Nanochemistry has sparked much excitement in the recent years and a large amount of research has bee — Physical Chemistry — Thermodynamics Chemistry Question
Nanoparticles and nanophases
Nanochemistry has sparked much excitement in the recent years and a large amount of research has been dedicated to understanding of nanomaterials. Single-walled carbon nanotubes (SWNTs) are a universally known example of such materials. SWNT can be thought of as a sheet of graphite rolled into a seamless cylinder (d ≈ 1.5 nm). These cylindrical carbon “molecules” might provide components for molecular electronic devices of the future.
The properties of nanometer-scale materials are size- and shape-dependent.
Saturated vapor pressure of a small spherical particle (crystalline or liquid) is higher than that of the bulk phase of the same material. At equilibrium the molar Gibbs functions (G) of the condensed phase (Gbulk) and vapor (Gvap) are equal. Equation (1) determines the saturated vapor pressure, p, above a bulk phase
Gbulk = Gvap = G°vap + RT ln p, (1)
G°vap is the standard molar Gibbs energy of vapor at standard pressure p = 1 bar.
The substance inside a small spherical sample is under excess pressure, caused by surface tension:
∆Pin = 2σ / r
r – the radius of the spherical sample, σ – the surface tension at the “condensed phase-vapor” interface. The increase of the internal pressure results in a change in the molar Gibbs energy of the substance inside the spherical sample. This molar Gibbs energy Gsph is larger than Gbulk. The difference in the Gibbs energy of the spherical sample and the bulk phase is equal to ∆PinV:
Gsph = Gbulk + ∆PinV = Gbulk + 2σV / r, (2)
V is the molar volume of the liquid or solid substance. Therefore from equation (1)
Gsph = Gbulk + 2σV / r = Gvap = G°vap + RT ln p* (3)
p* is the saturated vapor pressure of the spherical sample with the radius r.
The saturated vapor pressure of water at T = 298 K is 3.15×10–2 bar. Calculate the saturated vapor pressure of the spherical droplets of water with the radius of:
i) 1 µm,
ii) 1 nm.
The surface tension at the liquid-vapor interface of water is 0.072 J m–2.
Model Answer
From equations (1) and (3) one gets
p* / p = exp(2σV / rRT)
p* = p exp(2σV / rRT) (5)
Knowing p we get p*.
For r = 1 µm:
p* = 3.15×10^-2 × exp(2 × 0.072 × 18×10^-6 / (1.00×10^-6 × 8.314 × 298)) = 3.15×10^-2 bar
For r = 1 nm:
p* = 3.15×10^-2 × exp(2 × 0.072 × 18×10^-6 / (1.00×10^-9 × 8.314 × 298)) = 8.97×10^-2 bar
Assuming that the substance retains properties of a bulk while the difference between its saturated vapor pressure and the saturated pressure of the bulk is less than 1 %, what is the minimum radius of the spherical sample that can still be considered as a bulk phase? How many molecules of water are there in such a droplet?
Model Answer
The minimum size of the spherical sample that can still be considered as a bulk phase can be calculated from the inequality
exp(2σV / rRT) ≤ 1.01
exp(2 × 0.072 × 18×10^-6 / (r × 8.314 × 298)) ≤ 1.01
r ≥ 1.05×10–7 m = 105 nm.
r = 105 nm may be considered as the minimum radius.
The number of water molecules N in the drop with r = 105 nm can be calculated from the formula
V = (4/3)πr^3 = N / NA × V,
V = 18×10–6 m3 is the molar volume of water, NA = 6.02×1023 mol–1 is the Avogadro number.
N = (4/3)π(1.05×10^-7)^3 / (18×10^-6) × 6.02×10^23 = 1.62×10^8
Few droplets of mercury were put inside a SWNT maintained at 400 K. What is the minimum vapor pressure of mercury inside the tube? The saturated vapor pressure of bulk mercury is 1.38×10–3 bar, the density of mercury ρ(Hg) = 13.5 g cm–3, the surface tension at the liquid-vapor interface of mercury is 0.484 J m–2 at the given temperature.
Model Answer
The maximum radius of the droplet is equal to the internal radius of the nanotube. The saturated pressure goes up while the radius of the droplet goes down. Therefore, the maximum radius corresponds to the minimum vapor pressure of mercury inside the tube. One has to calculate the saturated vapor pressure above the droplet with r = 0.75 nm (d = 1.5 nm). From eq. 5 one gets:
p* = 1.38×10–3 exp(2 × 0.484 × 200.5×10–3 / (13.5 × 0.75×10–9 × 8.314 × 400)) = 0.440 bar.
This pressure is approximately three hundred times higher than the one of the bulk liquid mercury.
Comment. The droplets of mercury are so small, that the whole basis of calculation is suspect. There is an experimental evidence of a validity of the equation at least for r ≥ 3 nm. For smaller values it is believed that the orders of the magnitude of the vapor pressures are approximately correct.
The boiling point of benzene at the standard atmospheric pressure is Tb = 353.3 K. The temperature dependence of the saturated vapor pressure of benzene near the boiling point is given by the equation
∆ln(p) = -∆Hvap / RT + const (4)
where ∆Hvap = 30720 J mol–1 is the enthalpy of vaporization of benzene. Estimate the boiling point (T*) of the finely dispersed liquid benzene at the standard atmospheric pressure if the sample consists of droplets with the radius r = 50 nm. The surface tension of benzene near the boiling point is 0.021 J m–2 and its density is 0.814 g cm–3.
Model Answer
The boiling temperature of the dispersed benzene is T*. At this temperature the saturated vapor pressure p* is equal to the atmospheric pressure 1 bar. So,
ln p*(T*) = ln p(T*) + ln(p*(T*) / p(T*)) = 0
From equations (4) and (5)
– ∆Hvap / RT* + const + 2σV / rRT* = 0
The const can be calculated from the boiling point of bulk benzene:
ln p(Tb) = – ∆Hvap / RTb + const = 0
const = ∆Hvap / RTb
Thus
– ∆Hvap / RT* + ∆Hvap / RTb + 2σV / rRT* = 0
T* = Tb(1 – 2σV / r∆Hvap) = 353.3 (1 – 2 × 0.021 × 78×10–6 / (0.814 × 50×10–9 × 30720)) = 352.4 K
In general, properties of the bulk and nano-sized material composed by one and the same substance A are different. Which of the following thermodynamic constants will decrease when passing from the bulk to the nano-scaled material?
1) Solubility of A in any solvent;
2) the boiling temperature at atmospheric pressure;
3) the saturated vapor pressure over solid substance A;
4) the equilibrium constant of a chemical reaction, where A is a reagent;
5) the equilibrium constant of a chemical reaction, where A is a product.
Model Answer
The molar Gibbs energy of liquid A increases when passing from the bulk phase to the small droplet (see equation 2).
Increase of the molar Gibbs energy leads to the decrease of the boiling temperature at atmospheric pressure and the equilibrium constant of the chemical reaction (A is a product).
The decrease of the boiling temperature was demonstrated above.
The equilibrium constant K can be calculated from the standard reaction Gibbs energy, ∆rG0:
–RT ln K = ∆rG0 = G0prod – G0react
G0prod, G0react are molar Gibbs energies for products and reactants, respectively. If G0prod increases, the equilibrium constant K goes down.