How many SCSA Chemistry questions cover Chemical equilibrium systems?

AusGrader has 157 SCSA Chemistry questions on Chemical equilibrium systems, all with instant AI grading and detailed marking feedback.

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

Q15

2024

SCSA

1 mark

Q15

1 mark

The precipitation reaction between solutions of lead(II) nitrate and potassium iodide is very rapid at room temperature. This can be explained by the

A

low activation energy of the reaction.

B

reaction being very exothermic.

C

large number of types of particles involved in the reaction.

D

need for a flame to cause the reaction to occur.

Q3

2022

SCSA

1 mark

Q3

1 mark

Which of the following may occur in a closed chemical system?

A

Energy and matter are exchanged with the surroundings.

B

Energy, but not matter, is exchanged with the surroundings.

C

Matter, but not energy, is exchanged with the surroundings.

D

Neither energy nor matter are exchanged with the surroundings.

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

Q34

2023

SCSA

9 marks

Q34

Sorbic acid is a monoprotic weak acid that occurs widely in nature and is used as a food preservative due to its antimicrobial properties. The ionisation of sorbic acid in water to the sorbate ion and hydronium ion is shown in the equation below:

\text{CH}_3\text{(CH)}_4\text{COOH(aq)} + \text{H}_2\text{O}(\ell) \rightleftharpoons \text{CH}_3\text{(CH)}_4\text{COO}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}

Q34a

2 marks

Write the equilibrium constant K expression for the ionisation of sorbic acid in water.

\text{K} = \frac{[\text{CH}_3(\text{CH})_4\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3(\text{CH})_4\text{COOH}]}

Marking Criteria

Descriptor	Marks
K = [CH3(CH)4COO-][H3O+] / [CH3(CH)4COOH]	2
Partially correct equilibrium constant expression	1
None of the above	0

Q34b

4 marks

Under certain conditions, a $0.250\text{ mol L}^{-1}$ aqueous solution of sorbic acid has a pH of 2.23. Calculate the concentration of $\text{H}_3\text{O}^+$ to determine the percentage yield of the sorbate ion at equilibrium in $1.00\text{ L}$ of the solution.

$[\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-2.23} = 5.89 \times 10^{-3} \text{ mol L}^{-1}$

For 1.00 L solution, $n(\text{H}_3\text{O}^+) = n(\text{CH}_3(\text{CH})_4\text{COO}^-) = 5.89 \times 10^{-3} \text{ mol}$
For 1.00 L solution, $n(\text{CH}_3(\text{CH})_4\text{COOH}) = 0.250 \text{ mol}$

\begin{align*} \text{\% yield sorbate ion} &= \frac{n(\mathrm{CH_3CH(OH)COO^-})}{n(\mathrm{CH_3CH(OH)COOH})}\times 100\\ &= \frac{5.89\times 10^{-3}}{0.250}\times 100\\ &= 2.36\% \end{align*}

Marking Criteria

Descriptor	Marks
$[\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-2.23} = 5.89 \times 10^{-3} \text{ mol L}^{-1}$	1
For 1.00 L solution, $n(\text{H}_3\text{O}^+) = n(\text{CH}_3(\text{CH})_4\text{COO}^-) = 5.89 \times 10^{-3} \text{ mol}$	1
For 1.00 L solution, $n(\text{CH}_3(\text{CH})_4\text{COOH}) = 0.250 \text{ mol}$	1
Percentage yield is 2.36%	1

Q34c

3 marks

Explain the classification of sorbic acid as a weak acid with reference to both your answer to part (b) above and its acidity constant value $\text{K}_\text{a} = 1.73 \times 10^{-5}$ (20 °C).

Weak acids undergo partial or incomplete ionisation in water. The answer to part (b) is numerically small, indicating that only a small percentage of sorbic acid in solution is ionised. The value of $K_a$ is less than one, which indicates a greater proportion of reactants compared to products.

Marking Criteria

Descriptor	Marks
Recognises that weak acids undergo partial/incomplete ionisation in water	1
Explains that the answer to part (b) is numerically small, indicating that only a small percentage of sorbic acid in solution is ionised	1
Recognises that the value of $K_a$ is less than one, which indicates a greater proportion of reactants compared to products	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 equilibrium systems

Rates

Equilibrium

Economics

Compromise

Catalyst

Temperature

Pressure

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