(a) Calculate ⟨x⟩, the expectation value of position of a particle, in the ground state of one-dimensional box having length from…

(a) For a particle in a one-dimensional box of length l from 0 to l, the ground-state wavefunction is ψ₁(x) = √(2/l) sin(πx/l), 0 ≤ x ≤ l. The expectation value of position is defined by ⟨x⟩ = ∫₀ˡ x |ψ₁(x)|² dx. Substitute ψ₁: ⟨x⟩ = (2/l) ∫₀ˡ x sin²(πx/l) dx. Use sin²θ = (1 − cos 2θ)/2: ⟨x⟩ = (1/l) ∫₀ˡ x dx − (1/l) ∫₀ˡ x cos(2πx/l) dx. Now (1/l) ∫₀ˡ x dx = (1/l)(l²/2) = l/2. For the second integral, integrate by parts: ∫ x cos(ax) dx = (x/a) sin(ax) + (1/a²) cos(ax), with a = 2π/l. At x = l, sin(2π) = 0; at x = 0, x sin(ax) = 0. Thus ∫₀ˡ x cos(2πx/l) dx = (1/a²)[cos(2π) − cos 0] = 0. Therefore ⟨x⟩ = l/2. This agrees with the symmetry of the ground-state probability density about the centre of the box.

(b) Use the Born-Haber cycle, which applies Hess’s law to the formation of an ionic solid from its elements through sublimation, ionisation, dissociation, electron gain, and lattice formation.

For NaCl(s): Na(s) + ½Cl₂(g) → NaCl(s) Steps:

• Na(s) → Na(g): ΔHₛᵤᵦ = +108 kJ mol⁻¹
• Na(g) → Na⁺(g) + e⁻: IE₁ = +495 kJ mol⁻¹
• ½Cl₂(g) → Cl(g): ½ΔH_diss = +121 kJ mol⁻¹
• Cl(g) + e⁻ → Cl⁻(g): EA = −348 kJ mol⁻¹
• Na⁺(g) + Cl⁻(g) → NaCl(s): U₀ = −757 kJ mol⁻¹

Hence ΔH_f(NaCl) = 108 + 495 + 121 − 348 − 757 = −381 kJ mol⁻¹. The negative value shows that NaCl formation is thermodynamically favourable and NaCl is stable.

For NaCl₂(s), the formation reaction is Na(s) + Cl₂(g) → NaCl₂(s). Since sodium must become Na²⁺, both first and second ionisation enthalpies are needed. Taking the given IE₂ as the second ionisation step Na⁺ → Na²⁺ + e⁻:

• Na(s) → Na(g): +108 kJ mol⁻¹
• Na(g) → Na⁺(g) + e⁻: +495 kJ mol⁻¹
• Na⁺(g) → Na²⁺(g) + e⁻: +4561 kJ mol⁻¹
• Cl₂(g) → 2Cl(g): 2 × 121 = +242 kJ mol⁻¹
• 2Cl(g) + 2e⁻ → 2Cl⁻(g): 2 × (−348) = −696 kJ mol⁻¹
• Na²⁺(g) + 2Cl⁻(g) → NaCl₂(s): U₀ = −2155 kJ mol⁻¹

Thus ΔH_f(NaCl₂) = 108 + 495 + 4561 + 242 − 696 − 2155 = +2555 kJ mol⁻¹. This is highly endothermic. Although the lattice enthalpy of NaCl₂ is much more negative than that of NaCl, it cannot compensate for the enormous second ionisation enthalpy of sodium. Also, the formation entropy change for a solid from a solid and a gas is generally unfavourable. Hence NaCl₂ is not expected to form under normal conditions; NaCl is the stable salt.

(c) For a rubber band, the given thermodynamic relation is (∂T/∂l)ₛ = −(T/Cᵥ)(∂S/∂l)ₜ. When a rubber band is stretched, the long polymer chains become more aligned and ordered. Therefore entropy decreases with increasing length: (∂S/∂l)ₜ < 0. Since T and Cᵥ are positive, the right-hand side becomes positive: (∂T/∂l)ₛ > 0. Thus, under adiabatic conditions, stretching a rubber band raises its temperature, and sudden contraction cools it.

To answer the length change when temperature is increased, keep tension/load constant. For an elastic system, the relevant Maxwell relation from dG = −S dT − l df is (∂l/∂T)_f = (∂S/∂f)ₜ. Increasing tension f stretches the band and decreases entropy, so (∂S/∂f)ₜ < 0. Therefore (∂l/∂T)_f < 0. Hence, at constant tension, an increase in temperature causes the rubber band to contract; a decrease in temperature makes it lengthen. Physically, heating increases the entropy, and the entropically favoured state is the more coiled, disordered, shorter state. This is why a stretched rubber band contracts when warmed and expands when cooled.

(d) Use the integrated Clausius-Clapeyron equation, assuming ΔHᵥₐₚ is constant over the range and water vapour behaves ideally: ln(P₂/P₁) = (ΔHᵥₐₚ/R)(1/T₁ − 1/T₂). Given: P₁ = 634 mm at T₁ = 95 °C = 368.15 K, T₂ = 100 °C = 373.15 K, ΔHᵥₐₚ = 40593 J mol⁻¹, R = 8.314 J K⁻¹ mol⁻¹.

Compute ΔHᵥₐₚ/R = 40593/8.314 = 4882.49 K. Also 1/T₁ − 1/T₂ = 1/368.15 − 1/373.15 = (373.15 − 368.15)/(368.15 × 373.15) = 5/137375.17 = 3.6397 × 10⁻⁵ K⁻¹.

Therefore ln(P₂/P₁) = 4882.49 × 3.6397 × 10⁻⁵ = 0.1777. Thus P₂/P₁ = e⁰·¹⁷⁷⁷ = 1.1945. P₂ = 634 × 1.1945 = 757.3 mm.

The vapour pressure of water at 100 °C is approximately 757.3 mm Hg. The value is close to the standard 760 mm Hg; the small difference arises because the heat of vapourization is assumed constant over the interval.

(e) Definition: The electrochemical series is the arrangement of elements, electrodes, or half-reactions in order of their standard electrode potentials, usually standard reduction potentials E°, measured relative to the standard hydrogen electrode, whose potential is taken as 0.00 V at 298 K, 1 M activity, and 1 bar pressure. A half-reaction with a more positive E° has a greater tendency to undergo reduction; a more negative E° indicates a greater tendency to undergo oxidation.

Significance:

• It predicts the relative strength of oxidising and reducing agents. More positive E° means a stronger oxidising agent; more negative E° means a stronger reducing agent.
• It predicts the feasibility of a redox reaction. For a cell, E°cell = E°cathode − E°anode. If E°cell is positive, the reaction is spontaneous and ΔG° = −nFE°cell is negative.
• It helps calculate the equilibrium constant using log K = nE°cell/0.0591 at 298 K.
• It explains displacement reactions: a metal with more negative E° can displace a metal with more positive E° from its salt solution.
• It is used to choose electrodes in galvanic cells, batteries, electrolysis, electroplating, and electrochemical refining.
• It indicates corrosion tendency. Metals with highly negative E° are easily oxidised and can be used as sacrificial electrodes to protect iron.
• It explains reactions with acids: metals above hydrogen in the series displace H₂ from dilute acids, while those below hydrogen generally do not.