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

Paper 2Medium6 marks

Chlorine reacts with propane when irradiated with ultraviolet (UV) light. Figure 1 shows the displayed structural formula of propane.

(a)

Determine the number of different monochlorinated structural isomers that can form from propane.

[1]

(b)

Write the equation for the initiation step, using single-barbed arrows to show the movement of electrons.

[2]

(c)

For the formation of 2-chloropropane:

(i)

Write the first propagation equation, in which a hydrogen atom is transferred from propane to a chlorine radical.

[1]

(ii)

Using the propan-2-yl radical, CH₃CH•CH₃, write the second propagation equation.

[1]

(d)

Write one equation for a termination step in this chain reaction.

[1]

Question 4

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 5

Paper 2Medium6 marks

Boron trifluoride forms a 1:1 adduct with ammonia:

Reaction I: BF₃ + NH₃ → BF₃·NH₃

A proton reacts with water to form a hydronium ion:

Reaction II: H⁺ + H₂O → H₃O⁺

(a)

Identify the electrophile in reaction I and state which reactant supplies both electrons in the new B–N bond.

[2]

(b)

Identify the electrophile in reaction II and explain the origin of both electrons in the new O–H bond.

[2]

(c)

A student claims that an electrophile must be positively charged. Assess this claim using both reactions.

[2]

Question 6

Paper 2Hard12 marks

Boron trifluoride reacts with ammonia to form a boron-containing adduct. A sample initially contains $1.00\ \mathrm{mmol}$ of $\mathrm{BF_3}$ Ammonia is added in measured amounts, and the amount of adduct formed is determined.

The following two stages occur when propanone reacts with cyanide ions in the presence of hydrogen cyanide:

\mathrm{(CH_3)_2C{=}O + CN^- \longrightarrow (CH_3)_2C(O^-)(C{\equiv}N)}

\mathrm{(CH_3)_2C(O^-)(C{\equiv}N) + H{-}C{\equiv}N \longrightarrow (CH_3)_2C(OH)(C{\equiv}N) + CN^-}

In the product of the first stage, the carbon atom of $\mathrm{CN^-}$ is bonded to the carbon atom that was part of the carbonyl group.

(a)

Use Figure 1 to estimate the amount of $\mathrm{NH_3}$ at which the amount of adduct formed first becomes approximately constant. Hence, deduce the stoichiometric ratio of $\mathrm{BF_3}$ to $\mathrm{NH_3}$ in the adduct.

[2]

(b)

Write an equation for the formation of the adduct, showing the coordinate bond, and identify the Lewis acid and Lewis base.

[2]

(c)

State why the formation of the adduct in (b) is not a Brønsted–Lowry acid–base reaction.

[1]

(d)

A student claims, “Since each stage in the propanone reaction involves an electron-pair donor and an electron-pair acceptor, both stages must also be Brønsted–Lowry acid–base reactions.”

Annotate the two stages with curly arrows to show electron-pair movement. Identify the Lewis acid and Lewis base in each stage, and evaluate the student’s claim.

[7]

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