Nitrogen occurs mainly in the atmosphere. Its abundance in Earth`s Crust is only 0.002 % by mass. Th — Physical Chemistry — Kinetics Chemistry Question
Nitrogen oxides and oxoanions
Nitrogen occurs mainly in the atmosphere. Its abundance in Earth`s Crust is only 0.002 % by mass. The only important nitrogen containing minerals are sodium nitrate (Chile saltpeter) and potassium nitrate (saltpeter). Sodium nitrate, NaNO3, and its close relative sodium nitrite, NaNO2, are two food preservatives with very similar chemical formulae, but different chemical properties. Sodium nitrate helps to prevent bacterial colonization of food. Sodium nitrite is a strong oxidizing agent used as a meat preservative. As in the case of almost any food additive or preservative, sodium nitrate is linked to several adverse reactions in susceptible people. Consuming too much sodium nitrate can cause allergies. Excessive ingestion of the preservative can also cause headaches.
Draw the Lewis structures for the anions of these two salts including all possible resonance forms. Which one of these two anions has shorter N-O bond distance?
Model Answer
Nitrate anion:
N(=O)(-O^-)2 resonance hybrid with 3 equivalent structures showing a central nitrogen bonded to three oxygens (one double bond, two single bonds with negative charges, positive charge on nitrogen).
Nitrite anion:
N(=O)(-O^-) resonance hybrid with 2 equivalent structures showing a central nitrogen with a lone pair bonded to two oxygens (one double bond, one single bond with negative charge).
Nitrite has a shorter N-O bond distance (bond order of 1.5) compared to nitrate (bond order of 1.33).
Zn reduces NO3– ions to NH3 in basic solution forming tetrahydroxozincate(II) ion. Write a balanced equation for the reaction between zinc and ammonia in basic solution.
Model Answer
NO3–(aq) + 4 Zn(s) + 7 OH–(aq) + 6 H2O(l) → 4 [Zn(OH)4]2–(aq) + NH3(g)
When a strong base is gradually added to a solution containing Zn2+ ions a white precipitate of Zn(OH)2 first forms (Ksp = 1.2 ⋅ 10–17 for Zn(OH)2). To a 1.0 dm3 solution of 5.0 ⋅ 10–2 mol Zn2+ ions, 0.10 mol OH– is added. Calculate the pH of this solution.
Model Answer
Zn(OH)2(s) <=> Zn2+(aq) + 2 OH– (aq)
Ks = 1.2 ⋅ 10^-17 = 4x^3
[OH–] = 2.89 ⋅ 10–6 M
Thus, pOH = 5.54 and pH = 8.46
When more base is added to the solution, the white precipitate of Zn(OH)2 dissolves forming the complex ion Zn(OH)4 2–. The formation constant for the complex ion is 4.6 ⋅1017. Calculate the pH of the solution in part 5.3 when 0.10 mol of OH– ion is added (assuming the total volume does not change).
Model Answer
Zn(OH)2(s) <=> Zn2+(aq) + 2 OH– (aq) Ks = 1.2 ⋅ 10^-17
Zn2+(aq) + 4 OH– (aq) <=> [Zn(OH)4]2–(aq) Kf = 4.6 ⋅ 10^17
Zn(OH)2(s) + 2 OH– (aq) <=> [Zn(OH)4]2–(aq) K = Ks * Kf = 5.52
Using equilibrium concentrations: K = [Zn(OH)4^2-] / [OH-]^2 = 5.52
If x mol of Zn(OH)2 dissolves:
[Zn(OH)4^2-] = x = 0.030 M
[OH–] = 2x = 0.060 M
Thus, pOH = 1.22 and pH = 12.78
A mixture containing only NaCl and NaNO3 is to be analyzed for its NaNO3 content. In an experiment, 5.00 g of this mixture is dissolved in water and solution is completed to 100 cm3 by addition of water; then a 10 cm3 aliquot of the resulting solution is treated with Zn under basic conditions. Ammonia produced during the reaction is passed into 50.0 cm3 of 0.150 mol dm-3 HCl solution. The excess HCl requires 32.10 cm3 of NaOH solution (0.100 mol dm-3) for its titration. Find the mass % of NaNO3 in the solid sample.
Model Answer
n(HCl) = 0.0500 dm3 × 0.150 mol dm-3 = 7.50 ⋅ 10−3 mol
n(NaOH) = 0.0321 dm3 × 0.100 mol dm-3 = 3.21 ⋅ 10−3 mol
n(NH3) = (7.50 − 3.21) ⋅ 10−3 mol = 4.29 ⋅ 10−3 mol produced in 10.0 cm3 aliquot
In the total 100.0 cm3 solution, 4.29 ⋅ 10−2 mol NH3 is produced.
n(NaNO3)used = n(NH3)formed = 4.29 ⋅ 10−2 mol
mass of NaNO3 present in the solution = 4.29 ⋅ 10−2 mol × 85.0 g mol−1 = 3.65 g
% NaNO3 in the mixture = (3.65 g / 5.00 g) × 100 = 72.9 % by mass
Both NaCl and NaNO3 are strong electrolytes. Their presence in solution lowers the vapor pressure of the solvent and as a result freezing point is depressed. The freezing point depression depends not only on the number of the solute particles but also on the solvent itself. The freezing point depression constant for water is Kf = 1.86 K kg mol-1. Calculate the freezing point of the solution prepared by dissolving 1.50 g of the mixture described in 5.4 consisting of NaCl and NaNO3 in 100.0 cm3 of water. Density of this solution is ρ = 0.985 g·cm-3.
Model Answer
Since ρ = 0.985 g cm-3, the mass of 100.0 cm3 solution is 98.5 g (which consists of 1.50 g mixture and 97.0 g of water).
n(NaNO3) = 1.29 ⋅ 10–2 mol
n(NaCl) = (1.50 - 1.09) / 58.5 = 6.94 ⋅ 10–3 mol
Both NaCl and NaNO3 dissociate completely to form 2 ions each.
DTf = Kf * m = 1.86 * [2 * 1.29 ⋅ 10^-2 + 2 * 6.94 ⋅ 10^-3] / 0.0970 = 0.761 °C
Freezing point of this solution Tf = −0.761 °C
N2H4 is one of the nitrogen compounds which can be used as a fuel in hydrazine fuel cell. Calculate the standard free energy change for the fuel cell reaction given below.
N2H4(g) + O2(g) → N2(g) + 2 H2O(l)
The standard potentials are given below:
O2(g) + 2 H2O(l) + 4 e– → 4 OH–(aq) E° = 1.23 V
N2(g) + 4 H2O(l) + 4 e– → N2H4(g) + 4 OH–(aq) E° = – 0.33 V
Model Answer
E°cell = E°(cathode) - E°(anode) = (+1.23) – (–0.33) = 1.56 V
∆G° = –n F E°cell = –4 × 1.56 × 96485 = –602 kJ
The free energy change is related to the maximum amount of work wmax that can be obtained from a system during a change at constant temperature and pressure. The relation is given as -∆G = wmax. Calculate the maximum amount of work that can be obtained from the fuel cell which consumes 0.32 g N2H4(g) under standard conditions.
Model Answer
The maximum work that can be obtained from 1 mole of N2H4 = –∆G° = 602 kJ
For 0.32 g (0.010 mol) N2H4, the maximum work will be: 0.010 mol × 602 kJ/mol = 6.0 kJ