(a) By using the following given data, find that under what conditions is H₂ in the state corresponding to N₂ at 126 K and 1 atm…

(a) The law of corresponding states states that real gases obey the same reduced equation of state; hence they are in corresponding states when their reduced temperatures Tᵣ = T/T_c and reduced pressures Pᵣ = P/P_c are equal.

For N₂ at 126 K and 1 atm: T_c(N₂) = 126 K, so Tᵣ(N₂) = 126/126 = 1. P_c(N₂) = 39 atm, so Pᵣ(N₂) = 1/39 = 0.02564.

For H₂ to be in the corresponding state, it must have the same reduced variables: Tᵣ(H₂) = 1 and Pᵣ(H₂) = 1/39.

For H₂, T_c = 33 K and P_c = 13 atm. Thus, T(H₂) = Tᵣ × T_c = 1 × 33 K = 33 K. P(H₂) = Pᵣ × P_c = (1/39) × 13 atm = 13/39 atm = 1/3 atm = 0.333 atm.

Therefore, H₂ is in the state corresponding to N₂ at 126 K and 1 atm when T = 33 K and P = 0.333 atm (1/3 atm).

This result is based on the classical two-parameter law of corresponding states. Because H₂ is a light quantum gas at low temperature, small deviations from this law are expected. A third parameter such as the acentric factor would be needed for better accuracy.

(b) Use the integrated Clausius–Clapeyron equation for liquid–vapour equilibrium: ln(P₂/P₁) = −(ΔH̄/R)(1/T₂ − 1/T₁) = (ΔH̄/R)(1/T₁ − 1/T₂).

Assumptions: vapour behaves ideally, molar volume of liquid is negligible compared with vapour, and ΔH̄ is constant over the interval. The equation is valid when these conditions are met. The result is an average value because ΔH̄ varies slightly with temperature.

Given: T₁ = 363.2 K, P₁ = 529 torr. At the normal boiling point, T₂ = 373.2 K and P₂ = 760 torr.

Then ln(P₂/P₁) = ln(760/529) = ln(1.43667) = 0.36233.

Also, 1/T₁ − 1/T₂ = (T₂ − T₁)/(T₁T₂) = 10.0/(363.2 × 373.2) = 10.0/135546.24 = 7.3776 × 10⁻⁵ K⁻¹.

Therefore, ΔH̄ = R × ln(P₂/P₁)/(1/T₁ − 1/T₂) = 8.314 × 0.36233/(7.3776 × 10⁻⁵) J mol⁻¹ = 3.0124/(7.3776 × 10⁻⁵) J mol⁻¹ = 40830 J mol⁻¹ = 40.83 kJ mol⁻¹.

Average molar heat of vaporization of water between 363.2 K and 373.2 K = 4.08 × 10⁴ J mol⁻¹ = 40.8 kJ mol⁻¹.

(c) The capillary rise is given by h = 2γ cosθ/(ρgr), where γ is surface tension, θ is contact angle, ρ is density, g is acceleration due to gravity, and r is capillary radius.

For the same capillary and the same contact angle θ, r, g and θ are constant. Hence h ∝ γ/ρ.

Let the surface tension and density of liquid B be γ_B and ρ_B. For liquid A: γ_A = (1/2)γ_B and ρ_A = 2ρ_B.

Thus, h_A ∝ γ_A/ρ_A = (γ_B/2)/(2ρ_B) = γ_B/(4ρ_B). For liquid B, h_B ∝ γ_B/ρ_B.

Therefore, h_A/h_B = (γ_B/(4ρ_B))/(γ_B/ρ_B) = 1/4, so h_B = 4h_A. Given h_A = 2.0 cm, h_B = 4 × 2.0 cm = 8.0 cm.

Liquid B will rise to a height of 8.0 cm in the same capillary.

This assumes the contact angle is the same for both liquids in the capillary. If the contact angles differ, the result would change accordingly.

(d) Polarography is a voltammetric technique in which current is measured as a function of applied potential using a dropping mercury electrode (DME) as the working electrode. The DME gives a fresh mercury surface at each drop, which prevents accumulation of products and gives reproducible diffusion-controlled currents. A supporting electrolyte is added to suppress migration of the electroactive species and reduce solution resistance. The analyte reaches the electrode mainly by diffusion. When the applied potential becomes sufficiently cathodic or anodic, the analyte is reduced or oxidized. The current rises and then reaches a nearly constant plateau because the rate of diffusion of the analyte to the electrode surface becomes the limiting factor.

A typical polarogram is a plot of current (μA) on the y-axis against applied potential (V) on the x-axis. It shows:

• a small residual current before the wave,
• a rising portion when the electroactive species begins to undergo electron transfer,
• a limiting-current plateau, and
• a midpoint potential called the half-wave potential.

The limiting current is the total current at the plateau. After subtracting the residual current, the remaining current is the diffusion current, i_d. Under supporting-electrolyte conditions, the limiting current is essentially the sum of residual current and diffusion current.

Significance of the terms: (i) Half-wave potential, E½: It is the potential at which the current is exactly half of the diffusion current. For a reversible diffusion-controlled wave, E½ is independent of concentration and is characteristic of the reducible or oxidizable species in a given medium. Therefore, it is used for qualitative identification and for studying complex formation, pH effects, and reversibility. (ii) Diffusion current, i_d: It is the plateau current corrected for residual current. It is directly proportional to the concentration of the electroactive species, as expressed by the Ilkovic equation: i_d = 708 n C D^(1/2) m^(2/3) t^(1/6), where C is concentration, n is number of electrons, D is diffusion coefficient, m is mass flow rate of mercury, and t is drop time. Hence i_d is used for quantitative analysis. (iii) Limiting current: It is the maximum steady current attained at the plateau. It indicates that the electrode reaction is controlled by mass transport, usually diffusion. Its height above the residual current gives the diffusion current used for analytical measurement.

Thus, E½ gives qualitative identity, while i_d gives quantitative concentration; the limiting current plateau represents diffusion control.

(e) The percentage ionic character of a diatomic bond is calculated by comparing the observed dipole moment with the dipole moment expected for complete transfer of one electron: % ionic character = (μ_observed/μ_ionic) × 100, where μ_ionic = e × d.

The question writes the dipole moment as 1.42×10⁻³⁰ cm; this is read as 1.42×10⁻³⁰ C m, the SI unit, because otherwise the data are dimensionally inconsistent. Also the bond length 214×10⁻¹⁴ m is a typographical error; the physical bond length is 214×10⁻¹² m = 2.14×10⁻¹⁰ m.

Given: μ_observed = 1.42 × 10⁻³⁰ C m, e = 1.6 × 10⁻¹⁹ C, d = 2.14 × 10⁻¹⁰ m.

Then μ_ionic = e × d = (1.6 × 10⁻¹⁹ C)(2.14 × 10⁻¹⁰ m) = 3.424 × 10⁻²⁹ C m.

Therefore, % ionic character = (1.42 × 10⁻³⁰)/(3.424 × 10⁻²⁹) × 100 = (1.42/34.24) × 100 = 4.15%.

Percentage ionic character of BrCl ≈ 4.15%.

The partial charge is therefore about 0.0415e on each atom. Since Cl is more electronegative than Br, the bond is polarised as Br^δ+–Cl^δ−. The very low ionic character shows that the Br–Cl bond is predominantly covalent, with only slight polarity, consistent with the small electronegativity difference between Br and Cl.