Module 3 'Reactive Chemistry' Practice Test | Year 11 Chemistry

Test your exam-readiness for Year 11 Chemistry Module 3 'Reactive Chemistry' with 10 must-know exam-style questions with solutions.

Practice test for Year 11 Chemistry Module 3 ‘Reactive Chemistry’

The most effective way to finalise your preparation for a chemistry exam is to practise as many exam-style questions as possible. That’s why we have put together the Module 3 Reactive Chemistry Practice Test consisting a 10 challenging exam-style questions. Your exposure to this practice test will help you gain a competitive edge on your next exam. 

Module 3 ‘Reactive Chemistry’ practice test covers topics on:

 

Module 3 ‘Reactive Chemistry’ Practice Test

Instructions

General instructions for sitting the Module 3 Reactive Chemistry Practice Test are outlined below.

  • Reading Time: 5 minutes
  • Working Time: 55 minutes
  • Total marks: 37 marks
  • Write using black pen
  • Draw diagrams using pencil
  • Calculators approved by NESA may be used
  • A formulae sheet, data sheet and Periodic Table can be used
  • There are 10 questions with marks ranging from 1 to 7 marks

In the final HSC Chemistry exam, you will have approximately 1.8 minutes per mark.

In practice exams, you should aim to answer questions at a pace of 1.5 minutes per mark.

To simulate real exam conditions:

  • Put away your notes.
  • Close all unnecessary tabs and turn off notifications on your device.
  • Set up a timer for 55 minutes.
  • Complete this exam in a quiet environment where you will not be disturbed for the duration of the exam.

Once you have completed the Module 3 ‘Reactive Chemistry’ practice test, mark each question carefully using the provided solutions and note any areas for improvement. Remember that marking practice exams is just as important as answering the questions.

 

Chemistry formulas

n=\dfrac{m}{MM}c=\dfrac{n}{v}PV=nRT

 

Volume of 1 mole of ideal gas: at 100kPa and
at 0˚C (273.15K)22.71 L
  at 25˚C (298.15K)24.79 L
Gas constant8.314 J mol-1 K-1

 

 

Question 1: Periodicity

Consider the following first ionisation energy for successive elements in the periodic table denoted by the letters P–V. All values are in kJ/mol.

ElementPQRSTUV
First ionisation energy1407132016872087502744584

 

a)Predict which element is likely to form an ionic compound with bromine, with the ratio of 1:2. Write the formula of this compound.1
b)Predict the type of bonding present in the chloride of R.1
c)Explain the importance of first ionisation energy in determining the reactivity of elements like that of T.2
d)Do the elements P–V all have the same number of electron shells? Justify your answer.3

See Question 1 Solution

 

Question 2: Ionic equations

Write balanced full formula, full ionic and net ionic forms of the following equation:

a)silver nitrate(aq) + barium chloride(aq) → silver chloride(s) + barium nitrate(aq)3
b)aluminium metal + nitric acid → aluminium nitrate + hydrogen gas3

See Question 2 Solution

 

Question 3: Empirical formula

A student performed a first hand investigation to measure and identify the mass of the elements in magnesium oxide.

Year 11 Chemistry Experiment - Determining empirical formula of magnesium oxide

The information she record is as follows:

Mass of dry crucible + lid32.14 g
Mass of dry crucible + lid + magnesium32.63 g
 

 

Mass of dry crucible + lid + magnesium oxide32.95 g
a)Determine the empirical formula of magnesium oxide (show all working).3
b)Calculate the volume of oxygen (O2) taken from the air during this experiment. Assume the air temperature and pressure was 25 °C and 100 kPa respectively.2

See Question 3 Solution

 

Question 4: Oxidation states

Which of the following shows the greatest change in oxidation state of the metal atom?

(A)Mn(OH)3 to MnO4– 
(B)V2O5 to VO2+
(C)MnO2 to Mn2+
(D)Cr2O72- to Cr2O3

See Question 4 Solution

 

Question 5: Redox reactions

The following experiment was performed to investigate the relative activity of metals. The beaker initially contained 100 mL of 0.120 mol L-1 copper(II) sulfate solution.

Year 11 Module 3 Reactive Chemistry - Iron in copper(II) sulfate  

After several hours the dark blue colour of the solution had become lighter and a red-brown deposit had formed on the piece of iron metal. Iron(II) ions were present in the solution.

a)Write the full formula equation.1
b)Write the half-equations and label them as oxidation or reduction.2

See Question 5 Solution

 

Question 6: Galvanic cells

An experiment was set up as shown.

Module 3 Reactive Chemistry - Galvanic cell

Which of the following statements is correct?

(A)The chlorine gas is the anode.
(B)The copper electrode is the anode.
(C)The platinum electrode is the anode.
(D)There is no anode because chlorine is a gas.

See Question 6 Solution

 

Question 7: Cell potential

The diagram below shows a galvanic cell.

Module 3 Reactive Chemistry - Galvanic cell

a)Write half-equations and the overall equation for the reaction occurring above.2
b)Calculate the standard potential of the cell.1
c)The lead electrode gained 0.652 g in mass throughout the course of a reaction. If the zinc half-cell initially contained 200 mL of 0.100 M zinc nitrate solution, what is the final concentration of zinc ions?3

See Question 7 Solution

 

Question 8: Metal activity

A student performed three tests to investigate the relative activity of metals. In each test a metal strip was placed in a solution containing ions of a different metal. The results are shown in the diagrams.

Module 3 Reactive Chemistry - Activity of metals

What is the order of activity of the metals, based on these results?

(A)A > C > B
(B)B > A > C
(C)C > B > A
(D)C > A > B

See Question 8 Solution

 

Question 9: Catalyst

Assess the accuracy of the following statement: ‘Catalysts speed up a reaction but take no part in the reaction’. (3 marks)

See Question 9 Solution

 

Question 10: Collision theory

The graphs shown below illustrate the relationship between the number of collisions and kinetic energy at three different temperatures in a chemical reaction.

Explain, in terms of collision theory, what causes reaction rates to increase with increasing temperature. (4 marks)

See Question 10 Solution

 

Solutions for Module 3 Reactive Chemistry Practice Test

QAnswer
1In this question, it is important to notice the large decrease in first ionisation energy from S to T. This indicates that S is a group 18 noble gas and T is a group 1 metal. From this, we can deduce what groups the other elements belong to.

Part a)

UBr2

Explanation: U is a group 2 metal

Part b)

Covalent.

Explanation: Both R and Cl are non-metals and hence form covalent bonds.

Part c)

Element T is a metal. For metals, reactivity depends upon how readily it donates electrons to form cations (as they have few electrons in their valence shell). Hence the lower the first ionisation energy, the easier it is to remove an electron from that element, i.e. the higher the reactivity.

Part d)

No. There is a large jump in ionisation energies between S and T which means that S is a noble gas and therefore requires the most energy to remove an electron from its valence shell, while T is in Group 1 and has the lowest ionisation energy. Hence T must be in the next period, so elements T, U and V have one more electron shell than elements P, Q, R and S.

Back to Question 1 

2Part a)

Full formula: 2AgNO3(aq) + BaCl2(aq) → 2AgCl(s) + Ba(NO3)2(aq)

Full ionic: 2Ag+(aq) + 2NO3(aq) + Ba2+(aq) + 2Cl(aq) → 2AgCl(s)+ 2NO3(aq) + Ba2+(aq)

Net ionic: Ag+(aq) + Cl(aq) → AgCl(s)

Part b)

Full formula: 2Al(s) + 6HNO3(aq) → 2Al(NO3)3(aq) + 3H2(g)

Full ionic: 2Al(s) + 6H+(aq) + 6NO3(aq) → 2Al3+(aq) + 6NO3(aq) + 3H2(g)

Net ionic: 2Al(s) + 6H+(aq) → 2Al3+(aq) + 3H2(g)

Back to Question 2

3Part a)

Step 1: Calculate the mass of Mg

m(Mg) = 32.63-32.14 = 0.49\ g

 

Step 2: Calculate the moles of Mg

n(Mg) = \dfrac{0.49}{24.31} = 0.020156\ mol

 

Step 3: Calculate the mass of O

m(O) = 32.95-32.63 = 0.32\ g

 

Step 4: Calculate the moles of O

n(O) = \dfrac{0.32}{16} = 0.020\ mol

 

Step 5: Convert the mole ratio to the simplest whole number ratio

n(Mg) : n(O) = 1:1

Therefore, the empirical formula is MgO

 

Part b)

Step 1: Write a balanced chemical reaction for the synthesis of magnesium oxide

2Mg_{(s)} + O_{2(g)} → 2MgO_{(s)}

 

Step 2: Calculate the number of moles of O2

\begin{aligned} n(O_2) &= \dfrac{1}{2}\ n(O)\\ &= 0.010 mol \end{aligned}

 

Step 3: Calculate the volume of O2 using the molar volume of gas at 25°C and 100kPa. (You can find this value in the HSC Chemistry Formula and Data sheet)

V(O_2) = 0.010 \times 24.79 = 0.25\ L (2 sig. fig.)

 

Back to Question 3

4Answer (A)

Mn in Mn(OH)3 has oxidation state of +3, while Mn in MnO4 has oxidation state of +7.

Back to Question 4

5Part a)

Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)

Part b)

Oxidation: Fe(s) → Fe2+(aq) + 2e

Reduction: Cu2+(aq) + 2e → Cu(s)

Back to Question 5

6Answer (B)

Copper is undergoing oxidation reaction which occurs at the anode.

Back to Question 6

7Part a)

Oxidation: Zn(s) → Zn2+(aq) + 2e

Reduction: Pb2+(aq) + 2e → Pb(s)

Net ionic equation: Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s)

 

Part b)

E\degree=E_{red} + E_{ox} = -0.13 V + 0.76 V =0.63 \ V

 

Part c)

Step 1: Calculate the moles of lead formed. This is equal to the moles of Zn2+ formed.

\begin{aligned}n(Pb)_{formed} &= \dfrac{m}{mm} \\&= \dfrac{0.652}{207.2} \\&= 3.146718 \times 10^{-3}\ mol \\&= n(Zn^{2+})_{formed} \end{aligned}

 

Step 2: Calculate the original number of moles of Zn2+ present.

\begin{aligned}n(Zn^{2+})_{original} &= c \times V \\&= 0.1 \times 0.2 \\&= 0.02\ mol\end{aligned}

 

Step 3: Calculate the total number of moles of Zn2+ present.

\begin{aligned}n(Zn^{2+})_{final} &= 0.02 + 3.146718 \times 10^{-3} \\&= 0.0231467\ mol\end{aligned}

 

Step 4: Calculate the final concentration of Zn2+.

\begin{aligned}c(Zn^{2+})_{final} &= \dfrac{0.0231467}{ 0.2} \\&= 0.116\ M (3 sig.fig.)\end{aligned}

Back to Question 7

8Answer (D)

More active metals displace less active metals from solution.

Metal C displaced Metal B in solution. Hence, Metal C is more active than Metal B (C > B).

Metal A displaced Metal B in solution. Hence, Metal A is more active than Metal B (A > B).

Metal A didn’t displace Metal C in solution. Hence, Metal A is less reactive than Metal C (A < C).

Back to Question 8

9Catalysts provide an alternative pathway of lower activation energy for a reaction, e.g. by acting as a surface on which the reactants can meet in the correct orientation, or forming intermediates with the reactants. This decreased activation energy allows the reaction to occur at lower temperatures, thus increasing the rate of reaction. While they are not used up in the reaction (and are therefore not reactants), they do most certainly take part in the reaction. Thus the statement is inaccurate.

Back to Question 9

10According to collision theory:

Particles must collide in the correct orientation, with enough kinetic energy in order to react.

When the temperature is increased, the kinetic energy is increased.

This causes a large number of collisions to have the required activation energy for the reaction to occur (as seen in the above diagram – the areas of the shaded regions are proportional to the number of particles which can react)

As more collisions have the required activation energy with increased temperature, more reactants will react and hence reaction rates increase.

Back to Question 10

Back to Module 3 Reactive Chemistry Practice Test

 

Do you know all your chemical reactions for Module 3 ‘Reactive Chemistry’?

Check out our cheatsheet containing all the chemical reactions you need to know for Year 11 Chemistry.

 

Written by Varisara Laosuksri

Varisara is a 2019 St George Girls High School graduate who achieved Band 6 in her HSC Chemistry and Physics.

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