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Worked Solutions

Module 3: Reactive Chemistry — Worked Solutions (Preliminary Chemistry)

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Worked examples for Preliminary Chemistry Module 3: Reactive Chemistry. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.

How to use these

Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.

Example 1 — Single displacement and metal reactivity

Standard 3 marks

Question

A strip of zinc metal is placed into a colourless solution of copper(II) sulfate. Predict whether a reaction occurs, write a balanced full equation and the net ionic equation, and explain your prediction using the activity series of metals.

Solution

Zinc is above copper in the activity series, so zinc is the more reactive metal and will displace copper from solution. A reaction occurs.

Full equation:

$\text{Zn}_{(s)} + \text{CuSO}_{4(aq)} \rightarrow \text{ZnSO}_{4(aq)} + \text{Cu}_{(s)}$

Sulfate is a spectator ion, so the net ionic equation is:

$\text{Zn}_{(s)} + \text{Cu}^{2+}_{(aq)} \rightarrow \text{Zn}^{2+}_{(aq)} + \text{Cu}_{(s)}$

Zinc is oxidised, copper(II) is reduced. The blue colour fades as $\text{Cu}^{2+}$ leaves solution and reddish copper deposits on the zinc. Always justify with the activity series — "more reactive metal displaces the less reactive one".

Where the marks go

  • 1 mark: Predicts a reaction occurs and justifies using the activity series (Zn more reactive than Cu)
  • 1 mark: Correct balanced full equation with state symbols
  • 1 mark: Correct net ionic equation with sulfate identified as a spectator

Key idea

In a single displacement reaction, a more reactive metal (higher in the activity series) displaces a less reactive metal from solution; spectator ions are cancelled to give the net ionic equation.

Example 2 — Factors affecting reaction rate

Standard 4 marks

Question

Marble chips ($\text{CaCO}_3$) react with dilute hydrochloric acid. The balanced equation is $\text{CaCO}_{3(s)} + 2\text{HCl}_{(aq)} \rightarrow \text{CaCl}_{2(aq)} + \text{H}_2\text{O}_{(l)} + \text{CO}_{2(g)}$. Explain, in terms of collision theory, how (a) increasing the temperature of the acid and (b) crushing the marble chips into a powder each increase the rate of this reaction.

Solution

Collision theory: a reaction occurs only when particles collide with sufficient energy (the activation energy) and correct orientation. Rate depends on the frequency of successful collisions.

(a) Higher temperature. Particles gain kinetic energy, so they move faster and collide more often. More importantly, a greater proportion of collisions now have energy $\geq$ the activation energy. Both effects increase the rate.

(b) Crushing into powder. Powdering increases the surface area of the solid $\text{CaCO}_3$ exposed to the acid. More exposed particles means more frequent collisions between acid particles and the solid surface, so the rate increases.

Link every factor back to collision frequency or collision energy — that's the marking language the examiner wants.

Where the marks go

  • 1 mark: States collision theory: reaction needs collisions with sufficient energy ($E_a$) and correct orientation
  • 1 mark: Explains higher temperature increases collision frequency
  • 1 mark: Explains higher temperature increases the proportion of collisions reaching the activation energy
  • 1 mark: Explains crushing increases surface area, raising collision frequency at the solid surface

Key idea

Reaction rate depends on the frequency of successful collisions: temperature raises both collision frequency and collision energy, while increasing surface area raises collision frequency at the solid–liquid boundary.