Level 4 — ApplicationPhysical Chemistry (Advanced)

Physical Chemistry (Advanced)

60 marksprintable — key stays hidden on paper

Time: 60 minutes Total marks: 60 Instructions: Answer all questions. No hints given. Use standard constants: $h = 6.626\times10^{-34}\,\mathrm{J\,s}$ , $\hbar = 1.055\times10^{-34}\,\mathrm{J\,s}$ , $k_B = 1.381\times10^{-23}\,\mathrm{J\,K^{-1}}$ , $c = 2.998\times10^{10}\,\mathrm{cm\,s^{-1}}$ , $N_A = 6.022\times10^{23}\,\mathrm{mol^{-1}}$ , $F = 96485\,\mathrm{C\,mol^{-1}}$ , $R = 8.314\,\mathrm{J\,mol^{-1}K^{-1}}$ .

Q1. (Quantum chemistry + variational method) [14 marks]

A conjugated dye electron is modelled as a particle confined in a 1-D box of length $L$ .

(a) Using the trial function $\phi(x) = x(L-x)$ on the interval $[0,L]$ (which vanishes at both walls), apply the variational principle to estimate the ground-state energy. Show that $E_{\text{trial}} = \dfrac{5\hbar^2}{mL^2}$ . (7)

(b) Compare this to the exact ground-state energy $E_1 = \dfrac{\pi^2\hbar^2}{2mL^2}$ . Compute the percentage error. (3)

(c) The dye absorbs at $\lambda = 620\,\mathrm{nm}$ for a transition $n=1\to n=2$ . Treating the electron mass as $m_e = 9.109\times10^{-31}\,\mathrm{kg}$ , determine the box length $L$ in nm. (4)

Q2. (Spectroscopy + statistical thermodynamics) [14 marks]

For gaseous $^{12}\mathrm{C}^{16}\mathrm{O}$ : rotational constant $B = 1.931\,\mathrm{cm^{-1}}$ , vibrational wavenumber $\tilde\nu = 2143\,\mathrm{cm^{-1}}$ .

(a) Compute the bond length (in pm). The reduced mass of CO is $\mu = 1.139\times10^{-26}\,\mathrm{kg}$ . (4)

(b) At $T = 300\,\mathrm{K}$ , estimate the rotational partition function $q_{\text{rot}} = \dfrac{k_B T}{\sigma\, h c B}$ (symmetry number $\sigma=1$ ). (4)

(c) Compute the vibrational partition function $q_{\text{vib}}$ (measured from the ground vibrational level) at $300\,\mathrm{K}$ , and state whether vibration contributes significantly to $C_{V,m}$ at this temperature. Justify. (6)

Q3. (Surface chemistry) [10 marks]

Adsorption of a gas on a catalyst follows the Langmuir isotherm $\theta = \dfrac{Kp}{1+Kp}$ .

(a) Two data points are measured: at $p_1 = 20\,\mathrm{kPa}$ , $\theta_1 = 0.40$ ; at $p_2 = 60\,\mathrm{kPa}$ , $\theta_2 = 0.60$ . Determine whether these data are consistent with a single Langmuir constant $K$ . Show your working with the two implied $K$ values. (6)

(b) A student instead fits the low-coverage data to a Freundlich isotherm $\theta = c\,p^{1/n}$ . Given $\theta = 0.10$ at $p = 5\,\mathrm{kPa}$ and $\theta = 0.20$ at $p = 20\,\mathrm{kPa}$ , find the exponent $1/n$ . (4)

Q4. (Advanced electrochemistry) [12 marks]

For an electrode reaction the Tafel equation (anodic branch) is $\eta = a + b\log_{10} j$ , with transfer coefficient $\alpha_a = 0.5$ at $T = 298\,\mathrm{K}$ and $n=1$ .

(a) Derive the Tafel slope $b$ (in mV per decade) from the Butler–Volmer equation in its high-overpotential limit. (6)

(b) A current density of $j = 10\,\mathrm{mA\,cm^{-2}}$ requires an overpotential $\eta = 120\,\mathrm{mV}$ . Find the exchange current density $j_0$ . (4)

Q5. (Photochemistry + solid-state) [10 marks]

(a) A photochemical reaction is irradiated at $\lambda = 400\,\mathrm{nm}$ with an incident power of $10\,\mathrm{mW}$ for $60\,\mathrm{s}$ (all absorbed). $3.5\times10^{-6}\,\mathrm{mol}$ of product forms. Compute the quantum yield $\Phi$ . (6)

(b) An intrinsic semiconductor has band gap $E_g = 1.1\,\mathrm{eV}$ . Explain qualitatively how its conductivity changes with rising temperature, contrasting this with a metal. (4)

Answer keyMark scheme & solutions

Q1 [14]

(a) Variational energy $E = \dfrac{\langle\phi|\hat H|\phi\rangle}{\langle\phi|\phi\rangle}$ , with $\hat H = -\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}$ inside the box.

$\phi = xL - x^2$ , $\phi'' = -2$ . (1)
Numerator: $-\dfrac{\hbar^2}{2m}\displaystyle\int_0^L \phi\phi''\,dx = -\dfrac{\hbar^2}{2m}(-2)\int_0^L (xL-x^2)dx = \dfrac{\hbar^2}{m}\cdot\dfrac{L^3}{6} = \dfrac{\hbar^2 L^3}{6m}$ . (2) ( $\int_0^L(xL-x^2)dx = \tfrac{L^3}{2}-\tfrac{L^3}{3}=\tfrac{L^3}{6}$ ).
Denominator: $\displaystyle\int_0^L x^2(L-x)^2 dx = \dfrac{L^5}{30}$ . (2)
$E = \dfrac{\hbar^2 L^3/6m}{L^5/30} = \dfrac{30\hbar^2}{6mL^2} = \dfrac{5\hbar^2}{mL^2}$ . (2)

(b) Exact $E_1 = \dfrac{\pi^2\hbar^2}{2mL^2} = \dfrac{4.9348\,\hbar^2}{mL^2}$ . Error $= \dfrac{5-4.9348}{4.9348}\times100\% = 1.32\%$ (trial is an upper bound, as required). (3)

(c) $\Delta E = E_2 - E_1 = \dfrac{(4-1)h^2}{8mL^2} = \dfrac{3h^2}{8mL^2} = \dfrac{hc}{\lambda}$ . (2) $L^2 = \dfrac{3h\lambda}{8mc}$ . $L^2 = \dfrac{3(6.626\times10^{-34})(620\times10^{-9})}{8(9.109\times10^{-31})(2.998\times10^{8})}$ $= \dfrac{1.2324\times10^{-39}}{2.1847\times10^{-21}} = 5.641\times10^{-19}\,\mathrm{m^2}$ . $L = 7.51\times10^{-10}\,\mathrm{m} = 0.751\,\mathrm{nm}$ . (2)

Q2 [14]

(a) $B = \dfrac{h}{8\pi^2 c\, I}$ (in cm⁻¹ with $c$ in cm/s), $I = \mu r^2$ . $I = \dfrac{h}{8\pi^2 c B} = \dfrac{6.626\times10^{-34}}{8\pi^2(2.998\times10^{10})(1.931)}$ $= \dfrac{6.626\times10^{-34}}{4.574\times10^{12}} = 1.449\times10^{-46}\,\mathrm{kg\,m^2}$ . (2) $r = \sqrt{I/\mu} = \sqrt{1.449\times10^{-46}/1.139\times10^{-26}} = \sqrt{1.272\times10^{-20}} = 1.128\times10^{-10}\,\mathrm{m} = 112.8\,\mathrm{pm}$ . (2)

(b) $q_{\text{rot}} = \dfrac{k_B T}{hcB} = \dfrac{(1.381\times10^{-23})(300)}{(6.626\times10^{-34})(2.998\times10^{10})(1.931)}$ $= \dfrac{4.143\times10^{-21}}{3.836\times10^{-23}} = 108$ . (4)

(c) $\dfrac{hc\tilde\nu}{k_BT} = \dfrac{(6.626\times10^{-34})(2.998\times10^{10})(2143)}{(1.381\times10^{-23})(300)}$ $= \dfrac{4.257\times10^{-20}}{4.143\times10^{-21}} = 10.27$ . (2) $q_{\text{vib}} = \dfrac{1}{1-e^{-10.27}} = \dfrac{1}{1-3.5\times10^{-5}} \approx 1.00003 \approx 1.0$ . (2) Since $q_{\text{vib}}\approx1$ , essentially all molecules are in the vibrational ground state; the vibrational mode is "frozen" and contributes negligibly to $C_{V,m}$ at 300 K (need $T\gtrsim \theta_{\text{vib}}=3084$ K). (2)

Q3 [10]

(a) From $\theta=\dfrac{Kp}{1+Kp}$ , solve $K = \dfrac{\theta}{p(1-\theta)}$ . (1)

Point 1: $K_1 = \dfrac{0.40}{20(0.60)} = \dfrac{0.40}{12} = 0.0333\,\mathrm{kPa^{-1}}$ . (2)
Point 2: $K_2 = \dfrac{0.60}{60(0.40)} = \dfrac{0.60}{24} = 0.0250\,\mathrm{kPa^{-1}}$ . (2) $K_1 \neq K_2$ (0.0333 vs 0.0250, ~25% difference), so the data are not consistent with a single Langmuir constant — the simple Langmuir model does not describe both points. (1)

(b) $\dfrac{1}{n} = \dfrac{\ln(\theta_2/\theta_1)}{\ln(p_2/p_1)} = \dfrac{\ln(0.20/0.10)}{\ln(20/5)} = \dfrac{\ln 2}{\ln 4} = \dfrac{0.6931}{1.3863} = 0.500$ . (4)

Q4 [12]

(a) Butler–Volmer: $j = j_0\left[e^{\alpha_a f\eta} - e^{-\alpha_c f\eta}\right]$ , $f=\dfrac{F}{RT}$ . High anodic overpotential: second term negligible, $j = j_0 e^{\alpha_a f\eta}$ . (2) $\ln j = \ln j_0 + \alpha_a f\eta \Rightarrow \eta = \dfrac{RT}{\alpha_a F}\ln\dfrac{j}{j_0}$ . (2) In base-10: $\eta = \dfrac{2.303RT}{\alpha_a F}\log_{10}\dfrac{j}{j_0}$ , so $b = \dfrac{2.303RT}{\alpha_a F} = \dfrac{2.303(8.314)(298)}{0.5(96485)} = \dfrac{5706.6}{48242.5} = 0.1183\,\mathrm{V} = 118\,\mathrm{mV/decade}$ . (2)

(b) $\eta = b\log_{10}(j/j_0) \Rightarrow \log_{10}(j/j_0) = \dfrac{120}{118.3} = 1.0144$ . $j/j_0 = 10^{1.0144} = 10.34$ . (2) $j_0 = 10/10.34 = 0.967\,\mathrm{mA\,cm^{-2}} \approx 0.97\,\mathrm{mA\,cm^{-2}}$ . (2)

(c) Large $j_0$ ⇒ facile electron transfer, so the same current is delivered at much lower overpotential (lower energy loss / higher efficiency — a better electrocatalyst). (2)

Q5 [10]

(a) Photon energy $E = \dfrac{hc}{\lambda} = \dfrac{(6.626\times10^{-34})(2.998\times10^{8})}{400\times10^{-9}} = 4.966\times10^{-19}\,\mathrm{J}$ . (1) Total energy absorbed $= P\,t = (0.010)(60) = 0.60\,\mathrm{J}$ . (1) Photons $= 0.60/4.966\times10^{-19} = 1.208\times10^{18}$ . Moles of photons $= 1.208\times10^{18}/6.022\times10^{23} = 2.006\times10^{-6}\,\mathrm{mol}$ . (2) $\Phi = \dfrac{\text{mol product}}{\text{mol photons}} = \dfrac{3.5\times10^{-6}}{2.006\times10^{-6}} = 1.74$ . (2) ( $\Phi>1$ ⇒ a chain/secondary process amplifies product beyond the Stark–Einstein 1:1 limit.)

(b) In an intrinsic semiconductor, raising $T$ thermally promotes more electrons across the gap ( $n\propto e^{-E_g/2k_BT}$ ), increasing carrier concentration and hence conductivity — conductivity rises with $T$ . In a metal, carrier concentration is fixed; increased lattice vibrations enhance scattering, so conductivity falls with $T$ . Opposite temperature dependences. (4)

[
  {"claim":"Q1a variational energy is 5 hbar^2/(m L^2)","code":"x,L,m,hbar=symbols('x L m hbar',positive=True); phi=x*(L-x); num=integrate(-hbar**2/(2*m)*phi*diff(phi,x,2),(x,0,L)); den=integrate(phi**2,(x,0,L)); E=simplify(num/den); result = simplify(E - 5*hbar**2/(m*L**2))==0"},
  {"claim":"Q1c box length approx 0.751 nm","code":"h=6.626e-34; lam=620e-9; me=9.109e-31; c=2.998e8; L=(3*h*lam/(8*me*c))**0.5; result = abs(L-7.51e-10) < 1e-11"},
  {"claim":"Q2b q_rot approx 108","code":"kB=1.381e-23; T=300; h=6.626e-34; c=2.998e10; B=1.931; q=kB*T/(h*c*B); result = abs(q-108) < 2"},
  {"claim":"Q4a Tafel slope approx 0.1183 V","code":"R=8.314; T=298; F=96485; a=0.5; b=2.303*R*T/(a*F); result = abs(b-0.1183) < 0.001"},
  {"claim":"Q5a quantum yield approx 1.74","code":"h=6.626e-34; c=2.998e8; lam=400e-9; E=h*c/lam; Nphot=0.6/E; mol_phot=Nphot/6.022e23; Phi=3.5e-6/mol_phot; result = abs(Phi-1.74) < 0.05"}
]