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IOM 2018, Теория — Задача 3. Oil refining within the city

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Всички материали от IOM 2018

3rd Olympiad of Metropolises — Chemistry, Theoretical Problems, Official English version · 5 септември 2018 г.

Original cover

Country ______

Participant code ______

3rd Olympiad of Metropolises

Chemistry

Theoretical Problems

September 5, 2018

Moscow, Russia

Chemistry. Theoretical problems. Official English version

Указания

Begin only when the START command is given. You have 4 hours to work on the problems.
Use only the pen and calculator provided.
All results must be written in the appropriate boxes within the text. Anything written elsewhere will not be graded. Use the reverse of the problem pages if you need scratch paper.
Write relevant calculations in the appropriate boxes when necessary. If you provide only correct end results for complicated questions, you will receive no score.
Raise your hand if you have any questions concerning the text of the problems.
Raise your hand if you need a restroom break.
The official English version of this examination is available on request only for clarification.

Recommendations

Read the text carefully. Try to understand properly: a) what is given to you, b) what is required from you.
All the problems contain questions of various complexity including very simple ones. Try to answer as many questions as you can. Try to leave as few empty spaces in the answer sheets as you can.
In the calculations, use the atomic masses from the Periodic table given to you.

Good luck!

Original key cover

3rd Olympiad of Metropolises

Chemistry

Theoretical Problems

Answers, solutions, and grading scheme

September5,2018

Moscow,Russia

Chemistry. Theoretical problems. Official English version

Условие

Problem 3. Oil refining within the city (12 marks)

Question	1	2	3	4	5	6a	6b	7a	7b	Total
Points	2	2	1	1	1	3	2	3	3	18
Result

Moscow is a pretty clean city, because in recent decades many industrial enterprises have been taken out of its borders. However, there is one large enterprise that operates within the city and is able to strongly pollute the surrounding water and atmosphere. This is the Moscow Refinery, which is the largest supplier of fuel and other petroleum products to the capital of Russia.

The plant processes about 10 million tons of crude oil per year. The main products are motor gasoline (2 million tons) and diesel fuel (1.7 million tons). Both types of fuel meet the emission standard "Euro-5". One of the requirements of the standard is a very low sulfur content, not more than 0.001 wt.%.

1 In the "Euro-4" standard, the permissible sulfur content is 5 times higher than that in "Euro-5". If the plant produced fuel according to the previous standard, how many tons of sulfur dioxide would additionally be released into the atmosphere during the combustion of this fuel?

Calculations

$m(\mathrm{SO_2})=$ ______ [2 т.]

To remove sulfur, oil products are subjected to hydrodesulfurization – they are treated by an access of hydrogen at temperature of about $300^\circ\mathrm C$ and pressure of tens of atmospheres. Under such conditions, all organic substances are converted into alkanes and inorganic non-metal hydrides.

2 Write down the reactions of pyridine $\mathrm{C_5H_5N}$ and thiophene $\mathrm{C_4H_4S}$ with hydrogen during hydrotreatment. Use only molecular formulas, structures are not necessary.

Reaction of pyridine: ______

Reaction of thiophene: ______ [2 т.]

3 Hydrogen required for hydrotreatment is not produced separately, because it is formed during one of the chemical stages of oil refining. Write down the equation of any reaction which gives hydrogen in the oil refinery (use the structural formulas for organic substances).

Reaction: ______ [1 т.]

4 During hydrotreatment, sulfur is converted to hydrogen sulfide, the maximum permissible content of which in the urban atmosphere is $0.008\ \mathrm{mg/m^3}$ . How many $\mathrm{H_2S}$ molecules are found in a billion molecules of air at this concentration ( $25^\circ\mathrm C$ , $1\ \mathrm{atm}$ )? Round the result to the integer value.

Calculation:

$N(\mathrm{H_2S})=$ ______ [1 т.]

5 For separation of hydrogen sulfide from the gas mixture after hydrotreatment, the gases are passed through a solution of monoethanolamine (2-aminoethanol), in which hydrocarbons are insoluble. What happens to hydrogen sulfide in this solution? Write down the reaction.

Reaction: ______ [1 т.]

To prevent the release of hydrogen sulfide into the atmosphere, it is converted into elemental sulfur using the so called Claus process involving two stages. First, one third of hydrogen sulfide is burned in a stream of air at a high temperature ( $1000^\circ\mathrm C$ ), and then the gases formed are mixed with the remaining hydrogen sulfide in the presence of a catalyst at $250^\circ\mathrm C$ . As a result, more than 99% of hydrogen sulfide are consumed and the emissions to the atmosphere do not exceed the permissible content.

6a Write the equations of both reactions and the total oxidation equation for hydrogen sulphide. In the equations, specify the aggregate states of the substances.

a) Reactions: ______

The total reaction: ______ [3 т.]

6b Calculate the enthalpy of the total reaction, assuming that it does not depend on temperature.

b) Calculation

$\Delta H=$ ______ $\mathrm{kJ/mol}$ [2 т.]

7a The second stage of the Claus process is reversible. Calculate its equilibrium constant $K_p$ at $250^\circ\mathrm C$ .

a) Calculation

$K_p=$ ______ [3 т.]

7b Assuming that the initial mixture is stoichiometric, determine at what initial pressure of the hydrogen sulfide its degree of conversion at this stage will reach 99%. Assume that entropy and enthalpy of this reaction do not depend on temperature and the reaction takes place at a constant volume and temperature of $250^\circ\mathrm C$ .

b) Calculation

$p_0(\mathrm{H_2S})=$ ______ [3 т.]

Reference data.

Sulfur: melting point $113^\circ\mathrm C$ , boiling point $445^\circ\mathrm C$ .

Substance	$\mathrm{H_2S_{(g)}}$	$\mathrm{S_{(l)}}$	$\mathrm{SO_{2(g)}}$	$\mathrm{H_2O_{(g)}}$
$\Delta_\mathrm fH^\circ_{298}$ , kJ/mol	−21	2	−297	−242
$S^\circ_{298}$ , J/(mol·K)	206	37	248	189

Решение

Official answers and grading scheme

1 Calculations

$m(\text{fuel})=3.7\cdot10^6\ \mathrm t$ ,

$m(\mathrm S,\text{Euro-4})=3.7\cdot10^6\cdot0.005\%/100\%=185\ \mathrm t$

$m(\mathrm S,\text{Euro-5})=3.7\cdot10^6\cdot0.001\%/100\%=37\ \mathrm t$

$\Delta m(\mathrm S)=185-37=148\ \mathrm t$ : 1 pt.

\mathrm{S+O_2=SO_2}

$\Delta m(\mathrm{SO_2})=148/32\cdot64=296\ \mathrm t$ : 1 pt. [2 т.]

2 Reaction of pyridine:

\mathrm{C_5H_5N+5H_2\to C_5H_{12}+NH_3}

1 pt (0.5 pts, if cycloalkane is written).

Reaction of thiophene:

\mathrm{C_4H_4S+4H_2\to C_4H_{10}+H_2S}

1 pt (0.5 pts, if cycloalkane is written). [2 т.]

3 Reaction:

Original structural oil-refinery hydrogen-production example, full alkane chain to benzene and 4H2

1 pt (any reasonable reaction is accepted). [1 т.]

4 Calculation:

\frac{n(\mathrm{H_2S})}{n(\text{air})}=\frac{0.008\cdot10^{-3}/34}{1000\cdot101.3/(8.314\cdot298)}\approx6\cdot10^{-9}

$N(\mathrm{H_2S})=6$ : 1 pt.

(0 pts if the answer is given without any calculations) [1 т.]

5 Reaction:

\mathrm{HOCH_2CH_2NH_2+H_2S\to HOCH_2CH_2NH_3^+HS^-}

1 pt. [1 т.]

6a a) Reactions:

\mathrm{H_2S_{(g)}}+\tfrac32\mathrm{O_{2(g)}}\to\mathrm{SO_{2(g)}+2H_2O_{(g)}}

1 pt.

\mathrm{SO_{2(g)}+2H_2S_{(g)}\to3S_{(l)}+2H_2O_{(g)}}

1 pt.

The total reaction:

\mathrm{H_2S_{(g)}}+\tfrac12\mathrm{O_{2(g)}}\to\mathrm{S_{(l)}+H_2O_{(g)}}

1 pt. [3 т.]

6b b) Calculation

$\Delta H=2+(-242)-(-21)=-219\ \mathrm{kJ/mol}$ : 2 pts.

(1 pt if solid sulfur is used in the calculation) [2 т.]

7a a) Calculation

\mathrm{SO_{2(g)}+2H_2S_{(g)}\to3S_{(l)}+2H_2O_{(g)}}

$\Delta H^\circ=3\cdot2+2\cdot(-242)-(-297)-2\cdot(-21)=-139\ \mathrm{kJ/mol}$ : 0.5 pts.

(0 pts if solid sulfur is used)

$\Delta S^\circ=3\cdot37+2\cdot189-248-2\cdot206=-171\ \mathrm{J/(mol\mathord\cdot K)}$ : 0.5 pts.

$\Delta G^\circ_{523}=\Delta H^\circ-T\Delta S^\circ=-139-523\cdot(-171\cdot10^{-3})=-49.57\ \mathrm{kJ/mol}$ : 1 pt.

$K_p=\exp(-\Delta G^\circ/(RT))=\exp(49570/(8.314\cdot523))=8.93\cdot10^4$ : 1 pt. [3 т.]

7b b) Calculation

Denote $p_0(\mathrm{H_2S})=p$ , then $p_0(\mathrm{SO_2})=p/2$ .

Equilibrium pressures: $p(\mathrm{H_2S})=0.01p$ , $p(\mathrm{SO_2})=0.005p$ , $p(\mathrm{H_2O})=0.99p$

K_p=\frac{p(\mathrm{H_2O})^2}{p(\mathrm{SO_2})p(\mathrm{H_2S})^2}=\frac{(0.99p)^2}{0.005p(0.01p)^2}=8.93\cdot10^4

$p_0(\mathrm{H_2S})=22.0\ \mathrm{bar}$ : 3 pts. [3 т.]

Бележка от източника

Each native problem heading says 12 marks, while its printed score grid totals 65, 30, 18, 50 or 30. Problem1 request3 is 10 points in the grid but5 in the key; its printed permanganate equation and sodium-containing answer to request2 remain literal. Problem3 first Claus-stage equation prints2H2O. Problem4 requests Fischer projections A–C and Haworth D–G, while its key shows the opposite projection types; request2 awards2points for each of three structures despite its4-point grid. Problem5 methane price is0.75 $/kg in the question and0.70$ /kg in the repeated key request and calculations. These original inconsistencies are retained. Molecular drawings, reaction schemes, reference tables and answer grids reproduce the native originals. Empty calculation spaces retain their printed labels and blank response fields.

Оригинал в Архива: Problems.pdf · официални решения: Solutions.pdf