Part of PC-10 — Surface Chemistry

Formula Summary

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Freundlich Adsorption Isotherm: xm=kP1/n,0<1n<1\frac{x}{m} = k P^{1/n}, \quad 0 < \frac{1}{n} < 1

Freundlich Isotherm — Logarithmic (Linear) Form: log(xm)=logk+1nlogP\log\left(\frac{x}{m}\right) = \log k + \frac{1}{n} \log P

  • Slope of log(x/m) vs. log P = 1n\frac{1}{n}
  • Y-intercept = logk\log k, so k=10interceptk = 10^{\text{intercept}}

Langmuir Adsorption Isotherm: xm=aP1+bP\frac{x}{m} = \frac{aP}{1+bP}

  • Low P limit (bP << 1): xmaP\frac{x}{m} \approx aP (linear)
  • High P limit (bP >> 1): xmab\frac{x}{m} \approx \frac{a}{b} (monolayer saturation)

Thermodynamic Criteria for Adsorption: ΔG=ΔHTΔS<0 (spontaneous)\Delta G = \Delta H - T\Delta S < 0 \text{ (spontaneous)} ΔS<0 (entropy decreases, 3D → 2D)\Delta S < 0 \text{ (entropy decreases, 3D → 2D)} ΔH<0 (adsorption is always exothermic)\Rightarrow \Delta H < 0 \text{ (adsorption is always exothermic)}

Michaelis-Menten Enzyme Kinetics: v=Vmax[S]Km+[S]v = \frac{V_{max}[S]}{K_m + [S]} At [S]=Km:v=Vmax2\text{At } [S] = K_m: \quad v = \frac{V_{max}}{2} At [S]Km:vVmax\text{At } [S] \gg K_m: \quad v \to V_{max} At [S]Km:vVmaxKm[S] (first order)\text{At } [S] \ll K_m: \quad v \approx \frac{V_{max}}{K_m}[S] \text{ (first order)}

Hardy-Schulze Empirical Coagulating Power: Coagulating Powerz6\text{Coagulating Power} \propto z^6 where zz = valency of the coagulating ion (opposite sign to colloid)

Coagulation value 1coagulating power1z6\propto \frac{1}{\text{coagulating power}} \propto \frac{1}{z^6}

Gold Number (Zsigmondy): Gold number=mg of protective colloid needed10 mL standard gold sol+1 mL 10% NaCl\text{Gold number} = \frac{\text{mg of protective colloid needed}}{10 \text{ mL standard gold sol} + 1 \text{ mL } 10\% \text{ NaCl}}

Protective ColloidGold Number
Gelatin0.005 (best)
Albumin0.1
Starch25 (worst)

Freundlich Numerical Examples:

Example 1: x/m = 0.5P^(1/3) at P = 27 atm: xm=0.5×271/3=0.5×3=1.5 units/g\frac{x}{m} = 0.5 \times 27^{1/3} = 0.5 \times 3 = 1.5 \text{ units/g}

Example 2: x/m = kP^(1/n); at P = 1: x/m = 4; at P = 4: x/m = 8. Find 1/n: 4=k(1)1/nk=44 = k(1)^{1/n} \Rightarrow k = 4 8=4×41/n41/n=21n=log2log4=0.3010.602=0.58 = 4 \times 4^{1/n} \Rightarrow 4^{1/n} = 2 \Rightarrow \frac{1}{n} = \frac{\log 2}{\log 4} = \frac{0.301}{0.602} = 0.5

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