VCAA Chemistry How can the rate and yield of chemical reactions be optimised?

15 sample questions with marking guides and sample answers

Q14
2025
VCAA
1 mark
Q14
1 mark

The reaction to produce methanal, CH2OCH_2O, is shown below.

2CH3OH(g)+O2(g)→catalyst2CH2O(g)+2H2O(g)2CH_3OH(g) + O_2(g) \xrightarrow{\text{catalyst}} 2CH_2O(g) + 2H_2O(g)

The primary role of the catalyst in the production of CH2OCH_2O is to increase the

A

speed of all particles.

B

number of collisions per unit time.

C

proportion of particles that react.

D

overall kinetic energy of the system.

Reveal Answer
A

speed of all particles.

A catalyst does not change the speed of the particles. Increasing the temperature of the system would increase particle speed.

B

number of collisions per unit time.

A catalyst does not increase the frequency of collisions. Factors like increased concentration, pressure, or temperature would increase the number of collisions per unit time.

C

proportion of particles that react.

Correct Answer

A catalyst provides an alternative reaction pathway with a lower activation energy, meaning a greater proportion of particles have sufficient energy to react upon collision.

D

overall kinetic energy of the system.

The overall kinetic energy of the system is determined by its temperature, not by the presence of a catalyst.

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:

CH3(CH)4COOH(aq)+H2O(ℓ)⇌CH3(CH)4COO−(aq)+H3O+(aq)\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.

Reveal Answer

K=[CH3(CH)4COO−][H3O+][CH3(CH)4COOH]\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
DescriptorMarks

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 mol L−10.250\text{ mol L}^{-1} aqueous solution of sorbic acid has a pH of 2.23. Calculate the concentration of H3O+\text{H}_3\text{O}^+ to determine the percentage yield of the sorbate ion at equilibrium in 1.00 L1.00\text{ L} of the solution.

Reveal Answer

[H3O+]=10−pH=10−2.23=5.89×10−3 mol L−1[\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(H3O+)=n(CH3(CH)4COO−)=5.89×10−3 moln(\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(CH3(CH)4COOH)=0.250 moln(\text{CH}_3(\text{CH})_4\text{COOH}) = 0.250 \text{ mol}

% yield sorbate ion=n(CH3CH(OH)COO−)n(CH3CH(OH)COOH)×100=5.89×10−30.250×100=2.36%\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
DescriptorMarks

[H3O+]=10−pH=10−2.23=5.89×10−3 mol L−1[\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(H3O+)=n(CH3(CH)4COO−)=5.89×10−3 moln(\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(CH3(CH)4COOH)=0.250 moln(\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 Ka=1.73×10−5\text{K}_\text{a} = 1.73 \times 10^{-5} (20 °C).

Reveal Answer

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 KaK_a is less than one, which indicates a greater proportion of reactants compared to products.

Marking Criteria
DescriptorMarks

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 KaK_a is less than one, which indicates a greater proportion of reactants compared to products

1
Q5
2022
QCAA
Paper 2
8 marks
Q5

One step in the electrolytic refining of copper uses impure copper anodes and high purity copper cathodes in an electrolyte solution of copper(II) sulfate.

Q5a
4 marks

Predict whether the concentration of the copper(II) sulfate solution will change during the purification process. Provide appropriate half-equations to support your reasoning.

Reveal Answer

Copper ion is reduced and Cu is plated onto the cathode:
Cu2+(aq)+2e−→Cu(s)Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)

Copper anode is oxidised to Cu2+(aq)Cu^{2+}(aq) and is released into solution:
Cu(s)→Cu2+(aq)+2e−Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-

Therefore, for every copper ion that is reduced at the cathode, in principle, another one is oxidised at the anode.
Therefore, the concentration of the copper(II) sulfate solution should stay the same.

Marking Criteria
DescriptorMarks

Identifies copper ions are reduced to Cu metal at the cathode and reduction half-equation is Cu2+(aq)+2e−→Cu(s)Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)

1

Identifies copper metal is oxidised to Cu ions at the anode and oxidation half-equation is Cu(s)→Cu2+(aq)+2e−Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-

1

Predicts no change in concentration of copper(II) sulfate solution

1

Identifies that copper ions are reduced to copper and copper is oxidised to copper ions at the same rate

1
Q5b
4 marks

If the copper anodes contain silver and zinc impurities, determine whether either metal could be produced as a by-product of the electrolytic refining of copper. Explain your reasoning.

Reveal Answer

Silver is below copper in the reactivity series and therefore doesn't go into solution, as ions are not oxidised and could be found in the sludge.
Zinc impurities are above copper in the electrochemical series and will form ions at the anode and go into solution. However, they won't get discharged at the cathode, provided their concentration doesn't get too high.

Marking Criteria
DescriptorMarks

Identifies Ag is less reactive than Cu and Zn is more reactive than Cu

1

Deduces Ag metal is not oxidised (or reduced) and remains as metal

1

Deduces Zn metal is oxidised to form ions and found in the solution

1

Explains that Zn2+Zn^{2+} ions remain in solution at low concentration but are reduced to Zn metal at the cathode if concentration becomes too high

1
Q19
2021
QCAA
Paper 1
1 mark
Q19
1 mark

To form ethanol biofuel in the fermentation of glucose, a catalyst is used because

A

less energy is required and the rate of reaction is increased.

B

less energy is required and the rate of reaction is decreased.

C

more energy is required and the rate of reaction is increased.

D

more energy is required and the rate of reaction is decreased.

Reveal Answer
A

less energy is required and the rate of reaction is increased.

Correct Answer

Catalysts (such as the enzymes in yeast) provide an alternative reaction pathway with a lower activation energy, which allows the reaction to proceed faster.

B

less energy is required and the rate of reaction is decreased.

While catalysts do lower the activation energy required, they function to speed up the reaction, not slow it down.

C

more energy is required and the rate of reaction is increased.

Catalysts lower the activation energy required for the reaction to occur, rather than requiring more energy.

D

more energy is required and the rate of reaction is decreased.

This is incorrect because catalysts lower the activation energy and increase the rate of reaction.

Q5
2023
QCAA
Paper 1
1 mark
Q5
1 mark

The question refers to the decomposition of hydrogen iodide gas (HI) to produce hydrogen gas (H2\text{H}_2) and iodine gas (I2\text{I}_2) in a sealed 1-litre container.

2HI(g)⇌H2(g)+I2(g)ΔH=+53.6 kJ mol−12\text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \quad \Delta H = +53.6 \text{ kJ mol}^{-1}
Colourless Colourless Purple

Determine the equilibrium expression (KcK_c) for the reaction.

A

Kc=[H2][I2]2[HI]K_c = \frac{[\text{H}_2][\text{I}_2]}{2[\text{HI}]}

B

Kc=[H2][I2][HI]2K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}

C

Kc=2[H]2[I]2[HI]K_c = \frac{2[\text{H}]2[\text{I}]}{2[\text{HI}]}

D

Kc=2[H]2[I][HI]2K_c = \frac{2[\text{H}]2[\text{I}]}{[\text{HI}]^2}

Reveal Answer
A

Kc=[H2][I2]2[HI]K_c = \frac{[\text{H}_2][\text{I}_2]}{2[\text{HI}]}

This is incorrect because the stoichiometric coefficient of the reactant (2) must be used as an exponent, not a multiplier. The correct term is [HI]2[\text{HI}]^2, not 2[HI]2[\text{HI}].

B

Kc=[H2][I2][HI]2K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}

Correct Answer

This is correct based on the law of mass action for the reaction 2HI⇌H2+I22\text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2. The concentration of products is in the numerator, the reactant is in the denominator, and the coefficient 2 for HI becomes an exponent.

C

Kc=2[H]2[I]2[HI]K_c = \frac{2[\text{H}]2[\text{I}]}{2[\text{HI}]}

This is incorrect because it uses atomic species (H, I) instead of the molecular species (H2\text{H}_2, I2\text{I}_2) actually present in the reaction, and it treats coefficients as multipliers rather than exponents.

D

Kc=2[H]2[I][HI]2K_c = \frac{2[\text{H}]2[\text{I}]}{[\text{HI}]^2}

This is incorrect because it substitutes atomic concentrations for molecular products and uses coefficients as multipliers in the numerator.

Q18
2025
VCAA
1 mark
Q18
1 mark

In artificial photosynthesis

A

water is oxidised and hydrogen gas, H2H_2, is produced.

B

the same products are produced as in natural photosynthesis.

C

hydrogen ions, H+H^+, are reduced to produce hydrogen gas, H2H_2, at the anode.

D

electrical energy from an external power supply is required to oxidise water.

Reveal Answer
A

water is oxidised and hydrogen gas, H2H_2, is produced.

Correct Answer

This is correct because artificial photosynthesis uses solar energy to split water, oxidizing it to produce oxygen gas while reducing hydrogen ions to produce hydrogen gas (H2H_2).

B

the same products are produced as in natural photosynthesis.

This is incorrect because natural photosynthesis produces glucose and oxygen, whereas artificial photosynthesis typically produces hydrogen gas or other synthetic fuels.

C

hydrogen ions, H+H^+, are reduced to produce hydrogen gas, H2H_2, at the anode.

This is incorrect because the reduction of hydrogen ions (H+H^+) to hydrogen gas (H2H_2) occurs at the cathode, not the anode.

D

electrical energy from an external power supply is required to oxidise water.

This is incorrect because artificial photosynthesis relies on solar energy (light) to drive the oxidation of water, rather than electrical energy from an external power supply.

Q4
2021
QCAA
Paper 2
9 marks
Q4

5.00×10−45.00 \times 10^{-4} moles of hydrogen gas is mixed with 1.00×10−31.00 \times 10^{-3} moles of iodine vapour in a sealed 1.00 L vessel at 455.0 vC. The concentration of hydrogen iodide gas formed at equilibrium is 9.30×10−49.30 \times 10^{-4} M.
The balanced equation for the reaction is shown.

H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g)

Q4a
1 mark

Write the equilibrium law expression for the reaction.

Reveal Answer

Kc=[HI]2[H2]×[I2]K_{\text{c}} = \frac{[\text{HI}]^2}{[\text{H}_2]\times[\text{I}_2]}

Marking Criteria
DescriptorMarks

provides Kc=[HI]2[H2]×[I2]K_{\text{c}} = \frac{[\text{HI}]^2}{[\text{H}_2]\times[\text{I}_2]}

1
Q4b
5 marks

Calculate the equilibrium constant (KcK_c) for the reaction at 455.0 vC. Show your working. (three significant figures)

Reveal Answer

Change in [H2]=[I2]=9.30×10−4 molL×1 mol H22 mol HI=4.65×10−4 M[\text{H}_2] = [\text{I}_2] = \frac{9.30\times10^{-4} \text{ mol}}{\text{L}} \times \frac{1 \text{ mol H}^2}{2 \text{ mol HI}} = 4.65 \times 10^{-4} \text{ M}

[H2]eq=5.00×10−4−4.65×10−4=3.50×10−5 M[\text{H}_2]_{\text{eq}} = 5.00 \times 10^{-4} - 4.65 \times 10^{-4} = 3.50 \times 10^{-5} \text{ M}

[I2]eq=1.0×10−3−4.65×10−4=5.35×10−4 M[\text{I}_2]_{\text{eq}} = 1.0 \times 10^{-3} - 4.65 \times 10^{-4} = 5.35 \times 10^{-4} \text{ M}

Kc=(9.30×10−4)23.50×10−5×5.35×10−4=46.2K_{\text{c}} = \frac{(9.30\times10^{-4})^2}{3.50\times10^{-5} \times 5.35\times10^{-4}} = 46.2

Kc=46.2K_{\text{c}} = 46.2 (to three significant figures)

Marking Criteria
DescriptorMarks

correctly determines change in [H2]=[I2]=4.65×10−4[H_2] = [I_2] = 4.65 × 10^{-4}

1

determines [H2]eq=3.50×10−5[H_2]_{eq} = 3.50 × 10^{-5}

1

determines [I2]eq=5.35×10−4[I_2]_{eq} = 5.35 × 10^{-4}

1

shows substitution correctly performed

1

determines Kc=46.2K_c = 46.2

1
Q4c
3 marks

Predict the effect that adding a catalyst would have on the reaction rates, position of the equilibrium and value of KcK_c.

Reveal Answer

A catalyst will speed up both the forward and the reverse reactions.
Therefore, the position of the equilibrium will not change.
Therefore, there will be no change in the value of the equilibrium constant, KcK_{\text{c}}.

Marking Criteria
DescriptorMarks

identifies that a catalyst speeds up both the forward and reverse reactions

1

identifies that a catalyst has no effect on the position of the equilibrium

1

determines that a catalyst has no effect on the Kc value

1
Q15
2025
VCAA
1 mark
Q15
1 mark

Consider the following two reactions that are at equilibrium at 500 °C.

N2(g)+3H2(g)⇌2NH3(g)Kc=2.86×10−1 M−2N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \qquad K_c = 2.86 \times 10^{-1}\ \mathrm{M^{-2}}

4NH3(g)⇌2N2(g)+6H2(g)4NH_3(g) \rightleftharpoons 2N_2(g) + 6H_2(g)

The magnitude of the value of KcK_c for the second reaction is

A

8.18×10−28.18 \times 10^{-2}

B

5.72×10−15.72 \times 10^{-1}

C

1.75×1001.75 \times 10^0

D

1.22×1011.22 \times 10^1

Reveal Answer
A

8.18×10−28.18 \times 10^{-2}

This value is (Kc)2(K_c)^2, which would be the equilibrium constant if the first reaction were multiplied by 2 but not reversed.

B

5.72×10−15.72 \times 10^{-1}

This value is 2×Kc2 \times K_c. When a reaction is multiplied by a coefficient, the equilibrium constant must be raised to that power, not multiplied by it.

C

1.75×1001.75 \times 10^0

This value results from incorrect mathematical manipulation of the equilibrium constant. The correct operation is to take the inverse square of the original KcK_c.

D

1.22×1011.22 \times 10^1

Correct Answer

The second reaction is the reverse of the first reaction multiplied by 2. Therefore, its equilibrium constant is (Kc)−2=(2.86×10−1)−2=1.22×101(K_c)^{-2} = (2.86 \times 10^{-1})^{-2} = 1.22 \times 10^1.

Q23
2020
QCAA
Paper 1
6 marks
Q23a (i)
1 mark

Identify the products of the electrolysis of molten sodium chloride:

Reveal Answer

Na(l) and Cl2(g)Cl_2(g)

Marking Criteria
DescriptorMarks

provides Na(l) and Cl2(g)Cl_2(g)

1
Q23a (ii)
1 mark

Identify the products of the electrolysis of dilute aqueous sodium chloride solution:

Reveal Answer

H2(g)H_2(g) and O2(g)O_2(g)

Marking Criteria
DescriptorMarks

provides H2(g)H_2(g) and O2(g)O_2(g)

1
Q23b
4 marks

Explain how the nature of the electrolyte affects the products generated when a dilute aqueous solution of sodium chloride undergoes electrolysis.

Reveal Answer

In a dilute solution of aqueous sodium chloride, sodium ions, chloride ions, hydrogen ions, hydroxide ions and water molecules are present.
The concentration and the EoE^o value of the species create competition at the electrodes and affect the products formed.
Na+^+ and H+^+ compete to be reduced at the cathode. The EoE^o value for reducing H+^+ is more positive; therefore, H+^+ is preferentially reduced and H2_2 gas is formed rather than Na metal.
Cl−^- and OH−^- compete to be oxidised at the anode. As the concentration of Cl−^- is low in a dilute NaCl solution, OH−^- is preferentially oxidised and O2_2 gas is produced rather than Cl2_2 gas.

Marking Criteria
DescriptorMarks

Identifies that Na+^+, Cl−^-, OH−^-, H+^+, and H2_2O are present

1

Identifies that concentration and EoE^o values of the species affects products

1

Identifies that H+^+ is preferentially reduced, producing H2_2 gas due to a more positive EoE^o value

1

Identifies that OH−^- is preferentially oxidised, producing O2_2 gas due to a higher concentration of ions

1
Q7
2023
QCAA
Paper 2
4 marks
Q7

When heated in a sealed container, solid mercury(II) oxide (HgO) decomposed to form metallic mercury (Hg) and oxygen gas (O2_2).

2HgO(s)⇌2Hg(l)+O2(g)2\text{HgO(s)} \rightleftharpoons 2\text{Hg(l)} + \text{O}_2\text{(g)}
Orange \quad Silver \quad Colourless

Q7a
1 mark

Identify whether the reaction occurs in an open or closed system.

Reveal Answer

Closed system

Marking Criteria
DescriptorMarks

identifies closed system

1
Q7b
3 marks

Explain why the colour of the system does not change once equilibrium is established.

Reveal Answer

At equilibrium there is no net change in the concentration of the reactants and the products.
The forward and reverse reactions are still occurring.
As the forward reaction is equal to the rate of the reverse reaction the colour of the system at equilibrium remains constant.

Marking Criteria
DescriptorMarks

identifies the concentration of the reactants and products remain constant at equilibrium

1

explains that the forward and reverse reactions are occurring simultaneously

1

explains forward reaction is equal to reverse reaction and therefore there is no colour change

1
Q4
2025
SCSA
1 mark
Q4
1 mark

The equation for a system at equilibrium is given below.

2NO(g)+O2(g)⇋2NO2(g)+heat2 NO(g) + O_2(g) \leftrightharpoons 2 NO_2(g) + heat

At 25 `C, the value of K for this equilibrium is 2.19×10122.19 \times 10^{12}.

Which of the following statements about this system is true? Increasing the

A

partial pressure of NO(g) will increase the yield of NO2(g)NO_2(g) and will increase the rate of the forward reaction.

B

partial pressure of NO(g) will increase the yield of NO2(g)NO_2(g) but will decrease the rate of the forward reaction.

C

temperature will increase the yield of NO2(g)NO_2(g) but decrease the rate of the forward reaction.

D

temperature will increase the yield of NO2(g)NO_2(g) and increase the rate of the forward reaction.

Reveal Answer
A

partial pressure of NO(g) will increase the yield of NO2(g)NO_2(g) and will increase the rate of the forward reaction.

Correct Answer

According to Le Chatelier's principle, increasing the partial pressure of a reactant (NONO) shifts the equilibrium toward the products, increasing the yield. Additionally, a higher concentration of reactants increases the frequency of collisions, thereby increasing the forward reaction rate.

B

partial pressure of NO(g) will increase the yield of NO2(g)NO_2(g) but will decrease the rate of the forward reaction.

While increasing the partial pressure of a reactant does increase the yield, it also increases (rather than decreases) the reaction rate due to a higher frequency of molecular collisions.

C

temperature will increase the yield of NO2(g)NO_2(g) but decrease the rate of the forward reaction.

Increasing temperature increases the kinetic energy of the molecules, which always leads to an increase in the reaction rate, not a decrease.

D

temperature will increase the yield of NO2(g)NO_2(g) and increase the rate of the forward reaction.

The reaction forming NO2NO_2 is exothermic; therefore, increasing the temperature shifts the equilibrium toward the reactants (left), decreasing the yield of NO2NO_2.

Q23
2025
SCSA
1 mark
Q23
1 mark

Primary and secondary cells are both galvanic cells. Another similarity between these cells is that

A

reduction occurs at the negative electrode.

B

they act as an electrolytic cell when recharging.

C

they use spontaneous redox reactions as a source of energy.

D

they convert stored electrical energy into chemical energy.

Reveal Answer
A

reduction occurs at the negative electrode.

In galvanic cells, reduction always occurs at the cathode, which is the positive electrode, while oxidation occurs at the negative anode.

B

they act as an electrolytic cell when recharging.

Only secondary cells are rechargeable and act as electrolytic cells during the charging process; primary cells cannot be recharged.

C

they use spontaneous redox reactions as a source of energy.

Correct Answer

Both primary and secondary cells function as galvanic cells, meaning they generate electrical energy from spontaneous chemical redox reactions.

D

they convert stored electrical energy into chemical energy.

Galvanic cells convert chemical energy into electrical energy; the conversion of electrical energy into chemical energy occurs during recharging (electrolysis), which primary cells cannot undergo.

Q24
2021
VCAA
1 mark
Q24
1 mark

Which one of the following statements describes the effect that adding a catalyst will have on the energy profile diagram for an exothermic reaction?

A

The energy of the products will remain the same.

B

The shape of the energy profile diagram will remain the same.

C

The peak of the energy profile will move to the left as the reaction rate increases.

D

The activation energy will be lowered by the same proportion in the forward and reverse reactions.

Reveal Answer
A

The energy of the products will remain the same.

Correct Answer

A catalyst provides an alternative reaction pathway with a lower activation energy, but it does not alter the initial energy of the reactants or the final energy of the products.

B

The shape of the energy profile diagram will remain the same.

The shape of the energy profile diagram changes because the peak, which represents the activation energy, is lowered when a catalyst is added.

C

The peak of the energy profile will move to the left as the reaction rate increases.

The x-axis of an energy profile diagram represents reaction progress, not time. A catalyst lowers the peak vertically rather than shifting it horizontally.

D

The activation energy will be lowered by the same proportion in the forward and reverse reactions.

A catalyst lowers the activation energy of both the forward and reverse reactions by the same absolute amount, not the same proportion, since their initial activation energies are different.

Q1
2020
QCAA
Paper 1
1 mark
Q1
1 mark

A partly filled water bottle is sealed and left on a bench in a room with a constant temperature. After several minutes, it is noted that the water level in the bottle remains constant. In the water bottle, the rate of evaporation is

A

less than the rate of condensation.

B

greater than the rate of condensation.

C

equal to the rate of condensation and equal to zero.

D

equal to the rate of condensation but not equal to zero.

Reveal Answer
A

less than the rate of condensation.

If the rate of evaporation were less than the rate of condensation, the amount of liquid water would increase, causing the water level to rise.

B

greater than the rate of condensation.

If the rate of evaporation were greater than the rate of condensation, the amount of liquid water would decrease, causing the water level to drop.

C

equal to the rate of condensation and equal to zero.

While the rates are equal, they are not zero because molecules are constantly escaping and returning to the liquid surface in a state of dynamic equilibrium.

D

equal to the rate of condensation but not equal to zero.

Correct Answer

A constant water level in a closed system indicates dynamic equilibrium, where the rate of evaporation equals the rate of condensation, and both processes continue to occur simultaneously.

Q14
2021
VCAA
1 mark
Q14
1 mark

Use the following information to answer the question.

The overall discharge reaction for a lead-acid battery is

Pb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(s)+2H2O(l)\text{Pb(s)} + \text{PbO}_2\text{(s)} + 2\text{H}_2\text{SO}_4\text{(aq)} \rightarrow 2\text{PbSO}_4\text{(s)} + 2\text{H}_2\text{O(l)}

When the lead-acid battery is discharging, the oxidising agent is

A

Pb

B

PbO2\text{PbO}_2

C

PbSO4\text{PbSO}_4

D

H2SO4\text{H}_2\text{SO}_4

Reveal Answer
A

Pb

Pb undergoes oxidation at the anode (its oxidation state increases from 0 to +2), making it the reducing agent, not the oxidising agent.

B

PbO2\text{PbO}_2

Correct Answer

During discharge, PbO2\text{PbO}_2 undergoes reduction at the cathode (the oxidation state of Pb decreases from +4 to +2), meaning it acts as the oxidising agent.

C

PbSO4\text{PbSO}_4

PbSO4\text{PbSO}_4 is the product formed at both the anode and cathode during the discharge process, rather than a reactant acting as an oxidising agent.

D

H2SO4\text{H}_2\text{SO}_4

Sulfuric acid (H2SO4\text{H}_2\text{SO}_4) acts as the electrolyte that provides ions for the reaction, but it is not the species being reduced.

Frequently Asked Questions

How many VCAA Chemistry questions cover How can the rate and yield of chemical reactions be optimised??
AusGrader has 219 VCAA Chemistry questions on How can the rate and yield of chemical reactions be optimised?, all with instant AI grading and detailed marking feedback.

Ready to practise VCAA Chemistry?

Get instant AI feedback on past exam questions, aligned to the syllabus

Start Practising Free