IB Chemistry SL Reactivity 3: What are the mechanisms of chemical change? questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

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Question 1

Paper 2Medium5 marks

An aqueous sodium hydrogencarbonate solution is tested with a pH probe. At $25\,^\circ\mathrm{C}$ the pH is 8.30. At this temperature, neutral water has $[\mathrm{H^+}]=1.00\times10^{-7}\ \mathrm{mol\,dm^{-3}}$

A student concludes that $\mathrm{HCO_3^-}$ can act only as a Brønsted–Lowry base because the solution is alkaline.

(a)

Calculate $[\mathrm{H^+}]$ in the solution, in $\mathrm{mol\,dm^{-3}}$ to two significant figures.

[1]

(b)

Using your answer to (a), determine whether the solution is acidic, neutral or alkaline.

[1]

(c)

Write equations with state symbols to show $\mathrm{HCO_3^-}$ acting as a Brønsted–Lowry acid and as a Brønsted–Lowry base in water. Use the equations to assess the student's conclusion.

[3]

Question 2

Multiple choiceMedium1 mark

A voltaic cell operates by the reaction $\mathrm{Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)}$ Initially, $0.0040\ \mathrm{mol}$ Mg and $0.0060\ \mathrm{mol}$ HCl are available. What amount of $\mathrm{H_2}$ forms, and what is the sign of the electrode where it forms?

Show answer and explanation

Answer: D. The equation requires $2\ \mathrm{mol}$ HCl per $1\ \mathrm{mol}$ H₂, so HCl limits the reaction: $n(\mathrm{H_2})=0.0060/2=0.0030\ \mathrm{mol}$ H⁺ is reduced at the hydrogen electrode, making it the cathode; the cathode is positive in a voltaic cell.
A: gives the correct amount but assigns the hydrogen electrode the negative sign.
B: uses the amount of Mg as though it all reacts, despite insufficient HCl.
C: uses the Mg amount without checking the limiting reagent and also assigns the cathode a negative sign.
Remember: in a voltaic cell, reduction occurs at the positive cathode.

Question 3

Paper 2Medium6 marks

A student investigates displacement reactions between halogens and halide ions in aqueous solution. The results are shown below. Chlorine, bromine and iodine occur in this order from top to bottom in Group 17.

Halogen added	Halide ions present	Result
Cl₂(aq)	Br⁻(aq)	Reaction
Br₂(aq)	I⁻(aq)	Reaction
I₂(aq)	Br⁻(aq)	No reaction

(a)

Write the balanced ionic equation for the reaction between Br₂(aq) and I⁻(aq).

[2]

(b)

Using your equation in (a), identify the species oxidized.

[1]

(c)

A student predicts that I₂(aq) reacts with Cl⁻(aq). Predict whether this reaction occurs. Use the results and the positions of the elements in Group 17 to justify your answer.

[3]

Question 4

Multiple choiceMedium1 mark

Four oxygen-containing species are compared: OH, OH⁻, H₂O and H₃O⁺. Which analysis correctly identifies a radical and predicts its reactivity from its electron arrangement?

Show answer and explanation

Answer: C. Neutral OH has 6 + 1 = 7 valence electrons. Its Lewis representation has one unpaired electron, so it is a radical and is expected to be highly reactive.

A: omits the extra electron from the negative charge; OH⁻ has 8 valence electrons, all paired. B: counts H₂O correctly but mistakes paired lone-pair electrons for an unpaired electron. D: fails to subtract one electron for the positive charge; H₃O⁺ has 8 valence electrons, all paired.

Remember: add an electron for a negative charge and subtract one for a positive charge.

Question 5

Paper 2Medium6 marks

CFC-12, CF₂Cl₂, can reach the stratosphere. UV radiation can break one of its C—Cl bonds. In a simplified ozone-depletion sequence, a chlorine radical reacts with ozone to form a chlorine monoxide radical and oxygen. The chlorine monoxide radical then reacts with an oxygen radical: ClO• + O• → Cl• + O₂

(a)

Write the equation for the UV-induced breaking of one C—Cl bond in CF₂Cl₂, showing UV above the arrow. State whether this is an initiation, propagation or termination step.

[2]

(b)

Represent the bond being broken as C—Cl. Draw two single-barbed (fish-hook) arrows to show the movement of electrons during this bond breaking.

[1]

(c)

Write the balanced equation for the reaction of a chlorine radical with ozone described in the question.

[1]

(d)

Combine the reactions in the question and in part (c) to give the overall reaction. State why the chlorine radical can continue the sequence.

[2]

Question 6

Multiple choiceMedium1 mark

The displayed equation shows a reaction between hydrogen chloride and hydroxide ions. Which species donates both electrons to the new bond, and which atom receives the donated pair?

Show answer and explanation

Answer: D. A nucleophile supplies both electrons in the new bond. The figure shows an O—H bond in water and the loss of the H—Cl bond; therefore O in OH⁻ supplies the pair to H in HCl.

A: Reverses the donor role: HCl provides the H atom, but does not supply the pair that forms the new bond.
B: Identifies OH⁻ as the donor but mistakes the Cl atom, which leaves as chloride, for the atom receiving the pair.
C: Treats the negatively charged chloride product as the nucleophile; it is formed after the O—H bond is made.

Remember: identify a nucleophile by the source of the bonding pair, not merely by its charge.

Question 7

Paper 2Medium6 marks

In aqueous solution, bromoethane reacts with hydroxide ions. Figure 1 shows the reactants; the products are not shown.

(a)

On Figure 1, draw one curly arrow to show the movement of the bonding pair in the C—Br bond during the reaction.

[1]

(b)

Write a balanced ionic equation for the reaction, using structural formulae for the organic compounds and including state symbols.

[2]

(c)

Using Figure 1 and your equation in (b), deduce the type of reaction and explain the electron-pair changes at outcome level. Do not give a detailed mechanism.

[3]

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