VCE Chemistry Exam Revision: 20 Essential Practice Questions

If you’re preparing for the VCE Chemistry exam, you know it’s no walk in the park. With so much content to cover, it’s tempting to just memorise definitions and formulas. But the real success in Chemistry comes from understanding. You have to know why a reaction is occurring and how to apply that knowledge to solve problems.

The VCE Chemistry exam blends theory with application, and the best way to master it is through practice!

So, we’ve put together 20 essential practice questions. If you’re aiming for top marks, you should be able to work through these confidently. If not? That’s a sign to revisit your revision and keep going until it all clicks.

Table of contents:

• 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

Topics like atomic structure, bonding, and stoichiometry are the foundation for everything else. Every day, try practising the core formulas and balancing chemical equations until they’re second nature to you.

2. Prioritise understanding over memorisation

While formulas and definitions are important, the exam often tests how well you can apply them in different contexts.

Sure, you may know Le Chatelier’s principle, but do you know how it works to predict changes in equilibrium?

After reviewing a concept, see if you can explain it to someone in your own words.

3. Practise problem-solving

Practise regularly to improve your problem-solving skills. Use past VCE exam papers and complete questions without notes.

4. Mix up your questions

You need to move between theory and calculations smoothly. Focus on: Mole calculations, redox reactions, equilibrium questions, and spectroscopy problems.

To learn more about how to ace your exam, read our guide to the VCE Chemistry exam.

Test yourself: Redox reactions

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Common Chemistry questions students find difficult

These are the areas where VCE Chemistry students often lose marks. Focusing your revision on these difficult topics and sub-topics here can give you a big advantage.

Chemical equilibrium

• Calculating equilibrium concentrations and equilibrium constants.
• Interpreting equilibrium systems and their graphs. 

Analysis of organic compounds

• Spectroscopy (IR, NMR, mass spec) questions where you need to interpret data. Simple recall is not enough.

Organic chemistry reactions

• Identifying functional groups, and interpreting reaction pathways and multi-step synthesis.
• Predicting reaction outcomes, understanding reactivity, and remembering reaction pathways, especially for multi-step processes.

Redox reactions and electrochemistry

• Interpreting and balancing redox equations
• Understanding electrochemical cells
• Telling the different between oxidation and reduction, applying standard electrode potentials, and predicting cell potentials
• Identifying which species are oxidised or reduced

Thermochemistry and enthalpy

• Calculating enthalpy changes and bond enthalpy data. 

Practice questions

Here are 20 practice questions to help you cover key topics in your VCE Chemistry exam revision. There are 10 multiple-choice, 7 short-answer, and 3 extended-response questions. Scroll down for full solutions and explanations.

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.
   (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 H₂O.${}_{}$

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:

2H₂O(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

If you want to do well in the VCE Chemistry exam, you need to understand both the basics and the more complex topics. Practising questions like the ones in this guide will help you build confidence and improve your ability to apply what you’ve learnt in the exam.

Preparing effectively for the VCE Chemistry exam is much easier with a bit of help. In a Matrix Chemistry course, you’ll get access to expert teachers, structured lessons, and comprehensive resources that help you stay ahead.