Interleaved — Phase 5

Chemistry interleaved practice

printable — key stays hidden on paper

Instructions: Solve all problems in order. Each problem draws from a different subtopic — read carefully and decide which method/framework applies before computing. Show all work. Use $R = 8.314\ \mathrm{J\,mol^{-1}K^{-1}}$ , $h = 6.626\times10^{-34}\ \mathrm{J\,s}$ , $k_B = 1.381\times10^{-23}\ \mathrm{J\,K^{-1}}$ , $c = 2.998\times10^8\ \mathrm{m/s}$ , $N_A = 6.022\times10^{23}$ . Total: 60 marks.

1. (6 marks) A radioisotope sample has an activity of $8.0\times10^4\ \mathrm{Bq}$ . Its half-life is 5.0 hours. (a) Find the decay constant $\lambda$ in $\mathrm{s^{-1}}$ . (b) Find the number of radioactive atoms present. (c) What is the activity after 15 hours?

2. (6 marks) For the D–T fusion reaction $^2_1\mathrm{H} + {}^3_1\mathrm{H} \rightarrow {}^4_2\mathrm{He} + {}^1_0\mathrm{n}$ , the atomic masses are: $^2\mathrm{H} = 2.014102\ u$ , $^3\mathrm{H} = 3.016049\ u$ , $^4\mathrm{He} = 4.002603\ u$ , $n = 1.008665\ u$ . Compute the Q-value in MeV ( $1\ u = 931.5\ \mathrm{MeV}/c^2$ ). State whether it is exo- or endoergic.

3. (7 marks) An adsorbent follows the Langmuir isotherm. At pressures 2.0 kPa and 8.0 kPa the fractional coverages are $\theta = 0.40$ and $\theta = 0.72$ respectively. (a) Determine the Langmuir constant $K$ . (b) Verify both data points are consistent with a single $K$ . (c) Comment on what BET would add that Langmuir omits.

4. (6 marks) Estimate the rotational partition function $Q_\mathrm{rot}$ of $^{12}\mathrm{C}^{16}\mathrm{O}$ at 300 K given rotational constant $B = 1.931\ \mathrm{cm^{-1}}$ (symmetry number $\sigma = 1$ ). Use the high-temperature approximation and justify its validity.

5. (6 marks) A photochemical reaction has a quantum yield $\Phi = 0.25$ for product formation. A sample absorbs $3.0\times10^{18}$ photons per second of 400 nm light. (a) How many product molecules form per second? (b) What incident power (W) is absorbed? (c) State the Stark–Einstein law and explain why $\Phi$ can exceed 1 in some reactions.

6. (6 marks) For an electron in a 1-D box of length $L = 1.0\ \mathrm{nm}$ , compute the wavelength of the photon emitted in the $n=3 \to n=2$ transition. Which model assumption breaks down for real conjugated molecules?

7. (7 marks) Methane burns in air at the stoichiometric ratio: $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$ . (a) Write the balanced reaction including $\mathrm{N_2}$ for combustion in air (air = 21% $\mathrm{O_2}$ , 79% $\mathrm{N_2}$ by mole). (b) Compute the air-to-fuel mole ratio. (c) If the mixture is fuel-lean with 10% excess air, recompute the air-to-fuel mole ratio and state the effect on flame temperature qualitatively.

8. (5 marks) An electrochemical cell operates in the high-overpotential regime. Its Tafel plot has a slope of $0.118\ \mathrm{V/decade}$ . (a) Extract the transfer coefficient $\alpha$ at 298 K assuming a one-electron step. (b) The exchange current density is $j_0 = 1.0\times10^{-6}\ \mathrm{A/cm^2}$ ; find the current density at overpotential $\eta = 0.30\ \mathrm{V}$ (anodic).

9. (6 marks) Using the variational principle, a trial function $\psi = e^{-\alpha x^2}$ is used for the 1-D harmonic oscillator $\hat H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \frac12 m\omega^2 x^2$ . The energy expectation is $E(\alpha) = \frac{\hbar^2\alpha}{2m} + \frac{m\omega^2}{8\alpha}$ . (a) Minimize $E(\alpha)$ . (b) Show the result equals the exact ground-state energy. (c) Why does the variational estimate equal the exact value here?

10. (5 marks) A silicon semiconductor is doped with phosphorus. (a) Identify the dopant type (n or p) and the majority carrier. (b) Sketch (describe) where the donor level sits relative to the conduction band. (c) Explain qualitatively how conductivity changes with temperature versus a metal.

Answer keyMark scheme & solutions

1. Subtopic 5.2.3 (Decay kinetics). Why: "activity," "half-life," "Bq" → first-order decay law.

(a) $\lambda = \ln2/t_{1/2} = 0.6931/(5.0\times3600\ \mathrm{s}) = 0.6931/18000 = 3.85\times10^{-5}\ \mathrm{s^{-1}}$ .

(b) $A = \lambda N \Rightarrow N = A/\lambda = 8.0\times10^4/3.85\times10^{-5} = 2.08\times10^{9}$ atoms.

2. Subtopic 5.2.5 / 5.2.7 (Q-value, fusion). Why: mass defect of a nuclear reaction → Q-value.

$\Delta m = (2.014102 + 3.016049) - (4.002603 + 1.008665) = 5.030151 - 5.011268 = 0.018883\ u$ .

$Q = 0.018883 \times 931.5 = 17.59\ \mathrm{MeV}$ . Positive → exoergic (releases energy).

3. Subtopic 5.1.6 (Langmuir isotherm). Why: fractional coverage vs pressure → Langmuir.

Langmuir: $\theta = \dfrac{Kp}{1+Kp}$ .

(a) From $\theta=0.40$ at $p=2.0$ : $\frac{\theta}{1-\theta} = Kp \Rightarrow \frac{0.40}{0.60} = 0.6667 = K(2.0) \Rightarrow K = 0.333\ \mathrm{kPa^{-1}}$ .

(b) Check second point: $Kp = 0.333\times8.0 = 2.667$ ; $\theta = 2.667/3.667 = 0.727 \approx 0.72$ . ✓ Consistent.

(c) BET accounts for multilayer adsorption (physisorption beyond monolayer), whereas Langmuir assumes a single monolayer with uniform sites and no lateral interaction.

4. Subtopic 5.1.5 (Q_rot). Why: rotational constant + temperature → partition function.

High-T: $Q_\mathrm{rot} = \dfrac{k_B T}{\sigma h c B}$ .

$hcB = 6.626\times10^{-34}\times2.998\times10^8\times193.1\ \mathrm{m^{-1}} = 3.836\times10^{-23}\ \mathrm{J}$ .

$k_B T = 1.381\times10^{-23}\times300 = 4.143\times10^{-21}\ \mathrm{J}$ .

$Q_\mathrm{rot} = 4.143\times10^{-21}/(1\times3.836\times10^{-23}) = 108$ .

Valid because $k_BT/(hcB) = 108 \gg 1$ , so many rotational levels are populated (classical limit).

5. Subtopic 5.1.9 (Photochemistry, quantum yield). Why: photons absorbed + quantum yield.

(a) Products/s $= \Phi \times$ photons/s $= 0.25\times3.0\times10^{18} = 7.5\times10^{17}$ per second.

(b) Photon energy $E = hc/\lambda = 6.626\times10^{-34}\times2.998\times10^8/400\times10^{-9} = 4.966\times10^{-19}\ \mathrm{J}$ . Power $= 3.0\times10^{18}\times4.966\times10^{-19} = 1.49\ \mathrm{W}$ .

(c) Stark–Einstein: each absorbed photon activates one molecule (primary process). $\Phi>1$ arises when a secondary chain reaction (e.g. radical propagation, like $\mathrm{H_2 + Cl_2}$ ) produces many products per absorbed photon.

6. Subtopic 5.1.1 (Particle in a box). Why: box energy levels + transition wavelength.

$E_n = \dfrac{n^2 h^2}{8mL^2}$ . $\dfrac{h^2}{8mL^2} = \dfrac{(6.626\times10^{-34})^2}{8\times9.109\times10^{-31}\times(1.0\times10^{-9})^2} = 6.025\times10^{-20}\ \mathrm{J}$ .

$\Delta E = (9-4)\times6.025\times10^{-20} = 3.01\times10^{-19}\ \mathrm{J}$ .

$\lambda = hc/\Delta E = 1.986\times10^{-25}/3.01\times10^{-19} = 6.60\times10^{-7}\ \mathrm{m} = 660\ \mathrm{nm}$ .

Breakdown: real conjugated systems have a non-flat potential (electron–nuclear attraction, electron–electron repulsion), not infinite rigid walls.

7. Subtopic 5.3.1 (Stoichiometric/lean combustion). Why: air ratios and fuel-lean.

(a) $\mathrm{CH_4 + 2O_2 + 2\times\frac{79}{21}N_2 \rightarrow CO_2 + 2H_2O + 7.52\,N_2}$ . ( $2\times79/21 = 7.52$ .)

(b) Air per mole fuel $= 2\ \mathrm{O_2} + 7.52\ \mathrm{N_2} = 9.52$ mol air. A/F = 9.52 (mole).

(c) 10% excess air: $\times1.10 = 10.47$ mol air per mole fuel. Fuel-lean → excess air absorbs heat (extra $\mathrm{N_2/O_2}$ diluent), so flame temperature is lower than stoichiometric.

8. Subtopic 5.1.8 (Butler–Volmer / Tafel). Why: Tafel slope & overpotential.

(a) Tafel slope (anodic) $= \dfrac{2.303 RT}{\alpha F}$ . So $\alpha = \dfrac{2.303 RT}{F\times\text{slope}} = \dfrac{2.303\times8.314\times298}{96485\times0.118} = \dfrac{5706}{11385} = 0.50$ .

(b) $\eta/\text{slope} = 0.30/0.118 = 2.54$ decades. $j = j_0\times10^{2.54} = 1.0\times10^{-6}\times347 = 3.5\times10^{-4}\ \mathrm{A/cm^2}$ .

9. Subtopic 5.1.2 (Variational principle). Why: trial function + minimize energy.

(a) $\frac{dE}{d\alpha} = \frac{\hbar^2}{2m} - \frac{m\omega^2}{8\alpha^2} = 0 \Rightarrow \alpha^2 = \frac{m^2\omega^2}{4\hbar^2} \Rightarrow \alpha = \frac{m\omega}{2\hbar}$ .

(b) $E_\min = \frac{\hbar^2}{2m}\cdot\frac{m\omega}{2\hbar} + \frac{m\omega^2}{8}\cdot\frac{2\hbar}{m\omega} = \frac{\hbar\omega}{4} + \frac{\hbar\omega}{4} = \frac12\hbar\omega$ . ✓

(c) The Gaussian trial function has the exact functional form of the true HO ground state, so the variational bound is saturated (equality holds).

10. Subtopic 5.1.10 (Semiconductors/band theory). Why: doping and conductivity.

(a) P has 5 valence electrons vs Si's 4 → n-type; majority carrier = electrons.

(b) Donor level sits just below the conduction band (small ionization energy ~0.045 eV), easily thermally ionized.

(c) Semiconductor conductivity increases with T (more carriers excited across gap / ionized donors), opposite to a metal whose conductivity decreases with T (increased phonon scattering).

[
  {"claim":"Q-value of D-T fusion = 17.59 MeV", "code":"dm=(2.014102+3.016049)-(4.002603+1.008665); Q=dm*931.5; result=abs(Q-17.59)<0.02"},
  {"claim":"Langmuir K=0.333 gives theta=0.727 at 8 kPa", "code":"K=0.4/0.6/2.0; theta=K*8.0/(1+K*8.0); result=abs(theta-0.727)<0.005"},
  {"claim":"Variational HO ground state = hbar*omega/2", "code":"import sympy as sp; hbar,m,omega,a=sp.symbols('hbar m omega a',positive=