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

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Question 1

Multiple choiceMedium1 mark

An aqueous solution of sodium hydrogencarbonate is at 298 K. For $H_2CO_3$ $K_{\mathrm a1}=4.3\times10^{-7}$ and $K_{\mathrm a2}=4.7\times10^{-11}$ $K_{\mathrm w}=1.0\times10^{-14}$ Which option pairs the predominant hydrolysis of $HCO_3^-$ with the expected pH relative to 7?

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Answer: D. $HCO_3^-$ can act as an acid or a base. Its base strength is $K_{\mathrm b}=K_{\mathrm w}/K_{\mathrm a1}=2.3\times10^{-8}$ which is greater than its acid strength, $K_{\mathrm a2}=4.7\times10^{-11}$ The predominant hydrolysis produces $OH^-$ so the solution is basic.

A: Treats hydrogencarbonate as an acid and predicts the acidic effect of producing $H_3O^+$ although this hydrolysis is weaker.
B: Identifies the predominant hydrolysis but reverses the pH effect of producing $OH^-$
C: Predicts a basic solution but pairs it with acid hydrolysis, which produces $H_3O^+$

Remember: for an amphiprotic ion, compare its acid strength with its base strength.

Question 2

Multiple choiceMedium1 mark

A piece of magnesium is added to dilute hydrochloric acid. Which equation represents the reaction?

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Answer: B. Magnesium reacts with dilute hydrochloric acid to form magnesium chloride and hydrogen gas. Magnesium chloride has the formula MgCl₂; balancing the atoms gives Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).
A: identifies the products but leaves chlorine unbalanced by using one HCl.
C: balances the atoms but gives magnesium the wrong charge in the salt formula, writing MgCl instead of MgCl₂.
D: balances the atoms but represents the hydrogen as individual atoms rather than diatomic hydrogen, H₂.
Remember: hydrogen released in these reactions is H₂(g).

Question 3

Multiple choiceHard1 mark

The alkane shown reacts with $\mathrm{Cl_2}$ under UV light. Which sequence shows initiation followed by the two propagation steps in which one H atom at C2 is replaced by Cl? A dot denotes one unpaired electron.

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Answer: C. UV light causes homolytic fission of $\mathrm{Cl_2}$ producing two chlorine radicals. A chlorine radical abstracts H from C2, forming $\mathrm{HCl}$ and $\mathrm{CH_3CH\boldsymbol{\cdot}CH_3}$ This radical reacts with $\mathrm{Cl_2}$ to form $\mathrm{CH_3CHClCH_3}$ and regenerate $\mathrm{Cl\boldsymbol{\cdot}}$ sustaining the chain. A: abstracts H at a terminal carbon, so the product is 1-chloropropane. B: uses radical–radical combination, a termination step that consumes the radicals rather than regenerating one. D: shows heterolytic fission, producing ions rather than the radicals needed to initiate the chain. Remember: a propagation step consumes one radical and produces another.

Question 4

Multiple choiceEasy1 mark

In a reaction mechanism for bromomethane, a curly arrow starts at the C–Br bond and points to Br. Which equation represents the products of this bond cleavage?

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Answer: C. The curly arrow from the bond to Br shows the bonding electron pair moving to Br. Carbon loses its share of the pair, giving CH₃⁺; bromine gains both electrons, giving Br⁻.
A: splits the pair between the fragments, producing radicals by homolytic fission.
B: reverses the electron-pair movement, assigning both electrons to carbon.
D: represents cleavage of a C–H bond rather than the specified C–Br bond.
Remember: a curly arrow from a bond to an atom shows that the bonding pair moves to that atom.

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