How many SCSA Chemistry questions cover Chemical synthesis?

AusGrader has 99 SCSA Chemistry questions on Chemical synthesis, all with instant AI grading and detailed marking feedback.

SCSA (WACE) Chemistry — Chemical synthesis Questions & Answers | AusGrader

Q23

2024

SCSA

1 mark

Q23

1 mark

Which of the following is not a use of polytetrafluoroethene?

A

non-stick frypans

B

electrical wires

C

coatings for tanks

D

lubricants

Q4

2021

SCSA

1 mark

Q4

1 mark

Which of the following characteristics influence how a particular polymer might be used?

(i) The amount of cross-linking between the hydrogen atoms in the polymer.
(ii) The length of the carbon chains in the polymer.
(iii) The functional groups present in the monomer used to synthesise the polymer.
(iv) The melting point of the polymer.

A

ii, iii and iv only

B

i and ii only

C

ii and iii only

D

i, ii and iv only

Q38

2021

SCSA

12 marks

Q38

12 marks

Sulfuric acid is manufactured by the Contact process, the steps of which are outlined below.

Step One: Molten sulfur is burned in air at approximately 1000 °C:

S(l) + O_2(g) \rightarrow SO_2(g) + 297 kJ

Step Two: The resulting sulfur dioxide is converted to sulfur trioxide as shown in the following equilibrium reaction. It is conducted at a temperature of about 450 °C with a $V_2O_5$ catalyst at a pressure of between 100 and 200 kPa:

2 SO_2(g) + O_2(g) \rightleftharpoons 2 SO_3(g) + 198 kJ

Step Three: The resulting sulfur trioxide is absorbed into sulfuric acid, producing oleum ( $H_2S_2O_7$ ). Water is added to the oleum, producing 18 mol L $^{-1}$ sulfuric acid:

SO_3(g) + H_2SO_4(l) \rightarrow H_2S_2O_7(l)

H_2S_2O_7(l) + H_2O(l) \rightarrow 2 H_2SO_4(aq)

Use your understanding of collision theory and chemical equilibrium to discuss the reaction conditions for Steps 1 and 2 of the Contact process, given that the aim is to produce the greatest yield in the shortest time. In your discussion, also address economic concerns where appropriate.

High temperature increases the average kinetic energy of the particles, which means that the particles collide more frequently. Also, more of these collisions will have energy higher than the activation energy, which means a greater proportion of collisions are successful, and the reaction rate increases. The vanadium catalyst increases the rate of the forward reaction, and also the rate of the reverse reaction to an equal extent, as it provides an alternative pathway with a lower activation energy. Therefore, a greater proportion of the particles will have sufficient energy to react when they collide. High pressure or concentration has more particles per unit volume and so there is a higher frequency of collisions, and the reaction rate increases. As Step 1 is a combustion reaction, it essentially goes to completion at the high temperature and does not require a catalyst or high pressure. For Step 2, high temperature, high pressure and a catalyst would favour a high rate.

For equilibrium, which is only considered for Step 2, high temperature favours the reverse reaction because it is endothermic, and this decreases the SO $_3$ (g) yield, which is not desired. A low temperature decreases the rate of reaction, which is also not desired. A high pressure favours the forward reaction because there are a greater number of moles of gas reactants, increasing the SO $_3$ (g) yield which is desired.

Economically, high pressures are costly and dangerous.

Therefore, for Step 2, a compromise is required between the high temperature for rate and the low temperature for yield. A compromise is also required between the cost of higher pressures and the pressure that allows a satisfactory yield and rate.

Marking Criteria

Rates

Descriptor	Marks
1 mark each for any of the following, up to a maximum of 6 marks: Explains that high temperature increases the average kinetic energy of the particles, which means that the particles collide more frequently; Explains that more of these collisions will have energy higher than the activation energy, which means a greater proportion of collisions are successful, and the reaction rate increases; Describes that the vanadium catalyst increases the rate of the forward reaction (and also the rate of the reverse reaction to an equal extent) as it provides an alternative pathway with a lower activation energy; States that a greater proportion of the particles will have sufficient energy to react when they collide; Explains that high pressure (concentration) has more particles per unit volume and so there is a higher frequency of collisions, and the reaction rate increases; Identifies that as Step 1 is a combustion reaction, it essentially goes to completion at the high temperature (and does not require a catalyst or high pressure); States that for Step 2, high temperature, high pressure and catalyst would favour high rate.	6

Descriptor

Marks

1 mark each for any of the following, up to a maximum of 6 marks: Explains that high temperature increases the average kinetic energy of the particles, which means that the particles collide more frequently; Explains that more of these collisions will have energy higher than the activation energy, which means a greater proportion of collisions are successful, and the reaction rate increases; Describes that the vanadium catalyst increases the rate of the forward reaction (and also the rate of the reverse reaction to an equal extent) as it provides an alternative pathway with a lower activation energy; States that a greater proportion of the particles will have sufficient energy to react when they collide; Explains that high pressure (concentration) has more particles per unit volume and so there is a higher frequency of collisions, and the reaction rate increases; Identifies that as Step 1 is a combustion reaction, it essentially goes to completion at the high temperature (and does not require a catalyst or high pressure); States that for Step 2, high temperature, high pressure and catalyst would favour high rate.

6

Equilibrium

Descriptor	Marks
1 mark each for any of the following (only considered for Step 2), up to a maximum of 3 marks: Explains that high temperature favours the reverse reaction because it is endothermic, and this decreases the SO3(g) yield; States that a low temperature decreases the rate of reaction; Explains that a high pressure favours the forward reaction because there are a greater number of moles of gas reactants, increasing the SO3(g) yield.	3

Descriptor

Marks

1 mark each for any of the following (only considered for Step 2), up to a maximum of 3 marks: Explains that high temperature favours the reverse reaction because it is endothermic, and this decreases the SO3(g) yield; States that a low temperature decreases the rate of reaction; Explains that a high pressure favours the forward reaction because there are a greater number of moles of gas reactants, increasing the SO3(g) yield.

3

Economics

Descriptor	Marks
States that high pressures are costly (and dangerous).	1

Compromise

Descriptor	Marks
1 mark each for any of the following, up to a maximum of 2 marks: Identifies that for Step 2, a compromise is required between the high temperature for rate and the low temperature for yield; Identifies that a compromise is also required between the cost of higher pressures and the pressure that allows a satisfactory yield and rate.	2

Q39

2020

SCSA

12 marks

Q39

Fluorescent lights are glass tubes which are coated on the inside with rare earth metal phosphates (such as cerium, lanthanum and terbium phosphates) that provide light. Cerium, lanthanum and terbium are expensive, so are recovered once the fluorescent light is no longer functional.

The key steps in one method proposed for recovery of these rare earth metals are summarised below:

Step 1: Physical separation of the rare earth metal phosphates from the glass and any metallic components. This gives an impure powder consisting of cerium, lanthanum and terbium phosphates.
Step 2: Add excess solid sodium carbonate to the powder and heat, completely converting each rare earth metal phosphate to its corresponding oxide, as shown by the following balanced equations:
$2 \text{ LaPO}_4\text{(s)} + 3 \text{ Na}_2\text{CO}_3\text{(s)} \rightarrow \text{La}_2\text{O}_3\text{(s)} + 2 \text{ Na}_3\text{PO}_4\text{(s)} + 3 \text{ CO}_2\text{(g)}$
$4 \text{ CePO}_4\text{(s)} + 6 \text{ Na}_2\text{CO}_3\text{(s)} + \text{O}_2\text{(g)} \rightarrow 4 \text{ CeO}_2\text{(s)} + 4 \text{ Na}_3\text{PO}_4\text{(s)} + 6 \text{ CO}_2\text{(g)}$
$2 \text{ TbPO}_4\text{(s)} + 3 \text{ Na}_2\text{CO}_3\text{(s)} \rightarrow \text{Tb}_2\text{O}_3\text{(s)} + 2 \text{ Na}_3\text{PO}_4\text{(s)} + 3 \text{ CO}_2\text{(g)}$
Step 3: Wash the product from Step 2 with water.
Step 4: Add hydrochloric acid to the washed product from Step 3 to leach (dissolve) only the rare earth metal oxides.
Step 5: Use solvent extraction to separate the different rare earth metals from each other and create separate solutions of each of them.
Step 6: Add oxalic acid to the separated solutions to precipitate the rare earth metal ions as oxalate salts.
Step 7: Heat the oxalate salts to recover the rare earth metals as pure oxides, namely $\text{La}_2\text{O}_3$ , $\text{Tb}_4\text{O}_7$ and $\text{CeO}_2$ .

A chemist used the above procedure to determine the percentage by mass of lanthanum, terbium and cerium in some fluorescent lights and, after completing Step 1, had recovered 1.20 kg of the coating chemicals.

Q39b

The mass of the solid sent from Step 3 to Step 4 was 1.16 kg. This solid was leached with 6.00 mol L $^{-1}$ HCl at a solid to liquid ratio of 150 g per litre. Analysis of the solution at the end of leaching showed that it contained lanthanum, terbium and cerium, with its lanthanum concentration being $8.65 \times 10^{-3} \text{ mol L}^{-1}$ .

Q39a

3 marks

At the completion of Step 2, the mass of the mixture had decreased by 11.3 g. Calculate the mass of sodium carbonate that reacted with the rare earth metal phosphates.

n(CO₂) = 11.3/44.01 = 0.257 mol

n(CO₂) = n(C) = n(Na₂CO₃) = 0.257 mol

m(Na₂CO₃) = 0.257 × 105.99 = 27.2 g

Marking Criteria

Descriptor	Marks
n(CO₂) = 11.3/44.01 = 0.257 mol	1
n(CO₂) = n(C) = n(Na₂CO₃) = 0.257 mol	1
m(Na₂CO₃) = 0.257 × 105.99 = 27.2 g	1

Q39b

5 marks

Calculate the percentage, by mass, of lanthanum in the fluorescent light coating chemical, given that the leaching efficiency for lanthanum was 86%.

Note that the balanced equation for the leaching of lanthanum with hydrochloric acid is:
$\text{La}_2\text{O}_3\text{(s)} + 6 \text{ HCl(aq)} \rightarrow 2 \text{ LaCl}_3\text{(aq)} + 3 \text{ H}_2\text{O(l)}$

Volume of HCl used: 1160/150 = 7.73 L

n(La) = cV = 8.65 × 10⁻³ × 7.73 = 0.0669 mol La in solution

Taking into account the leaching efficiency,
n(La in the solid that was leached) = 0.0669/0.86 = 0.0778

m(La in the solid that was leached) = 0.0778 × 138.9 = 10.8 g

% La in the coating chemical = (10.8/1200) × 100 = 0.900%

Marking Criteria

Descriptor	Marks
Volume of HCl used: 1160/150 = 7.73 L	1
n(La) = cV = 8.65 × 10⁻³ × 7.73 = 0.0669 mol La in solution	1
Taking into account the leaching efficiency, n(La in the solid that was leached) = 0.0669/0.86 = 0.0778	1
m(La in the solid that was leached) = 0.0778 × 138.9 = 10.8 g	1
% La in the coating chemical = (10.8/1200) × 100 = 0.900%	1

Q39c

4 marks

Analysis of the cerium-containing solution produced in Step 5 showed that its cerium concentration was 0.146 mol L $^{-1}$ . This solution, which had a volume of 424 mL, was added to 110 mL of aqueous 1.15 mol L $^{-1}$ oxalic acid during Step 6, resulting in the precipitation of cerium oxalate, $\text{Ce(C}_2\text{O}_4\text{)}_2$ . The balanced equation for this reaction is:

$\text{CeCl}_4\text{(aq)} + 2 \text{ H}_2\text{C}_2\text{O}_4\text{(aq)} \rightarrow \text{Ce(C}_2\text{O}_4\text{)}_2\text{(s)} + 4 \text{ HCl(aq)}$

Did the chemist add enough oxalic acid solution to precipitate all of the cerium? Use calculations to support your answer.

n(cerium) = 0.146 × 0.424 = 0.0619 mol

n(oxalic acid needed to react with cerium) = 2 × 0.0619 = 0.124 mol

n(oxalic acid available) = 0.110 × 1.15 = 0.127 mol

comparison of the moles of oxalic acid shows that enough oxalic acid was added

Marking Criteria

Descriptor	Marks
n(cerium) = 0.146 × 0.424 = 0.0619 mol	1
n(oxalic acid needed to react with cerium) = 2 × 0.0619 = 0.124 mol	1
n(oxalic acid available) = 0.110 × 1.15 = 0.127 mol	1
comparison of the moles of oxalic acid shows that enough oxalic acid was added	1

Q35

2023

SCSA

14 marks

Q35

The Ostwald process is used in the conversion of ammonia to nitric acid according to the equations below.

Equation 1:
$4\text{ NH}_3(g) + 5\text{ O}_2(g) \rightarrow 4\text{ NO}(g) + 6\text{ H}_2O(g) \quad \Delta H = -905.2\text{ kJ mol}^{-1}$

Equation 2:
$2\text{ NO}(g) + \text{O}_2(g) \rightleftharpoons 2\text{ NO}_2(g) \quad \Delta H = -114.0\text{ kJ mol}^{-1}$

Equation 3:
$3\text{ NO}_2(g) + \text{H}_2O(\ell) \rightarrow 2\text{ HNO}_3(aq) + \text{NO}(g) \quad \Delta H = -117.0\text{ kJ mol}^{-1}$

Q35a

8 marks

The reaction in Equation 1 is carried out with a platinum-rhodium catalyst at approximately 850.0 °C and 1500 kPa. Using collision theory, account for these conditions.

The platinum-rhodium catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy. This means a greater proportion of collisions are successful.

The high temperature of about 850.0 °C increases the kinetic energy of the reacting particles. The particles move faster, so collisions occur more frequently, and a greater proportion of collisions have energy greater than the activation energy. Therefore, more collisions are successful.

The high pressure of 1500 kPa forces the gas particles closer together. This reduces the space between particles and increases the frequency of collisions. As a result, the number of successful collisions per second increases, so the reaction rate increases.

Marking Criteria

Catalyst

Descriptor	Marks
Recognises that rate is increased as the platinum-rhodium catalyst provides an alternate reaction pathway with a lower activation energy	1
Recognises that there are an increased proportion of reacting particles colliding with energy greater than the activation energy (resulting in an increased frequency of successful collisions)	1

Temperature

Descriptor	Marks
Recognises that increased temperature increases the average kinetic energy of particles and they collide more frequently	1
Recognises that the increased proportion of collisions will have energy higher than the activation energy (which has been lowered due to the catalyst)	1
Recognises that a greater proportion of collisions are therefore successful (leading to an increased rate of reaction)	1

Pressure

Descriptor	Marks
Recognises that increased pressure reduces the space between reacting particles	1
Recognises that at high pressure particles collide more frequently	1
Recognises that this leads to a higher frequency of successful collisions	1

Q35b

6 marks

A nitric acid plant requires a production of 1095 tonnes of nitric acid by means of the Ostwald process each day. If the conversion of ammonia to nitric acid is 77.65% efficient, calculate the volume of ammonia at standard temperature and pressure (STP) that must be fed into the process each day. Give your answer to the appropriate number of significant figures.

$m(\text{HNO}_3)\text{required} = 1.095 \times 10^9 \text{ g}$

$n(\text{HNO}_3)\text{required} = \frac{1.095 \times 10^9}{63.018} = 1.738 \times 10^7 \text{ mol}$

$n(\text{NH}_3)\text{required } 100\% = \frac{12}{8} \times 1.738 \times 10^7 = 2.606 \times 10^7 \text{ mol}$

$n(\text{NH}_3)\text{required } 77.65\% = \frac{100}{77.65} \times 2.606 \times 10^7 = 3.356 \times 10^7 \text{ mol}$

$v(\text{NH}_3)\text{STP} = 3.356 \times 10^7 \times 22.71 = 7.621 \times 10^8 \text{ L}$

Marking Criteria

Descriptor	Marks
$m(\text{HNO}_3)\text{required} = 1.095 \times 10^9 \text{ g}$	1
$n(\text{HNO}_3)\text{required} = \frac{1.095 \times 10^9}{63.018} = 1.738 \times 10^7 \text{ mol}$	1
$n(\text{NH}_3)\text{required } 100\% = \frac{12}{8} \times 1.738 \times 10^7 = 2.606 \times 10^7 \text{ mol}$	1
$n(\text{NH}_3)\text{required } 77.65\% = \frac{100}{77.65} \times 2.606 \times 10^7 = 3.356 \times 10^7 \text{ mol}$	1
$v(\text{NH}_3)\text{STP} = 3.356 \times 10^7 \times 22.71 = 7.621 \times 10^8 \text{ L}$	1
Answer to 4 significant figures	1

SCSA Chemistry Chemical synthesis

Rates

Equilibrium

Economics

Compromise

Catalyst

Temperature

Pressure

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