Compound A is a stable salt of metal H. It contains 11.97 % N, 3.45 % H and 41.03 % O (mass fraction — Physical Chemistry — Thermodynamics Chemistry Question
THEORETICAL PROBLEM 2
Compound A is a stable salt of metal H. It contains 11.97 % N, 3.45 % H and 41.03 % O (mass fractions), besides the metal. The following chart describes some reactions starting from A and H (∆ signifies heating). Above the arrows the necessary reactants are displayed. All substances tagged with a letter contain the metal, but none of the by-products do. (When a substance is labeled as dissolved in water, then it is ionic and you have to show only the ion containing the metal.)
Identify the substances A – K and write down all the equations 1 – 14.
Model Answer
The composition of A: n(N) : n (H) : n (O) = (11.97 / Ar(N)) : (3.45 / Ar(H)) : (41.03 / Ar(O)) = 1 : 4 : 3.
From these molar ratios it is clear that A contains ammonium ions.
The amount of the metal H in 100 g A is 43.55 g. Supposing that one mol of A contains one mol of both H and nitrogen, the molar mass of H is 43.55 g / 0.8550 mol = 50.94 g mol-1. That is vanadium. From here, the compounds (species) A – K are the following:
A: NH4VO3
B: V2O5
C: VO2+
D: VO3–
E: V2+
F: VO2+
G: V3+
H: V
I: VCl4
J: VCl3
K: VCl2
The reactions are as follows:
2 NH4VO3(s) → 2 NH3(g) + V2O5(s) + H2O(g)
V2O5(s) + 2 H3O+(aq) → 2 VO2+(aq) + 3 H2O(l)
2 VO2+(aq) + 3 Zn(s) + 8 H3O+(aq) → 2 V2+(aq) + 3 Zn2+(aq) + 12 H2O(l)
NH4VO3(s) → NH4+(aq) + VO3–(aq)
VO3–(aq) + 2 H3O+(aq) → VO2+(aq) + 3 H2O(l)
2 VO3–(aq) + SO2(g) + 4 H3O+(aq) → 2 VO2+(aq) + SO42–(aq) + 6 H2O(l)
VO2+(aq) + V2+(aq) + 2 H3O+(aq) → 2 V3+(aq) + 3 H2O(l)
V(s) + 2 Cl2(g) → VCl4(l)
2 VCl4(l) → 2 VCl3(s) + Cl2(g)
2 VCl3(s) + H2(g) → 2 VCl2(s) + 2 HCl(g)
VCl4(l) + 3 H2O(l) → VO2+(aq) + 4 Cl–(aq) + 2 H3O+(aq)
2 VCl3(s) → 2 V3+(aq) + 6 Cl–(aq)
VCl2(s) → V2+(aq) + 2 Cl–(aq)
2 VCl3(s) → VCl2(s) + VCl4(l)
Select the redox processes from the reactions.
Model Answer
3, 6, 7, 8, 9, 10, 14
Select those compounds from A – K that are not expected to have unpaired electrons.
Model Answer
Since vanad atom has a 4s23d3 configuration, the loss of 2, 3 or 4 electrons leads to species having unpaired d electrons. Only compounds in the oxidation state +V do not have these: A, B, C, D.
On the basis of the above chart propose a reaction to obtain G starting from F, but without using E.
Model Answer
The reduction of VO2+ ions with zinc in acidic medium can be stopped at V3+:
VO2+(aq) + Zn(s) + 4 H3O+(aq) → 2 V3+(aq) + Zn2+(aq) + 6 H2O(l)
Compound B is industrially very important. Show a reaction where its presence is indispensable. What role does it play?
Model Answer
Production of SO3, then sulfuric acid: 2 SO2 + O2 = 2 SO3. V2O5 is a catalyst.
What percentage of product I contains 35Cl if chlorine gas containing 99 % 37Cl and 1 % 35Cl is used in reaction 8?
Model Answer
The probability of no 35Cl atoms being in a VCl4 molecule is 0.994 = 96.06 %. Thus the probability of 35Cl being present, i.e. the mole fraction of 35Cl containing products is 3.94 %.
What percentage of J produced from this sample of I contains 35Cl?
Model Answer
The same reasoning gives the probability of a 35Cl atom being present in a VCl3 molecule as 2.97 %.