(a) Identify the least stable ion of the following ions and justify your answer. OCN⁻ ONC⁻ SCN⁻ (10 marks) (b) A 74·6 g ice cube…

(a) Least stable ion The least stable ion is ONC⁻. In OCN⁻ the carbon is central and the valence electrons can be arranged in octet-satisfying resonance forms, O(−)–C≡N and O=C=N(−). The negative charge is delocalized over oxygen and nitrogen, and the contributor with the charge on oxygen is especially favourable because oxygen is more electronegative. In SCN⁻ the analogous forms, S(−)–C≡N and S=C=N(−), delocalize the charge over sulfur and nitrogen; sulfur is larger and more polarizable, so it can accommodate negative charge without excessive electron-electron repulsion, and the C–S and C–N multiple bonds are strong. Both OCN⁻ and SCN⁻ keep the central carbon neutral and avoid large charge separation. In ONC⁻ the required octet structures are much less favourable: the best form, O(−)–N(+)≡C(−), has charge separation and places negative charge on carbon as well as oxygen, while O=N=C(2−) puts a large negative charge on carbon and a positive charge on nitrogen. Thus the negative charge is not placed mainly on the most electronegative atom, and the resonance stabilization is weak; hence ONC⁻ is the least stable.

(b) Entropy changes for melting ice Moles of ice = 74.6 g / 18.015 g/mol = 4.14 mol. At 0°C (273.15 K) melting is endothermic. The heat absorbed is q = 4.14 × 6.01 = 24.9 kJ. The molar entropy change is 6.01 kJ/mol / 273.15 K = 22.0 J/K/mol. For the whole cube, ΔS_syst = 4.14 × 22.0 = +91.1 J/K. The surroundings supply this heat at the same temperature, so ΔS_surr = −91.1 J/K. Therefore ΔS_univ = ΔS_syst + ΔS_surr = 0. On a molar basis the corresponding values are +22.0 J/K/mol and −22.0 J/K/mol. A zero total entropy change means the process is reversible and the ice-water system is at equilibrium at the melting point; there is no net driving force for melting or freezing.

(c) Speed distributions of Ar and He The Maxwell-Boltzmann peak is at u_mp = sqrt(2RT/M). At the same temperature the heavier gas has the smaller u_mp, so its curve is shifted to lower speed, is narrower and has a higher peak. For equal numbers of molecules the areas under the two curves are the same, so the heavier curve must be taller to compensate for its narrower width. Helium (M = 4 g/mol) is much lighter than argon (M = 40 g/mol); therefore He gives the broader, lower curve shifted to higher speed, and Ar gives the sharper, higher curve at lower speed. In the figure printed with the question, curve 1 is the narrow, tall curve peaking at low speed and curve 2 is the broad, flat curve extending to high speed.

(i) Curve 1 represents argon, the heavier gas, and curve 2 represents helium.

(ii) Curve 1. The heavier gas, argon, effuses more slowly, because by Graham’s law the effusion rate is proportional to 1/√M (He would effuse about √10 ≈ 3.2 times faster).

(iii) Curve 1. Fluorine, F₂, has M = 38 g/mol, close to argon (40 g/mol) and far from helium (4 g/mol), so at the same temperature its speed distribution almost coincides with curve 1.

(d) Phase diagram of water For a one-component system the phase diagram is plotted with pressure on the vertical axis and temperature on the horizontal axis. The regions are: solid at low temperature and high pressure, liquid between the fusion and vaporization curves, and gas at low pressure and high temperature.

[Diagram: P–T phase diagram of water with solid, liquid and vapour regions, fusion curve OB with negative slope, vaporization curve OA ending at critical point C (374 °C, 218 atm), sublimation curve OC′, triple point O (0.01 °C, 611 Pa)] The three curves are boundaries where two phases coexist in equilibrium because their chemical potentials are equal. The solid-liquid (fusion) curve starts at the triple point and goes upward and to the left for water. Its slope is given by the Clapeyron equation, dP/dT = ΔH_fus/(T ΔV_fus). For water, melting decreases volume (ice is less dense than liquid water), so ΔV_fus is negative and the fusion curve has a negative slope; increasing pressure therefore favours the denser liquid and lowers the melting point slightly. The liquid-vapour (vaporization) curve runs from the triple point upward and to the right; it separates liquid from vapour and ends at the critical point. Along it, liquid and vapour have equal chemical potential; as temperature and pressure rise the density difference disappears. The solid-vapour (sublimation) curve runs from the triple point downward and to the left; below the triple-point pressure the stable condensed phase is solid, and heating causes direct solid-vapour transition. The triple point of water is at 0.01°C (273.16 K) and 0.006 atm (611 Pa); at this unique condition solid, liquid and vapour coexist. The critical point is at 374°C (647 K) and 218 atm (22.06 MPa); above the critical temperature no amount of pressure can produce a distinct liquid, and above the critical pressure the fluid is supercritical. The triple point fixes the coexistence condition and was the reference point of the kelvin (and hence Celsius) scale until the 2019 SI redefinition, while the critical point marks the limit of the liquid-vapour boundary.

(e) Forces on a moving ion in a strong electrolyte In a strong electrolyte an ion under an applied field experiences several forces. The primary driving force is the electric force qE, which tends to accelerate the ion. In the solvent this acceleration is opposed by viscous drag; at the limiting low-field speed the drag balances the electric force, giving the limiting ionic mobility. The ion is surrounded by a solvation or hydration shell; the bound water increases the effective hydrodynamic radius and therefore increases friction, reducing speed. At finite concentration the ion is also surrounded by an ionic atmosphere of opposite charge. When the ion moves, the spherical atmosphere is distorted: it lags behind, producing a relaxation force that opposes the motion. In addition, the ionic atmosphere tends to move in the opposite direction to the ion and drags solvent with it; this electrophoretic effect increases the effective drag. Interionic attractions and, at higher concentrations, ion pairing or specific interactions further reduce the net mobility. Thus the observed speed is the result of the electric driving force reduced by solvation, viscous drag, relaxation, electrophoretic and interionic forces.