20 Essential Questions: Effective VCE Chemistry Exam Study

VCE Chemistry exam revision can be challenging as you must cover a broad range of topics and skills. Chemistry combines theory with practical problem-solving, meaning you need to understand both the ‘why’ and ‘how’ behind chemical processes. An effective way to tackle this is by practising specific questions to test your knowledge and pinpoint areas for improvement.

Success in VCE Chemistry hinges on your ability to use what you know in practical situations, like predicting reactions and interpreting data. This blog gives you 20 essential questions to help you revise key topics and improve your exam techniques across multiple question types.

Free VCE Study Planning Kit Download

Get VCE ready with this comprehensive study planner.

Download now

Free VCE Study Planning Kit Download

Fill out your details below to get this resource emailed to you.

"*" indicates required fields

First Name*

Last Name*

Email*

Postcode*

Grade in 2025*

Grade in 2025123456789101112

Hidden

Resource (DO NOT DELETE OR CHANGE)

By submitting this form, you agree to receive promotional messages from Matrix about its products and services. You can unsubscribe at any time by clicking on the link at the bottom of our emails.

Our VCE Chemistry study guide will cover:

• Tips to boost your VCE Chemistry exam revision
• Common Chemistry questions students find difficult
• Multiple-choice practice questions
• Short-answer practice questions
• Extended-response practice questions
• Multiple-choice solutions
• Short-answer solutions
• Extended-response solutions

Tips to boost your VCE Chemistry exam revision

Chemistry students must grasp key concepts and use them to solve problems. It’s a unique blend of theory and application. Here are some tips to help you prepare:

1. Master the basics: Chemistry builds on itself, so it’s important to understand the basics like atomic structure, bonding, and stoichiometry before moving on to more advanced topics. Daily practice of balancing chemical equations and applying stoichiometry principles will help you reduce mistakes and ensure accuracy in the exam.
2. Prioritise understanding over memorisation: While formulas and definitions are important, the exam often tests how well you can apply them in different contexts. For example, understanding how Le Chatelier’s principle works to predict changes in equilibrium is more important than just knowing the definition.
3. Practise problem-solving: Practise regularly to improve your problem-solving skills. This includes solving mole calculations and working on past VCE Chemistry exam questions.
4. Simulate exam conditions: Practise under timed conditions to improve your time management and reduce exam stress.

Common Chemistry questions students find difficult

Certain topics in Chemistry are consistently challenging for students. Focusing on these key areas will improve your VCE Chemistry exam revision and performance:

1. Chemical equilibrium:
   Many students struggle with calculations relating to equilibrium concentrations and equilibrium constants, as well as interpreting equilibrium systems, especially when trying to understand graphical representations of the systems. 
2. Analysis of organic compounds:
   Students often find interpreting spectra difficult since these are application-type questions that involve the analysis of data. Simple recall is not enough to perform well on these questions.
3. Organic chemistry reactions:
   It can be difficult to identify functional groups and interpret reaction pathways, and multi-step synthesis. Many students struggle with predicting reaction outcomes, understanding reactivity, and remembering reaction pathways, especially for more complex processes with multiple steps.
4. Redox reactions and electrochemistry:
   Interpreting and balancing redox equations and understanding electrochemical cells can be tricky. It can be difficult to tell the difference between oxidation and reduction, apply standard electrode potentials, and predict cell potentials. Small mistakes in identifying which species are oxidised or reduced can lead to big errors.
5. Thermochemistry and enthalpy:
   Students often have trouble calculating enthalpy changes and bond enthalpy data. 

Practice questions

Below are 20 essential questions designed to help you cover key topics in your VCE Chemistry exam revision. They are divided into 10 multiple-choice, 7 short-answer, and 3 extended-response questions, with detailed solutions provided afterwards.

Multiple-Choice Questions (1 mark each)

1. In a standard galvanic cell, which factor primarily determines the direction of electron flow?
   (a) The type of salt in the salt bridge.
   (b) The concentration of the electrolytes in the half-cells.
   (c) The size of the electrodes.
   (d) The difference in electrode potentials between the anode and the cathode.
2. In the complete combustion of methane (CH₄), which of the following occurs?
   (a) Energy is absorbed, and the system cools down.
   (b) Energy is released, and carbon dioxide is produced.
   (c) The system remains in a steady state with no net energy change.
   (d) The reaction reaches equilibrium and stops releasing energy.
3. In a fuel cell, methanol (CH₃OH) is used as a fuel in the presence of oxygen. Which is the correct overall reaction?
   (a) 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l)
   (b) CH₃OH(l) + 2O₂(g) → CO₂(g) + 2H₂O(l)
   (c) 2CH₃OH(l) + 3O₂(g) → CO(g) + 4H₂O(l)
   (d) CH₃OH(l) + O₂(g) → CO₂(g) + H₂O(l)
4. A solution containing Fe²⁺ and Fe³⁺ ions is titrated using a potassium permanganate solution. Which half-reaction represents the reduction at the endpoint?
   (a) Fe²⁺ → Fe³⁺ + e⁻
   (b) Fe³⁺ + e⁻ → Fe²⁺
   (c) MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
   (d) Mn²⁺ → MnO₄⁻ + 4e⁻
5. When designing a pathway to convert ethene to ethanol, which reagent and conditions will minimise the formation of unwanted by-products?
   (a) Addition of water with dilute sulphuric acid and high temperature
   (b) Substitution with NaOH and high temperatures
   (c) Hydration using H₂O and H₃PO₄ catalyst under high pressure
   (d) Direct hydrogenation using H₂ over a Pt catalyst
6. Which of the following compounds would most likely produce a positive Lucas’ reagent test?
   (a) Butan-2-ol
   (b) Ethanal
   (c) Cyclohexane
   (d) Bromoethane
7. Which of the following represents a correct IUPAC name for a compound with the formula C₄H₉OH?
   (a) Methanol
   (b) 2-methylpropanol
   (c) Propan-2-ol
   (d) 2-methylbutan-1-ol
8. Which of the following reactions will not increase the concentration of ammonia at equilibrium?
   N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92.4 kJ/mol
   (a) Increasing pressure
   (b) Decreasing temperature
   (c) Adding a catalyst
   (d) Removing NH₃
9. In an electrolytic cell, water is oxidised at the anode. What is the balanced half-reaction occurring at the anode?
   (a) H₂O(l) → O₂(g) + H⁺(aq)
   (b) H₂O(l) → H₂(g) + OH⁻(aq)
   (c) 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq)
   (d) 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻
10. Given the reaction: 2NO₂(g) ⇌ N₂O₄(g), how would the equilibrium constant, Kc, be affected by a decrease in temperature, given the forward reaction is exothermic?
    (a) Decrease
    (b) Increase
    (c) Remain constant
    (d) Fluctuate

Short-Answer Questions

1. Consider the exothermic reaction:
   2NO₂(g)↔N2O₄(g)
   Predict and explain the effect of increasing pressure and temperature on the position of equilibrium. How would the equilibrium constant change with increasing temperature? (4 marks)
2. Draw and label the structural isomers of C₅H₁₂. List the isomers in order of ascending boiling point and account for the differences in their boiling points. (4 marks)
3. A student conducts an experiment to determine the enthalpy of neutralisation by mixing 450.0 mL of 1.60 M HCl with 550.0 mL of 1.00 M NaOH in a calorimeter. The temperature of the solution rises from 22.5°C to 28.7°C. Assume the specific heat capacity of the solution is 4.18 J/g°C and the density of the solution  is 1.00 g/mL.
   a) Calculate the heat released during the reaction.
   (2 marks)
   b) Determine the enthalpy of neutralisation in kJ/mol of HCl.
   (2 marks)
4. Balance the following redox reaction in a basic solution (2 marks):
   Cr₂O₇²⁻(aq)+I⁻(aq)→Cr³⁺(aq)+I₂(aq)
5. A galvanic cell is constructed using zinc and copper electrodes. The zinc electrode is placed in a 1.0 M ZnSO₄ solution, and the copper electrode is placed in a 1.0 M CuSO₄ solution. Write the balanced half-reactions for the oxidation and reduction processes occurring in the cell and calculate the standard cell potential (E°cell) for the reaction. (3 marks)
6. Calculate the equilibrium constant for the following reaction at 300°C if only 0.50 M of CO₂ was initially present and 0.12 M O2 was present at equilibrium (3 marks):
   2CO₂(g)⇌2CO(g) + O₂(g)
7. Molecule V contains only carbon atoms, hydrogen atoms and one oxygen atom. The infrared (IR) spectrum of molecule V is shown below.
   a) Identify the bonds that are present in molecule V based on the spectrum. (1 mark)
   b) Explain why different frequencies of infrared radiation can be absorbed by the same molecule as shown in the spectrum above.
   (2 marks)

Extended-Response Questions

1. Green Chemistry and Atom Economy
   Evaluate the atom economy and environmental impact of the following reaction used to synthesise aspirin:
   C₇H₆O₃ + (CH₃CO)₂O → C₉H₈O₄ + CH₃COOH
   Discuss how the principles of green chemistry could be applied to improve this synthesis and calculate the atom economy of the reaction. (10 marks)
2. Organic Synthesis
   Design a reaction pathway to synthesise ethyl ethanoate (ethyl acetate) from ethene, using no more than four steps. Include reagents, conditions, and justify each step in terms of yield and/or selectivity.
   (9 marks)
3. Electrolysis (7 marks)
   In an electrolysis experiment of aqueous CuSO₄, 1.50 A of current is passed through a solution for 30 minutes.
   a) Identify and justify what is produced at the anode and the cathode. Support your answer with half-equations. (4 marks)
   b) Determine the mass of the product deposited at the cathode. (3 marks)

Detailed solutions

Multiple-choice solutions

1. Answer: d) The difference in oxidation potential between the anode and the cathode.
   The difference in electrode potentials drives the movement of electrons. Electrons move from side with high electrode potential to side with low electrode potential
2. Answer: b) Energy is released, and carbon dioxide is produced.
   Complete combustion of methane releases energy and produces CO₂ and H₂O.
3. Answer: (a) 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(g)
   In a methanol fuel cell, methanol reacts with oxygen to form carbon dioxide and water. Only (a) has the correct reactants and products and is correctly balanced.
4. Answer: (c) MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
   Reduction involves the gain of electrons. According to the electrochemical series, MnO₄⁻ has a larger reduction potential than Fe³⁺, hence the reduction of MnO₄⁻ to Mn²⁺ will occur spontaneously in a galvanic cell.
5. Answer: (c) Hydration using H₂O and H₃PO₄ catalyst under high pressure
   Conversion of ethene to ethanol requires addition of water in the presence of an acid catalyst. Hydration of ethene using water and a phosphoric acid (H₃PO₄) catalyst under high pressure is an industrial method that avoids unwanted by-products that could form when high temperatures are used.
6. Answer: (a) Butan-2-ol
   Lucas’ test identifies the presence of secondary or tertiary alcohols. Butan-2-ol is a secondary alcohol, hence will produce a positive Lucas’ test.
7. Answer: b) 2-methylpropanol
   2-methylpropanol correctly identifies the four-carbon alcohol.
8. Answer: (c) Adding a catalyst
   Adding a catalyst does not affect the position of equilibrium; it only speeds up the rate of both the forward and reverse reactions.
9. Answer: (d) 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻
   At the anode, water is oxidised to oxygen gas, releasing protons and electrons.
10. Answer: (b) Increase
    Decreasing temperature favours the exothermic reaction as heat is released which counteracts the removal of heat. This is the forward reaction, hence more products are formed when equilibrium shifts. This increases Kc.

Short-answer solutions

1. Le Chatelier’s Principle and Temperature (4 marks)

Solution:

According to Le Chatelier’s principle, the equilibrium will shift to minimise a disturbance. Increasing pressure shifts the equilibrium toward the side with fewer gas molecules, so the reaction will shift to the right, favouring N₂O₄ formation (1 mole of gas instead of 2).

Increasing temperature will shift the equilibrium to the left, as the reverse reaction is endothermic. This favours NO₂ formation, absorbing the added heat.

The equilibrium constant (K) will decrease with increasing temperature because higher temperatures favour the reverse reaction, resulting in greater NO₂ formation.

Explanation:
The student correctly applies Le Chatelier’s principle to both pressure and temperature changes, explaining the shifts in equilibrium position and how temperature impacts the equilibrium constant.

Tip:
For equilibrium questions, always balance the equation first, then consider the number of gas molecules on each side and whether the reaction is exothermic or endothermic.

2. Structural Formula (4 marks)

The structural isomers of C₅H₁₂ are:

• n-pentane (CH₃-CH₂-CH₂-CH₂-CH₃)
• 2-methylbutane (CH₃-CH(CH₃)-CH₂-CH₃)
• 2,2-dimethylpropane (C(CH₃)₄).

Branching lowers the boiling point because branched molecules have a smaller surface area and less efficient packing compared to straight-chain isomers, leading to weaker dispersion forces.

Explanation:
This response correctly lists all structural isomers and explains the impact of branching on boiling points, showing an understanding of both molecular structure and intermolecular forces.

Tip:
When asked to draw structural isomers, start by drawing the straight-chain version, then gradually add branching in different positions.

3. Solution Calorimetry. (4 marks)

Solution:

   a) Heat released during the reaction (2 marks)
Assuming no heat loss, the quantity of heat released by the reaction is equal to the total quantity of heat absorbed by the solution. To calculate the heat absorbed by the solution, we use the formula: q=mcΔT

Where:

• m = 100.0 g (since the volume is 100.0 mL and density of the solution is 1.00 g/mL)
• c = 4.18 J/g°C (specific heat capacity of the solution)
• ΔT = 28.7°C−22.5∘C = 6.2°C (temperature change)

Substituting the values:

q(solution) = 100.0 g × 4.18 J/g°C × 6.2 °C = 2591.6 J = 2.5916 kJ

Thus, the heat released is 2.6 kJ (2 sig. fig.).

   b) Enthalpy of neutralisation (ΔH) (2 marks)
First, determine the moles of HCl used:

moles of HCl total = concentration × volume = 1.6 M × 0.045 L = 0.072 mol

moles of NaOH total = concentration × volume = 1.0 M × 0.055 L = 0.055 mol

NaOH is the limiting reactant, so moles of HCl used = 0.055 mol

The enthalpy of neutralisation is calculated by dividing the heat released by the moles of HCl used:

Since the reaction is exothermic, the enthalpy of neutralisation is ΔH = −47 kJ/mol (2 sig. fig.).

Explanation:
This response clearly demonstrates the application of relevant formulas and concepts in a step-by-step manner. Each calculation is fully explained and the reasoning behind every step is clear. 

Tip:
For calorimetry calculations, apply the formula q=mcΔT, and then ΔH = -q/n. Remember that for exothermic processes, enthalpy change will be a negative value.

4. Balance the Redox Reactions (2 marks)

Solution:

1. Assign oxidation states: Cr in Cr₂O₇²⁻ is +6, I⁻ is -1, Cr³⁺ is +3, and I₂ is 0.
2. Write half-reactions:
   •   Oxidation: 2I⁻ → I₂
   •   Reduction: Cr₂O₇²⁻→ 2Cr³⁺
3. Balance oxygens by adding water. Then balance hydrogens by adding H+.
   •   Oxidation: 2I⁻ → I₂
   •   Reduction: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O
4. For basic conditions, add 14OH⁻ to both sides and cancel excess water. 
   •   Oxidation: 2I⁻ → I₂
   •   Reduction: Cr₂O₇²⁻ + 7H₂O → 2Cr³⁺ + 14OH⁻
5. Add electrons to balance charges.
   •   Oxidation: 2I⁻ → I₂ + 2e⁻
   •   Reduction: Cr₂O₇²⁻ + 7H₂O + 6e⁻ → 2Cr³⁺ + 14OH⁻
6. Combine the half-reactions: Cr₂O₇²⁻ + 6I⁻ + 7H₂O → 2Cr³⁺ + 3I₂ + 14OH⁻.

Explanation:
This response follows the correct steps for balancing redox reactions in a basic solution, systematically showing each step. The inclusion of OH⁻ in the final step demonstrates an understanding of reactions in basic conditions.

Tip:
Start by balancing atoms other than O and H, then balance oxygen with H₂O and hydrogen with H⁺ before accounting for basic conditions by adding OH⁻.

5. Galvanic Cell with Zinc and Copper (3 marks)

Solution:
Oxidation: Zn²⁺ + 2e⁻ → Zn (E° = -0.76 V)
Reduction: Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V)

E°cell = E°cathode – E°anode
E°cell = 0.34 V – (-0.76 V)
E°cell = 1.10 V

Explanation:

This response is exemplary because it correctly provides the chemical equations for oxidation and reduction and applies the correct formula for calculating cell potential. It demonstrates a solid understanding of galvanic cell function, and correct calculation methods, all conveyed concisely.

5. Equilibrium Concentrations (3 marks)
a) Construct an ICE table to set out your working. 

   b) Substitute the equilibrium concentrations into the equilibrium expression to determine the equilibrium constant. 

7. IR Spectrum: Absorption of Different Frequencies (3 marks)

(a)Identify the bonds that are present in molecule V based on the spectrum. (1 mark)

 C–H and C=O

(b) Explain why different frequencies of infrared radiation can be absorbed by the same molecule as shown in the spectrum above.
(2 marks) 

Different frequencies of IR radiation are absorbed by a molecule because various bonds vibrate at characteristic frequencies depending on their bond strength and atomic masses. For example, stretching vibrations of C=O bonds occur at a higher frequency (~1700 cm⁻¹) than O–H stretches, which occur between 3200-3600 cm⁻¹ due to different bond polarities and masses.

Extended-Response Solutions

1. Green Chemistry and Atom Economy (10 marks)

Atom economy:
For the reaction:

C₇H₆O₃ + (CH₃CO)₂O → C₉H₈O₄ + CH₃COOH

• Molar masses:
  • Aspirin (C₉H₈O₄): 180.16 g/mol
  • Reactants: C₇H₆O₃ (138.12 g/mol) and (CH₃CO)₂O (102.09 g/mol)
  • Total reactants = 240.21 g/mol

Green Chemistry improvements:

• Use renewable sources of acetic anhydride.
• Minimise waste by using catalytic processes that prevent the formation of acetic acid by-product.
• Implement solvent-free synthesis or use of green solvents.

2. Organic Synthesis of Ethyl Ethanoate from Ethene (9 marks)

1. Step 1: Hydration of ethene to form ethanol.
   •   Reagents: H₂O (steam), H₃PO₄​ catalyst.
   •   Conditions: High temperature and pressure.
     Justification: Direct addition of water forms ethanol selectively.
2. Step 2: Oxidation of ethanol to ethanoic acid.
   •   Reagents: Acidified potassium dichromate (K₂Cr₂O₇)
   •   Conditions: Reflux.
     Justification: Using a strong oxidant selectively yields ethanoic acid from ethanol.
3. Step 3: Esterification of ethanoic acid with ethanol.
   •   Reagents: Concentrated sulfuric acid.
   •   Conditions: Reflux.
     Justification: Esterification yields ethyl ethanoate and water as a by-product. Concentrated sulfuric acid acts as a catalyst and dehydrating agent which further improves the yield of the ester by shifting the equilibrium in the forward direction.

3. Electrolysis (7 marks)

a) Electrolysis of Aqueous Copper(II) Sulfate (4 marks)

Species present in aqueous copper sulfate include Cu²⁺$,\text{SO₄²⁻}$, and

${\text{H}}_2\text{O.}$

Oxidation occurs at the anode. Water or sulfate ions can be oxidised, but water is a stronger reductant. Hence water will be oxidised producing oxygen gas:

2

${\text{H}}_2\text{O}$

(l)→O₂​(g)+4H⁺(aq)+4e⁻

Reduction occurs at the cathode: Copper ions and water can be reduced, but copper ions are a stronger oxidant. Hence copper will be reduced to solid copper:

Cu²⁺(aq)+2e⁻→Cu(s)

Thus, oxygen gas is produced at the anode, and copper metal is deposited at the cathode.

b) Mass of Copper Deposited (3 marks)

Given:

• Current = 1.50 A
• Time = 30 minutes = 1800 seconds

1. Calculate total charge (Q):
   Q = It = 1.50A × 1800s = 2700 C
2. Using Faraday’s constant (F = 96 500 C/mol), calculate moles of electrons:
   Moles of electrons = 2700 ÷ 96,500 = 0.028 mol
3. Copper deposited (n_Cu):
   n_Cu= 0.028 ÷ 2 = 0.014 mol
4. Mass of copper deposited:
   m_Cu=0.014 mol × 63.5g/mol = 0.89g

Thus, the mass of copper deposited is 0.89 g.

Conclusion

For effective VCE Chemistry exam revision, you need to have a strong grasp of both simple and complex topics. Practising these key questions during your VCE Chemistry exam revision will help strengthen your understanding and sharpen your ability to apply concepts in the actual VCE Chemistry exam. Consistent practice in high-priority areas will give you the confidence and skills you need to succeed on exam day.

Preparing effectively for VCE exams is essential, but choosing the right subjects is just as important. Download our Subject Scaling Guide for a complete breakdown of VCE subject scaling and find out how to make your subject choices work to your advantage.

Free 2025 VCE ATAR & Scaling Guide Download

An in-depth explanation of VCE and scaling with strategies from top-performing students.

Download now

Free 2025 VCE ATAR & Scaling Guide Download

Fill out your details below to get this resource emailed to you.

"*" indicates required fields

First Name*

Last Name*

Email*

Postcode*

Grade in 2025*

Grade in 2025123456789101112

Hidden

Resource (DO NOT DELETE OR CHANGE)

By submitting this form, you agree to receive promotional messages from Matrix about its products and services. You can unsubscribe at any time by clicking on the link at the bottom of our emails.