Silicon tetrachloride, $\text{SiCl}_4$, is formed when silicon reacts with chlorine under suitable conditions. It is a colourless liquid with a low boiling point. (a) Explain why $\text{SiCl}_4$ has a low boiling point. (b) $\text{SiCl}_4$ reacts with water to produce an acidic solution. (i) Write an equation for this reaction. (ii) Describe two visual observations when silicon tetrachloride is added drop by drop to a small amount of water. 1.…

(a) Simple molecular; weak London forces; little energy needed to overcome them. · (b)(i) SiCl4(l) + 2H2O(l) -> SiO2(s) + 4HCl(aq) · (b)(ii) White solid; steamy/white/misty fumes. · (b)(iii) 1.6 · (c)(i) 223 J K^-1 mol^-1 · (c)(ii) Decrease in number of moles of gas (2 mol Cl2(g) to 0 mol gas). · (d) 2840 K

Chemistry 9701/42 — May/June 2018

Cambridge A-Level · A Level Structured Questions · worked solutions for every part, with the mark scheme

questions

100

marks

120

minutes

Topics Introduction to A Level Organic Chemistry · Equilibria · Electrochemistry · Hydroxy Compounds · Nitrogen Compounds · Chemical Energetics · +5 more

Q1EquilibriaChemical EnergeticsFree sample

Silicon tetrachloride, $\text{SiCl}_4$ , is formed when silicon reacts with chlorine under suitable conditions. It is a colourless liquid with a low boiling point.

(a)

Explain why $\text{SiCl}_4$ has a low boiling point.

DifficultyMedium-Easy

Worked solution

Answer

$\text{SiCl}_4$ is a simple molecular (simple covalent) substance. The molecules are held together by weak London (instantaneous induced dipole–induced dipole) forces. Only a small amount of energy is needed to overcome these weak intermolecular forces, so the boiling point is low.

Final answer

Simple molecular; weak London forces; little energy needed to overcome them.

Detailed explanation

Background Concept

Simple molecular substances consist of discrete molecules. The covalent bonds within each molecule are strong, but the forces between molecules (intermolecular forces) are weak. Boiling a liquid requires separating the molecules, so it is the intermolecular forces that must be overcome, not the covalent bonds. For non-polar molecules such as $\text{SiCl}_4$ , the only intermolecular forces are London (instantaneous induced dipole–induced dipole) forces, which are weak.

Understanding the Question

The question asks why $\text{SiCl}_4$ , a covalent liquid, has a low boiling point. It is an 'explain' question, so the answer must link structure to the energy needed to boil.

Approach

Identify the type of structure (simple molecular), name the intermolecular forces (weak London forces), and state that little energy is needed to overcome them.

Step-by-Step Reasoning

$\text{SiCl}_4$ is made of discrete tetrahedral molecules. Between these molecules there are only weak London forces. When the liquid boils, molecules are separated from each other, which requires breaking these weak intermolecular forces. Because the forces are weak, only a small amount of energy is needed, so the boiling point is low. The strong Si–Cl covalent bonds are not broken during boiling.

Key Takeaways

For molecular substances, boiling point is controlled by the strength of intermolecular forces, not by the strength of covalent bonds within molecules.

Common Mistakes

Saying the Si–Cl bonds are weak or are broken on boiling. Covalent bonds are strong and stay intact.
Saying hydrogen bonds are present. $\text{SiCl}_4$ has no H bonded to N, O or F.
Omitting the 'simple molecular' point, which is a separate mark.

Things to Be Careful About

The mark scheme credits 'simple molecular / simple covalent' and 'weak London / id-id / VDW forces' with a small amount of energy to break. Use the term 'London forces' or 'van der Waals forces' rather than just 'intermolecular forces'.

Techniques used

identify simple molecular structuredescribe weak London forcesrelate intermolecular forces to boiling point

(b)

$\text{SiCl}_4$ reacts with water to produce an acidic solution.

(i)

Write an equation for this reaction.

DifficultyEasy

Worked solution

Answer

$\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{aq})$

Final answer

SiCl4(l) + 2H2O(l) -> SiO2(s) + 4HCl(aq)

Detailed explanation

Background Concept

Silicon tetrachloride is a covalent chloride of a non-metal. When added to water, it undergoes hydrolysis: chlorine leaves as HCl and silicon combines with oxygen/hydroxide to form silicon dioxide or silicic acid. The HCl makes the solution acidic.

Understanding the Question

The question asks for a balanced equation for the reaction of $\text{SiCl}_4$ with water. The acidic product is HCl, so the equation must show HCl and either $\text{SiO}_2$ or $\text{Si(OH)}_4$ .

Approach

Write the reactants and likely products, then balance atoms: Si, Cl, H and O.

Step-by-Step Reasoning

Start with $\text{SiCl}_4 + \text{H}_2\text{O} \rightarrow \text{SiO}_2 + \text{HCl}$ . Balance Cl: 4 HCl. Balance H: 4 H on the right, so need 2 $\text{H}_2\text{O}$ . Check O: 2 O in 2 $\text{H}_2\text{O}$ and 2 O in $\text{SiO}_2$ . Balanced. The alternative $\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{Si(OH)}_4 + 4\text{HCl}$ is also accepted.

Key Takeaways

Covalent chlorides of non-metals hydrolyse to give an oxide/hydroxide and HCl. Balancing requires care with H and O.

Common Mistakes

Writing an unbalanced equation.
Using wrong products such as $\text{Si(OH)}_2$ or $\text{SiCl}_2$ .
Forgetting that HCl is the acidic product.

Things to Be Careful About

State symbols are not always required, but if used they must be correct: $\text{SiCl}_4(\text{l})$ , $\text{H}_2\text{O}(\text{l})$ , $\text{SiO}_2(\text{s})$ , $\text{HCl}(\text{aq})$ .

Techniques used

write balanced hydrolysis equationbalance atoms with state symbols

(ii)

Describe two visual observations when silicon tetrachloride is added drop by drop to a small amount of water.

$\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots$
$\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots$

DifficultyEasy

Worked solution

Answer

A white solid is formed.
Steamy/white/misty fumes are given off.

Final answer

White solid; steamy/white/misty fumes.

Detailed explanation

Background Concept

The hydrolysis of $\text{SiCl}_4$ produces a white solid, $\text{SiO}_2$ (or $\text{Si(OH)}_4$ ), and hydrogen chloride gas. Hydrogen chloride gas is colourless but reacts with moisture in the air to form a mist of hydrochloric acid droplets, seen as steamy/white/misty fumes.

Understanding the Question

The question asks for two visual observations when $\text{SiCl}_4$ is added dropwise to water. It is a recall/observation question, so give the visible changes, not the chemical explanation.

Approach

Use the products of the hydrolysis: a white solid forms and acidic fumes are given off.

Step-by-Step Reasoning

The solid product $\text{SiO}_2$ is white and insoluble, so a white solid is seen. The HCl produced is a gas; with water vapour it forms a white mist, so steamy/white/misty fumes are observed.

Key Takeaways

Hydrolysis of covalent chlorides gives a solid oxide/hydroxide and HCl fumes.

Common Mistakes

Giving 'effervescence' or 'bubbles' as an observation; this is not on the mark scheme.
Giving two versions of the same fume observation as two separate observations.
Forgetting to say the solid is white.

Things to Be Careful About

'Steamy fumes', 'white fumes' and 'misty fumes' are all accepted for one mark. The white solid is the other mark.

Techniques used

describe visual observations of hydrolysisidentify white solid and acidic fumes

(iii)

A sample of $0.8505\text{ g}$ of $\text{SiCl}_4$ is added to $800\text{ cm}^3$ of water. All of the soluble acidic product is dissolved in the water.

Calculate the pH of the solution obtained.

\text{pH} = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots

DifficultyMedium-Easy

Worked solution

Working

$M_r(\text{SiCl}_4) = 28.1 + 4(35.5) = 170.1$

$n(\text{SiCl}_4) = \frac{0.8505}{170.1} = 0.00500\ \text{mol}$

Each $\text{SiCl}_4$ gives 4 $\text{H}^+$ :

$n(\text{H}^+) = 4 \times 0.00500 = 0.0200\ \text{mol}$

$[\text{H}^+] = \frac{0.0200}{0.800} = 0.0250\ \text{mol dm}^{-3}$

$\text{pH} = -\log_{10}(0.0250) = 1.60$

Answer

$\text{pH} = 1.6$

Final answer

1.6

Detailed explanation

Background Concept

pH is defined as $\text{pH} = -\log_{10}[\text{H}^+]$ . Hydrochloric acid is a strong acid, so it fully dissociates and $[\text{H}^+] = [\text{HCl}]$ . The stoichiometry of the hydrolysis is important: each mole of $\text{SiCl}_4$ produces 4 moles of HCl and therefore 4 moles of $\text{H}^+$ .

Understanding the Question

A known mass of $\text{SiCl}_4$ is added to a known volume of water. All the soluble acidic product (HCl) dissolves. We need the pH of the resulting solution.

Approach

Convert mass to moles using $M_r$ ; multiply by 4 to get moles of $\text{H}^+$ ; divide by the volume in $\text{dm}^3$ to get concentration; take the negative log.

Step-by-Step Reasoning

$M_r(\text{SiCl}_4) = 28.1 + 4(35.5) = 170.1$ . Moles $= 0.8505/170.1 = 0.00500$ mol. Moles of $\text{H}^+ = 4 \times 0.00500 = 0.0200$ mol. Volume $= 800\ \text{cm}^3 = 0.800\ \text{dm}^3$ . $[\text{H}^+] = 0.0200/0.800 = 0.0250\ \text{mol dm}^{-3}$ . $\text{pH} = -\log_{10}(0.0250) = 1.60$ , which rounds to 1.6.

Key Takeaways

Strong acid pH calculations require moles of $\text{H}^+$ , not moles of the original compound, and volume in $\text{dm}^3$ .

Common Mistakes

Forgetting to multiply by 4.
Using 800 instead of 0.800 for the volume.
Writing $\text{pH} = \log[H^+]$ instead of $-\log[H^+]$ .
Using the mass of HCl rather than $\text{SiCl}_4$ .

Things to Be Careful About

The mark scheme gives pH = 1.6. Use at least two significant figures in intermediate steps. The volume of the solution is taken as 800 $\text{cm}^3$ ; the small volume of $\text{SiCl}_4$ added is ignored.

Techniques used

calculate moles from mass and molar massuse stoichiometry to find H+ concentrationcalculate pH from hydrogen ion concentration

(c)

(i)

Silicon tetrachloride can be prepared according to reaction 1.

\text{reaction 1} \quad \text{Si(s)} + 2\text{Cl}_2\text{(g)} \rightarrow \text{SiCl}_4\text{(l)} \quad \Delta S^\ominus = -225.7\text{ J K}^{-1}\text{ mol}^{-1}


standard entropy of silicon, $S^\ominus$ $\text{Si(s)}$	$18.7\text{ J K}^{-1}\text{ mol}^{-1}$
standard entropy of silicon tetrachloride, $S^\ominus$ $\text{SiCl}_4\text{(l)}$	$239.0\text{ J K}^{-1}\text{ mol}^{-1}$

Calculate the standard entropy of chlorine, $S^\ominus$ $\text{Cl}_2\text{(g)}$ . Show all your working.

S^\ominus \text{ Cl}_2\text{(g)} = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots \text{ J K}^{-1}\text{ mol}^{-1}

DifficultyMedium-Easy

Worked solution

Working

$\Delta S^\ominus = S^\ominus(\text{SiCl}_4) - [S^\ominus(\text{Si}) + 2S^\ominus(\text{Cl}_2)]$

Let $x = S^\ominus(\text{Cl}_2)$ .

$-225.7 = 239.0 - (18.7 + 2x)$

$18.7 + 2x = 464.7$

$2x = 446.0$

$x = 223.0$

Answer

$S^\ominus(\text{Cl}_2) = 223\ \text{J K}^{-1}\text{ mol}^{-1}$

Final answer

223 J K^-1 mol^-1

Detailed explanation

Background Concept

For a reaction, the standard entropy change is

$\Delta S^\ominus = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants})$

Each substance is multiplied by its stoichiometric coefficient. Standard entropies are usually quoted in $\text{J K}^{-1}\text{ mol}^{-1}$ .

Understanding the Question

We are given $\Delta S^\ominus$ for the formation of $\text{SiCl}_4$ and the standard entropies of Si and $\text{SiCl}_4$ . We need to find the standard entropy of $\text{Cl}_2$ .

Approach

Write the entropy change equation for the reaction, substitute the known values, let $x = S^\ominus(\text{Cl}_2)$ , and solve.

Step-by-Step Reasoning

For $\text{Si(s)} + 2\text{Cl}_2(\text{g}) \rightarrow \text{SiCl}_4(\text{l})$ :

$\Delta S^\ominus = S^\ominus(\text{SiCl}_4) - [S^\ominus(\text{Si}) + 2S^\ominus(\text{Cl}_2)]$

Substitute: $-225.7 = 239.0 - (18.7 + 2x)$ . Rearrange: $18.7 + 2x = 239.0 + 225.7 = 464.7$ . Then $2x = 446.0$ , so $x = 223.0\ \text{J K}^{-1}\text{ mol}^{-1}$ .

Key Takeaways

Entropy changes are calculated using products minus reactants, with coefficients included.

Common Mistakes

Forgetting the factor 2 for $\text{Cl}_2$ .
Getting the sign wrong when rearranging.
Mixing up $\Delta S$ with $S$ .

Things to Be Careful About

All values are in $\text{J K}^{-1}\text{ mol}^{-1}$ , not kJ. The answer is positive, about 223.

Techniques used

apply entropy change formularearrange for unknown entropysolve linear equation

(ii)

Explain why the entropy change for reaction 1 is negative.

DifficultyEasy

Worked solution

Answer

The number of moles of gas decreases: 2 mol of $\text{Cl}_2$ gas on the left become 0 mol of gas on the right. Gases have much higher entropy than solids/liquids, so the system becomes more ordered and $\Delta S^\ominus$ is negative.

Final answer

Decrease in number of moles of gas (2 mol Cl2(g) to 0 mol gas).

Detailed explanation

Background Concept

Entropy is a measure of disorder. Gases have much higher entropy than solids or liquids because their particles are far apart and move freely. A reaction that consumes gas and produces a condensed phase has a negative entropy change.

Understanding the Question

The reaction $\text{Si(s)} + 2\text{Cl}_2(\text{g}) \rightarrow \text{SiCl}_4(\text{l})$ has $\Delta S^\ominus = -225.7\ \text{J K}^{-1}\text{ mol}^{-1}$ . We need to explain the negative sign.

Approach

Compare the number of moles of gas on each side of the equation.

Step-by-Step Reasoning

On the left there are 2 mol of $\text{Cl}_2$ gas (Si is solid). On the right there is only liquid $\text{SiCl}_4$ , so no gas. The number of moles of gas decreases from 2 to 0. Since gases contribute far more to entropy than solids or liquids, the products are more ordered than the reactants, so $\Delta S^\ominus$ is negative.

Key Takeaways

A decrease in the number of moles of gas is a reliable indicator of a negative entropy change.

Common Mistakes

Saying 'the number of moles decreases' without specifying gas.
Treating solid Si as a gas.
Saying the products are more disordered.

Things to Be Careful About

The mark scheme accepts 'decrease in number of moles of gas' or 'more moles of gas on the left/reactants'. Either wording is fine.

Techniques used

compare moles of gas on each siderelate gas moles to entropy

(d)

The standard enthalpy change of formation of silicon tetrachloride, $\text{SiCl}_4\text{(l)}$ , is $-640\text{ kJ mol}^{-1}$ .

Reaction 1 is spontaneous at lower temperatures, but it is not spontaneous at very high temperatures.

Calculate the temperature above which reaction 1 is not spontaneous.

\text{temperature} = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots \text{ K}

DifficultyMedium-Easy

Worked solution

Working

At the threshold, $\Delta G = 0$ :

$\Delta G = \Delta H - T\Delta S = 0$

$T = \frac{\Delta H}{\Delta S}$

$\Delta H = -640\ \text{kJ mol}^{-1} = -640000\ \text{J mol}^{-1}$

$T = \frac{-640000}{-225.7} = 2835.6\ \text{K} \approx 2840\ \text{K}$

Above this temperature, $\Delta G > 0$ , so the reaction is not spontaneous.

Answer

$\text{temperature} = 2840\ \text{K}$

Final answer

2840 K

Detailed explanation

Background Concept

The Gibbs free energy change determines feasibility: $\Delta G = \Delta H - T\Delta S$ . A reaction is spontaneous when $\Delta G < 0$ . At the temperature where $\Delta G = 0$ , the reaction is just at the boundary between spontaneous and non-spontaneous. For reaction 1, $\Delta H$ is negative (exothermic) and $\Delta S$ is negative, so the $-T\Delta S$ term is positive and increases with temperature.

Understanding the Question

We are given $\Delta H_f^\ominus(\text{SiCl}_4) = -640\ \text{kJ mol}^{-1}$ and $\Delta S^\ominus = -225.7\ \text{J K}^{-1}\text{ mol}^{-1}$ . We need the temperature above which the reaction is not spontaneous.

Approach

Set $\Delta G = 0$ , rearrange to $T = \Delta H/\Delta S$ , convert $\Delta H$ to joules, and calculate.

Step-by-Step Reasoning

At the threshold, $\Delta H - T\Delta S = 0$ , so $T = \Delta H/\Delta S$ . Convert: $\Delta H = -640\ \text{kJ mol}^{-1} = -640000\ \text{J mol}^{-1}$ . Substitute: $T = (-640000)/(-225.7) = 2835.6\ \text{K} \approx 2840\ \text{K}$ . Above this temperature, the positive $-T\Delta S$ term is larger in magnitude than $\Delta H$ , so $\Delta G > 0$ and the reaction is not spontaneous.

Key Takeaways

For exothermic reactions with negative $\Delta S$ , spontaneity decreases as temperature increases; the threshold temperature is $T = \Delta H/\Delta S$ .

Common Mistakes

Not converting kJ to J.
Using positive values for $\Delta H$ or $\Delta S$ .
Forgetting that both are negative, so $T$ is positive.
Giving the answer in °C instead of K.

Things to Be Careful About

The mark scheme accepts 2836 or 2840 K. Show the equation with $\Delta G = 0$ . Use K, not °C.

Techniques used

apply Gibbs free energy equationset ΔG = 0 at thresholdconvert kJ to Jcalculate temperature

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