Practice paper 3: soft matter, kinetics, surfaces and MOFs (Lectures 26 to 39)

Working, in plain words. Heat can only do things that cost about $kT$ or less. Bond breaking costs 140 times that, so it never happens by heat alone. Twisting costs a few times $kT$, so heat can do it. For (ii) you do not calculate, you read the lecture’s table of $e^{-\text{gap}/kT}$. For (iii), one walker is unpredictable, but a crowd spreads out in a steady way, always from crowded to empty.

Model answer.

(i) $kT \approx 2.5$ kJ/mol, so a C–C bond (348 kJ/mol) is far above $kT$ and almost no molecule ever has enough thermal energy to break one. A twist (about 12 kJ/mol) is only a few $kT$, so heat can supply it and the chain bends and twists.

(ii) Using $N_i/N_j = e^{-(\varepsilon_i-\varepsilon_j)/kT}$ and the lecture table: $0.5\,kT$ gives $0.607$ and $10\,kT$ gives $0.000045$. At $0.5\,kT$ the upper state is almost as full as the lower one, so heat easily moves molecules between them. At $10\,kT$ almost none are excited, so heat cannot change anything.

(iii) The average of $x$ stays near 0 but the average of $x^2$ rises roughly linearly with time. Fick’s first law, $J_x = -DA\,dc/dx$, has a minus sign, so the net flux runs down the gradient, from high to low concentration, even though every step is random. Fick’s second law, $dc/dt = D\,d^2c/dx^2$, says a local maximum has falling concentration (it drains) and a local minimum rises, so bumps flatten.

Where this is taught.

• Bond, twist and $kT$ numbers, and the $N_i/N_j$ table: soft matter and thermal energy.
• Average of $x$ against $x^2$, flux direction, both Fick laws: random walks and diffusion.

Working, in plain words. Picture a crowded box of pencils. Tipped at random they jam each other. Lined up they leave each pencil a bit of room to slide about. Room to move is entropy, so lining up can win without any attraction. For (iii) just put $2L$ in place of $L$ and see what happens to the number.

Model answer.

(i) $S = 1$ for perfect alignment and $S = 0$ for isotropic. On heating $S$ falls, then drops suddenly to 0 at the transition. This is a first-order transition.

(ii) Hard rods have no interactions, so $U = 0$ and the free energy is just $-TS$, so the stable phase is the one with the most entropy. Isotropic rods exclude about $DL^2$ of volume from each other, but aligned rods exclude only about $D^2L$ (about 4 times a rod’s volume). At high concentration the aligned phase gains more translational entropy than it loses in orientational entropy, so the nematic is stable.

(iii) The boundary is $7.0/(DL^2)$. With $L \to 2L$ the denominator becomes $D(2L)^2 = 4DL^2$, so the concentration becomes $7.0/(4DL^2)$. It falls by a factor of 4. Longer rods need a four times lower concentration to go nematic.

Where this is taught.

• $S$, its end values, first-order transition, $U = 0$ so maximise entropy: order parameter and hard particles.
• Excluded volume $DL^2$ against $D^2L$, the entropy trade, the $7.0/(DL^2)$ boundary: excluded volume and Onsager theory.

Working, in plain words. Light shakes in a direction. Aligned rods hold electrons back more along the rod than across it, so the two shake-directions of light run at different speeds. A voltage twists the rods round like compass needles, and that changes what the analyser sees.

Model answer.

(i) Crystal, SmC, SmA, N (nematic), isotropic.

(ii) In an achiral aligned phase the electrons respond differently to light polarised across and along the molecules, so the two components travel at different speeds and drift out of phase. That is birefringence. The size of the drift depends on wavelength, so different colours go out of phase by different amounts, which is why the textures are colourful.

(iii) The charge distribution or polarisability of the molecules means the field turns the rods along $E$. With the voltage off, surface patterns set the alignment and light passes. With the voltage on, the new alignment means the analyser passes no light, so the pixel goes dark.

Where this is taught.

• Heating sequence: self-organisation and mesophases.
• Birefringence, colour, field switching: nematics, light and fields.

Working, in plain words. Each order has its own straight-line plot, so the plot tells you the order. For (iii) the only slow part is the meeting, so how fast and how big the partners are is all that matters. For (iv), a catalyst lowers the hill, it does not move the valleys.

Model answer.

(i) First order. The gradient is $-k$. The half-life is $\ln 2/k$, which does not depend on $[A]_0$.

(ii) The line means second order, $t_{1/2} = 1/(k[A]_0)$. Doubling $[A]_0$ halves the half-life.

(iii) The rate coefficient is $k = 4\pi(r_o + r_g)(D_o + D_g)N_A$. It depends on the diffusion coefficients of the partners and on the distance at which they touch. The rate law is second order, $dc_o/dt = -k\,c_oc_g$.

(iv) $K^\ddagger = \exp(-\Delta G^\ddagger/RT)$, so lowering $\Delta G^\ddagger$ puts more molecules at the transition state, and the rate coefficient rises. The products end at the same Gibbs free energy, so only the barrier has changed.

Where this is taught.

• Straight-line plots, half-lives, diffusion-controlled $k$ and rate law: rate laws and diffusion-controlled reactions.
• $K^\ddagger$, the barrier and catalysts: Arrhenius, transition states and catalysis.

Working, in plain words. A balanced equation is the finished dish, the mechanism is the recipe. At equilibrium every single step balances on its own. Write each step’s ratio $k/k_{-}$, multiply the three, and watch the middle species cancel.

Model answer.

(i) A balanced equation only gives the amounts of reactants and products, not how fast they react. The rate law needs the mechanism, which is the series of elementary steps. For an elementary step A + B → C the order equals the molecularity, so $d[A]/dt = -k[A][B]$, second order.

(ii) At equilibrium each step’s forward and backward rates are equal, so $k_1/k_{-1}$, $k_2/k_{-2}$ and $k_3/k_{-3}$ equal the concentration ratios of each step. Multiplying them gives $\frac{k_1k_2k_3}{k_{-1}k_{-2}k_{-3}} = \frac{[\mathrm{N_2}]_{eq}[\mathrm{H_2O}]^2_{eq}}{[\mathrm{NO}]^2_{eq}[\mathrm{H_2}]^2_{eq}} = K$ The intermediates $\mathrm{N_2O_2}$ and $\mathrm{N_2O}$ each appear once on top and once on the bottom, so they cancel.

Where this is taught. Elementary steps and mechanisms covers the balanced-equation point, the elementary rate laws and the whole detailed-balance working.

Working, in plain words. Surface molecules have lost neighbours, so more surface costs energy. For (iii), ask whether covering the solid with liquid lowers or raises the energy.

Model answer.

(i) The area goes from $6\ \text{cm}^2$ ($6\cdot10^{-4}\ \text{m}^2$) to $600\ \text{m}^2$, a factor of $10^6$. At that size most of the material is at a surface, so surface effects and intermolecular forces dominate.

(ii) A surface molecule has lost neighbours, so making surface costs energy, and the cost is bigger when the intermolecular forces are stronger. Stronger forces also mean a higher boiling point. So water has both the higher $\gamma$ and the higher boiling point.

(iii) Glass is a high-energy surface, so covering it with liquid lowers the energy and water spreads into a film ($\theta = 0$, $S = 0$ at the boundary). PTFE is a low-energy surface, so covering it would raise the energy: $S < 0$, which means partial wetting. Young-Dupre gives $S = \gamma(\cos\theta_E - 1)$, which is zero or negative for any drop with an angle, so water on PTFE forms a drop with $\theta$ well above 0.

Where this is taught.

• Area worked example, why surface costs energy, the table of $\gamma$ and boiling points: surface tension and surface energy.
• $S$, high and low energy surfaces, Young-Dupre: wetting, contact angle and Wilhelmy.

Working, in plain words. Two rough solids touch only at their peaks. Only those real contact spots matter, not the size of the visible patch.

Model answer.

(i) The two sources are adhesion at the real contact points and ploughing or deformation. The adhesive force is $F_{ad} = A\,s$, with $A$ the real contact area and $s$ the shear strength. For plastic deformation $A = W/\tau_0$, so more load flattens more bumps and raises the real contact area, giving more bonds to shear and so more friction. The real area depends on the load, not on the visible area.

(ii) $W_{AB} = \gamma_{AV} + \gamma_{BV} - \gamma_{AB}$. Real joints fall short because flaws in the interface concentrate stress so failure starts there (also acceptable: viscoelastic processes dissipate energy, or trapped air and dust cut the contact area).

Where this is taught. Friction and adhesion covers $F_{ad} = A\,s$, $A = W/\tau_0$, $W_{AB}$ and why real joints are weaker. The $W_{AB}$ definition is also in surface tension and surface energy.

Working, in plain words. Physisorption is weak sticking, chemisorption is a real bond. The BET line is $y = b + mx$, so slope over intercept removes the $c\,n_{mono}$ part.

Model answer.

(i) Physisorption is by van der Waals forces, about 20 kJ mol⁻¹ released. Chemisorption forms a chemical bond, about 200 kJ mol⁻¹. Chemisorption breaks bonds in the molecule as it binds, which is why surfaces act as catalysts.

(ii) Any two of: a maximum of one monolayer; every site is equivalent; whether a particle binds does not depend on neighbouring sites being occupied. A plot of $p/n_{ads}$ against $p$ is a straight line with slope $1/n_{mono}$.

(iii) Slope divided by intercept: $c - 1 = 319/3.35$, so $c = 96.2$. From the intercept, $n_{mono} = 1/(96.2\times3.35) = 3.10\times10^{-3}\ \text{mol}$. Area $= 3.10\times10^{-3}\times0.162\times10^{-18}\times6.02\times10^{23} = 302\ \text{m}^2$.

Where this is taught.

• Heats of physisorption and chemisorption, and the catalysis link: adsorption, films and SAMs.
• Langmuir assumptions and plots, BET line, the silica worked example: Langmuir and BET isotherms.

Working, in plain words. The lecture’s rule: the slope of $\gamma$ against $\ln c$ gives the surface excess, with a minus sign. For the area, one molecule gets $1/(N_A\Gamma)$ of surface. Remember $1\ \text{Å}^2 = 10^{-20}\ \text{m}^2$.

Model answer.

(i) For the surfactant, $d\gamma/d\ln c$ is negative, so $\Gamma$ is positive: it collects at the surface. For NaCl, $\gamma$ rises, so the slope is positive and $\Gamma$ is negative: the ions stay in the bulk.

(ii) No salt: $A = 1/(6.022\times10^{23}\times3.8\times10^{-6}) = 4.4\times10^{-19}\ \text{m}^2$, about $44\ \text{Å}^2$. With 0.20 M NaCl: $A = 1/(6.022\times10^{23}\times8.1\times10^{-6}) = 2.05\times10^{-19}\ \text{m}^2$, about $20\ \text{Å}^2$.

(iii) The charged heads repel each other, and salt screens that repulsion, so the heads sit closer and more molecules fit.

(iv) Above the cmc extra surfactant forms micelles, which are not surface active, and the free monomer concentration is fixed. So $\Gamma = 0$ and $d\gamma = 0$.

Where this is taught. Gibbs adsorption isotherm and surfactants covers every point: signs of $\Gamma$ for surfactant and NaCl, both $\Gamma$ values and the 45 and 20 Ų areas, salt screening and the cmc.

Working, in plain words. (i) Hubs and rods that click together. (ii) Think of a tap and a plug: it must hold the gas on filling and let go on emptying. (iii) The field either lets electrons spread out or forces them to pair.

Model answer.

(i) In Prussian blue the node is an octahedral metal centre with six connections, and the linker is the linear $CN^-$ with two connections. They are joined by coordinate bonds from the ligand to the metal ion, giving a cubic 3D framework whose structure can be designed.

(ii) Gas has to be released again. Very strong binding needs pressures below 1 bar to release the gas, and very weak binding gives little uptake even at high pressure. The ideal is about 14 kJ mol⁻¹, against typical values of 4 to 6 kJ mol⁻¹. Bare metal sites (as in HKUST-1) bind H₂ first because the interaction energy is more favourable, which raises the binding energy.

(iii) Fe(II) is d⁶. Weak field ($\Delta < P$) gives high spin $(t_{2g})^4(e_g)^2$. Strong field ($\Delta > P$) gives low spin $(t_{2g})^6$. In $[Fe(azpy)_2(NCS)_2]\cdot\tfrac12(EtOH)$ the structure depends on the guest in the pores. The empty material does not switch, but with MeOH, EtOH or 1-PrOH it does, with guest-dependent behaviour, so the response identifies the guest.

Where this is taught.

• Node, linker, coordinate bond, Prussian blue: MOF design, nodes and linkers.
• Optimum binding, 14 and 4 to 6 kJ mol⁻¹, bare metal sites: MOFs for hydrogen, CO₂ and methane.
• High and low spin Fe(II), the Fe(azpy) switch: MOF magnetism, conduction and spin switching.