Level 5 — MasteryPhysical Chemistry (Advanced)

Physical Chemistry (Advanced)

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Level 5 — Mastery (cross-domain: math + physics + coding, build/prove) Time limit: 2 hours 30 minutes Total marks: 60

Instructions: Answer all three questions. Show full derivations. Where code is requested, pseudo-Python with SymPy/NumPy semantics is acceptable. Physical constants: $h=6.626\times10^{-34}\,\mathrm{J\,s}$ , $k_B=1.381\times10^{-23}\,\mathrm{J\,K^{-1}}$ , $c=2.998\times10^{10}\,\mathrm{cm\,s^{-1}}$ , $N_A=6.022\times10^{23}\,\mathrm{mol^{-1}}$ , $R=8.314\,\mathrm{J\,mol^{-1}K^{-1}}$ .

Question 1 — Variational bound for the particle in a box (20 marks)

Consider a particle of mass $m$ confined to a 1-D box $0\le x\le L$ with infinite walls.

(a) State the exact ground-state energy $E_1^{\text{exact}}$ and wavefunction. (3)

(b) Use the normalized-free trial function $\phi(x)=x(L-x)$ (unnormalized) in the variational principle. Compute the variational energy $E_{\text{var}}=\frac{\langle\phi|\hat H|\phi\rangle}{\langle\phi|\phi\rangle}$ where $\hat H=-\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}$ . Show all integrals. (9)

(c) Express $E_{\text{var}}$ as a numerical multiple of $E_1^{\text{exact}}$ and prove it satisfies the variational theorem ( $E_{\text{var}}\ge E_1^{\text{exact}}$ ). Give the fractional error in %. (4)

(d) Write a short SymPy snippet that reproduces the variational integrals and the ratio $E_{\text{var}}/E_1^{\text{exact}}$ . (4)

Question 2 — Statistical thermodynamics of a diatomic (22 marks)

For gaseous $\mathrm{CO}$ at $T=298\,\mathrm{K}$ , $p=1\,\mathrm{bar}$ : rotational constant $\tilde B = 1.931\,\mathrm{cm^{-1}}$ , vibrational wavenumber $\tilde\nu = 2170\,\mathrm{cm^{-1}}$ , molar mass $M=28.0\,\mathrm{g\,mol^{-1}}$ , symmetry number $\sigma=1$ , non-degenerate ground electronic state.

(a) Derive the high-temperature rotational partition function $q_{\text{rot}}=\dfrac{k_BT}{\sigma\,h c\tilde B}$ from $q_{\text{rot}}=\sum_J(2J+1)e^{-hcB J(J+1)/k_BT}$ , stating the approximation used. Evaluate $q_{\text{rot}}$ . (6)

(b) Compute the vibrational partition function $q_{\text{vib}}$ (measure energy from the vibrational ground state) and the vibrational contribution to the molar internal energy $U_{\text{vib}}-U_{\text{vib}}(0)$ . (6)

(c) Derive the general expression for the molar constant-volume heat capacity of one vibrational mode, $C_{V,\text{vib}}=R\left(\frac{\theta_v}{T}\right)^2\frac{e^{\theta_v/T}}{(e^{\theta_v/T}-1)^2},\quad \theta_v=\frac{hc\tilde\nu}{k_B},$ starting from $U=Nk_BT^2(\partial\ln q/\partial T)$ . Evaluate $C_{V,\text{vib}}$ at 298 K. (6)

(d) State (with one-line justification) whether translational, rotational, or vibrational modes dominate $C_V$ at 298 K, and give the total $C_{V,m}$ (ideal-gas, rigid-rotor–harmonic). (4)

Question 3 — Electrode kinetics: Butler–Volmer and Tafel (18 marks)

An electrode reaction obeys the Butler–Volmer equation $j=j_0\left[\exp\!\left(\frac{\alpha_a F\eta}{RT}\right)-\exp\!\left(-\frac{\alpha_c F\eta}{RT}\right)\right].$

(a) Define $j_0$ , $\eta$ , $\alpha_a$ , $\alpha_c$ . State the constraint linking $\alpha_a$ and $\alpha_c$ for a single-step, single-electron transfer. (4)

(b) Derive the anodic Tafel equation from the Butler–Volmer equation, stating the limiting condition. Show that a plot of $\eta$ vs $\ln j$ is linear with slope $RT/(\alpha_a F)$ . (6)

(c) A Tafel plot ( $\eta$ vs $\log_{10} j$ ) at 298 K gives an anodic slope of $118\,\mathrm{mV/decade}$ and an intercept indicating $\eta=0$ at $j=j_0$ . Determine $\alpha_a$ . At $\eta=+0.20\,\mathrm{V}$ , find $j/j_0$ . (5)

(d) Write a NumPy snippet that computes $j/j_0$ over $\eta\in[-0.3,0.3]\,\mathrm{V}$ using the full BV equation ( $\alpha_a=\alpha_c=0.5$ ) and identifies the $\eta$ at which $|j|/j_0=100$ . (3)

Answer keyMark scheme & solutions

Question 1

(a) (3) $E_n=\frac{n^2h^2}{8mL^2}\Rightarrow E_1^{\text{exact}}=\frac{h^2}{8mL^2}=\frac{\pi^2\hbar^2}{2mL^2}.$ (1) $\psi_1(x)=\sqrt{\frac{2}{L}}\sin\frac{\pi x}{L}.$ (2)

(b) (9) Numerator: $\phi''=-2$ , so $\hat H\phi=-\frac{\hbar^2}{2m}(-2)=\frac{\hbar^2}{m}.$ (2) $\langle\phi|\hat H|\phi\rangle=\frac{\hbar^2}{m}\int_0^L x(L-x)\,dx=\frac{\hbar^2}{m}\cdot\frac{L^3}{6}.$ (3) Denominator: $\langle\phi|\phi\rangle=\int_0^L x^2(L-x)^2\,dx=\frac{L^5}{30}.$ (3) $E_{\text{var}}=\frac{(\hbar^2/m)(L^3/6)}{L^5/30}=\frac{5\hbar^2}{mL^2}.$ (1)

(c) (4) $\frac{E_{\text{var}}}{E_1^{\text{exact}}}=\frac{5\hbar^2/(mL^2)}{\pi^2\hbar^2/(2mL^2)}=\frac{10}{\pi^2}\approx1.01321.$ (2) Since $10/\pi^2>1$ , $E_{\text{var}}>E_1^{\text{exact}}$ — consistent with the variational theorem (any trial function gives an upper bound). (1) Fractional error $=1.321\%$ . (1)

(d) (4)

import sympy as sp
x, L, hbar, m = sp.symbols('x L hbar m', positive=True)
phi = x*(L-x)
H_phi = -hbar**2/(2*m)*sp.diff(phi, x, 2)
num = sp.integrate(phi*H_phi, (x, 0, L))
den = sp.integrate(phi**2, (x, 0, L))
Evar = sp.simplify(num/den)                 # 5*hbar**2/(m*L**2)
E1 = sp.pi**2*hbar**2/(2*m*L**2)
ratio = sp.simplify(Evar/E1)                # 10/pi**2

Question 2

(a) (6) Sum $q_{\text{rot}}=\sum_J(2J+1)e^{-hcB J(J+1)/k_BT}$ . When $k_BT\gg hcB$ (levels closely spaced) replace the sum by an integral in $J$ : let $u=J(J+1)$ , $du=(2J+1)dJ$ : (2) $q_{\text{rot}}\approx\int_0^\infty e^{-hcBu/k_BT}\,du=\frac{k_BT}{hcB},\ \text{÷}\ \sigma.$ (2) Numerics: $\dfrac{k_BT}{hc\tilde B}=\dfrac{(1.381\times10^{-23})(298)}{(6.626\times10^{-34})(2.998\times10^{10})(1.931)}$ $=\dfrac{4.115\times10^{-21}}{3.836\times10^{-23}}\approx107.3.$ So $q_{\text{rot}}\approx107$ . (2)

(b) (6) $\theta_v=\dfrac{hc\tilde\nu}{k_B}=\dfrac{(6.626\times10^{-34})(2.998\times10^{10})(2170)}{1.381\times10^{-23}}=3122\,\mathrm{K}.$ (2) $\dfrac{\theta_v}{T}=\dfrac{3122}{298}=10.48;\ e^{-\theta_v/T}=2.8\times10^{-5}.$ $q_{\text{vib}}=\frac{1}{1-e^{-\theta_v/T}}\approx1.0000.$ (2) $U_{\text{vib}}-U(0)=\frac{R\theta_v}{e^{\theta_v/T}-1}=\frac{(8.314)(3122)}{e^{10.48}-1}\approx\frac{25959}{35700}\approx0.73\,\mathrm{J\,mol^{-1}}.$ (2) (essentially unpopulated at 298 K.)

(c) (6) $q_{\text{vib}}=(1-e^{-\theta_v/T})^{-1}$ ; $\ln q=-\ln(1-e^{-\theta_v/T})$ . $U=Nk_BT^2\partial_T\ln q=\dfrac{Nk_B\theta_v}{e^{\theta_v/T}-1}=\dfrac{R\theta_v}{e^{\theta_v/T}-1}$ (molar). (2) $C_{V,\text{vib}}=\frac{dU}{dT}=R\frac{d}{dT}\!\left[\frac{\theta_v}{e^{\theta_v/T}-1}\right]=R\left(\frac{\theta_v}{T}\right)^2\frac{e^{\theta_v/T}}{(e^{\theta_v/T}-1)^2}.$ (2) At 298 K: $(\theta_v/T)^2=109.8$ ; $e^{\theta_v/T}=3.57\times10^4$ ; $(e^{\theta_v/T}-1)^2\approx1.275\times10^9$ . $C_{V,\text{vib}}=8.314\times109.8\times\frac{3.57\times10^4}{1.275\times10^9}\approx8.314\times109.8\times2.80\times10^{-5}\approx0.0256\,\mathrm{J\,mol^{-1}K^{-1}}.$ (2)

(d) (4) Translational + rotational modes dominate; vibrational is essentially frozen ( $\theta_v\gg T$ ). (2) $C_{V,m}=\tfrac32R+\R_{\text{rot}}+C_{V,\text{vib}}=\tfrac32R+R+0.026\approx\tfrac52R+0.03\approx20.81\,\mathrm{J\,mol^{-1}K^{-1}}.$ (2)

Question 3

(a) (4) $j_0$ = exchange current density (equal forward/backward rate at equilibrium). $\eta=E-E_{\text{eq}}$ = overpotential. $\alpha_a,\alpha_c$ = anodic/cathodic transfer coefficients (fraction of $\eta$ aiding each direction). Constraint: $\alpha_a+\alpha_c=1$ (single-electron step, $n=1$ ). (4)

(b) (6) For large positive $\eta$ , the cathodic term $\exp(-\alpha_c F\eta/RT)\to0$ : (2) $j\approx j_0\exp\!\left(\frac{\alpha_a F\eta}{RT}\right).$ (2) Take $\ln$ : $\ln j=\ln j_0+\dfrac{\alpha_a F}{RT}\eta$ , i.e. $\eta=-\frac{RT}{\alpha_a F}\ln j_0+\frac{RT}{\alpha_a F}\ln j,$ linear in $\ln j$ with slope $RT/(\alpha_a F)$ . (2)

(c) (5) Decade slope $b=\dfrac{2.303RT}{\alpha_a F}=0.118\,\mathrm V.$ $\alpha_a=\frac{2.303RT}{Fb}=\frac{2.303(8.314)(298)}{(96485)(0.118)}=\frac{5706}{11385}\approx0.501\approx0.5.$ (3) At $\eta=0.20$ V: $\dfrac{j}{j_0}=\exp\!\left(\dfrac{0.5\times96485\times0.20}{8.314\times298}\right)=\exp(3.896)\approx49.2.$ (2)

(d) (3)

import numpy as np
F, R, T = 96485., 8.314, 298.
aa = ac = 0.5
eta = np.linspace(-0.3, 0.3, 6001)
r = np.exp(aa*F*eta/(R*T)) - np.exp(-ac*F*eta/(R*T))   # j/j0
idx = np.argmin(np.abs(np.abs(r) - 100))
print(eta[idx])            # ~ +0.2367 V (|j|/j0 = 100)

[
 {"claim":"Variational ratio Evar/E1 = 10/pi**2",
  "code":"x,L,hbar,m=symbols('x L hbar m',positive=True); phi=x*(L-x); Hphi=-hbar**2/(2*m)*diff(phi,x,2); num=integrate(phi*Hphi,(x,0,L)); den=integrate(phi**2,(x,0,L)); Evar=simplify(num/den); E1=pi**2*hbar**2/(2*m*L**2); result=simplify(Evar/E1 - Rational(10)/pi**2)==0"},
 {"claim":"Evar = 5 hbar^2/(m L^2)",
  "code":"x,L,hbar,m=symbols('x L hbar m',positive=True); phi=x*(L-x); Hphi=-hbar**2/(2*m)*diff(phi,x,2); Evar=simplify(integrate(phi*Hphi,(x,0,L))/integrate(phi**2,(x,0,L))); result=simplify(Evar-5*hbar**2/(m*L**2))==0"},
 {"claim":"q_rot for CO approx 107",
  "code":"kB=1.381e-23;T=298;h=6.626e-34;c=2.998e10;B=1.931; q=kB*T/(h*c*B); result=abs(q-107.3)<1.0"},
 {"claim":"j/j0 at eta=0.20V, alpha=0.5 approx 49.2",
  "code":"F=96485.;R=8.314;T=298.;import sympy as _; val=exp(0.5*F*0.20/(R*T)); result=abs(float(val)-49.2)<0.6"}
]