Laumontite is a natural zeolite, a hydrated calcium... Some samples of laumontite are orange. This c — Physical Chemistry — Kinetics Chemistry Question
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Laumontite is a natural zeolite, a hydrated calcium...
Some samples of laumontite are orange. This coloration is caused by the presence of an impurity, an element E that partly substitutes calcium, yielding the compound of formula:
(EiCa(1‒i)O)x(A)y(B)z ∙ y H2O
The dissolution of a 0.500 g sample in nitric acid led to the formation of the same precipitate as before. The filtrate was separated. When a few drops of NH4SCN are added...
1,2-benzanthracene 1.00 colorless colorless
Zeolites are widely used as materials in heterogeneous catalysis because of their large specific surface area, their structural framework and their large number of acid sites. The structured porous system of zeolites provides a molecular sieve effect. This effect...
Data at T = 298 K:
Eo(E3+/E2+) = 0.53 V /SCE
Eo(NO3‒/NO2) = 0.56 V /SCE
Eo(Zn2+/Zn) = ‒1.00 V /SCE
Eo(Ce4+/Ce3+) = 1.09 V /SCE
E(SCE) = 0.24 V
Eo(Ox/Red) (V /SCE) = Eo(Ox/Red)(V /SHE) ‒ E(SCE) (V)
Model Answer
ρ = m / V = m / (abc sinβ) and n = m / M = Z / NA i.e. m = M Z / NA then
M = (ρ NA × abc sinβ) / Z = 470 g mol‒1
Model Answer
From the qualitative data (formation of a FeSCN2+ red complex) we can deduce that E is Fe (Fe3+ when oxidized and Fe2+ in the crystal).
Model Answer
Fe3+(aq) + SCN‒(aq) = Fe(SCN2+(aq)
Fe3+(aq) + 3 NH3(aq) + 3 H2O(l) = Fe(OH)3(s) + 3 NH4 +(aq)
2 Fe(OH)3(s) + 6 H+(aq) + Zn(s) = 2 Fe2+(aq) + Zn2+(aq) + 6 H2O(l)
Fe2+(aq) + Ce4+(aq) = Ce3+(aq) + Fe3+(aq)
Determine the amount of impurity E (mol. % compared to Ca).
Model Answer
Using the titration reaction Fe2+(aq) + Ce4+(aq) = Ce3+(aq) + Fe3+(aq) we find that:
n(Fe3+) = 5.15·10‒3 × 2.00·10‒3 = 1.03·10‒5 mol in the titrated solution and thus:
n(Fe3+) = 1.03·10‒5 × 100.0/20.0 = 5.15·10‒5 mol in the initial solution which corresponds to n(Fe2+) = 5.15·10‒5 mol in 0.500 g of the solid. In 0.500 g of the pure crystal, n(Ca2+) = 0.500 / 471 = 1.06·10‒3 mol, so the molar percentage of the impurity compared to calcium is 5.15·10‒5 / 1.06·10‒3 = 4.86%.
Model Answer
Ee.p. = E°(Fe3+/Fe2+) - (RT/F)ln([Fe2+]/[Fe3+]) = E°(Ce4+/Ce3+) - (RT/F)ln([Ce3+]/[Ce4+])
Ee.p. = 1/2 (E°(Fe3+/Fe2+) - (RT/F)ln([Fe2+]/[Fe3+]) + E°(Ce4+/Ce3+) - (RT/F)ln([Ce3+]/[Ce4+]))
The combination of these two expressions leads to:
Note: such a formula WILL NOT be expected to be known by heart for the competition exam but the simple use of Nernst equation as it is demonstrated here could be required.
Model Answer
According to the value of the potential at the equivalence point (0.81 V /SCE), we can use the following indicators that exhibit the standard potential the closest to this value: 5,6- dimethy-l,10-phenanthroline and 4-ethoxychrysoidine hydrochloride.
Draw the two main products F and G.
This reaction can also be catalyzed by laumontite. Determine which product will mainly be formed in the pore system of the mineral.
Model Answer
According to the values of the diameters given in the text of the problem, the product F seems to be smaller than G: F is then the main one that can be synthesized in laumontite.
F