N*m^-1 mol^-1 L = - 132.4 * 10^-6 N*m² mol^-1

  Γ = - c/RT * dγ/ dC = - 0.08 mol/L / (8.314 J/K-mol * 293 K) *- 132.4* 10^-6 N*m² mol^-1

  Γ = 5.33*10^-6 mol/m²

No. of molecules/mol : 6.023*10²³

Thus average area occupied by each adsorbed molecules : ( 1/ 5.33*10^-6 mol/m² ) / 6.023*10²³ mol-1

           = 3.12*10^-19 m2   or 0.312 nm²