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Question 1

Topic: 9.3

(a) (i) Explain the lack of reactivity of nitrogen gas, N₂(g).

(ii) Covalent bonds can be σ bonds or π bonds. Complete Table 1.1 to show the number of σ and π bonds in a molecule of N₂ and to describe how the orbitals overlap to form σ and π bonds

(b) (i) A sample of Al reacts with an excess of Cl₂. State the oxidation number of Al in the product of the reaction.
(ii) State what determines the maximum oxidation number of the Period 3 elements in their oxides.

(c) Separate samples of aluminium oxide, Al₂O₃, and phosphorus(V) oxide, P₄O₁₀, react with an excess of NaOH(aq) at room temperature.
(i) Give the state of Al₂O₃ and P₄O₁₀ at room temperature.
(ii) Write an equation for the reaction of each oxide with an excess of NaOH(aq) at room temperature.
Al₂O₃ + ……………………
P₄O₁₀ + ……………………

(d) The oxide of silicon reacts with calcium oxide in an addition reaction to produce calcium silicate, CaSiO₃. The oxidation number of calcium in CaSiO₃ is +II.

(i) Deduce the oxidation number of silicon in calcium silicate.
(ii) Calcium oxide can be made from calcium carbonate in a single-step reaction. Identify the type of reaction that occurs.

▶️ Answer/Explanation

Solution

(a)(i) Nitrogen gas (N₂) is unreactive due to its strong triple covalent bond (high bond dissociation energy) and non-polar nature, making it stable under standard conditions.

(a)(ii)

Explanation: In N₂, there is 1 σ bond (formed by head-on overlap of sp hybrid orbitals) and 2 π bonds (formed by sidewise overlap of p orbitals).

(b)(i) Oxidation number of Al: +III.

Explanation: Al forms AlCl₃ with Cl₂, where Cl has an oxidation state of -1, so Al must be +3 to balance the charges.

(b)(ii) The maximum oxidation number is determined by the number of valence electrons in the Period 3 element.

(c)(i) Both Al₂O₃ and P₄O₁₀ are in the solid state at room temperature.

(c)(ii)

Al₂O₃ + 2NaOH + 3H₂O → 2NaAl(OH)₄

P₄O₁₀ + 12NaOH → 4Na₃PO₄ + 6H₂O

Explanation: Al₂O₃ reacts to form sodium aluminate (NaAl(OH)₄), while P₄O₁₀ reacts to form sodium phosphate (Na₃PO₄).

(d)(i) Oxidation number of silicon: +IV.

Explanation: In CaSiO₃, Ca is +2, and each O is -2. Solving for Si: +2 + Si + 3(-2) = 0 ⇒ Si = +4.

(d)(ii) The reaction is thermal decomposition.

Explanation: CaCO₃ decomposes upon heating to form CaO and CO₂.

Question 2

Topic: 5.1

\(N_2(g)\) reacts with \(H_2(g)\) in the Haber process, as shown in reaction 1.

Table 2.1 shows the different conditions used to produce three equilibrium mixtures, A, B and C.

(a) Describe and explain the change, if any, to the percentage yield of NH₃(g) produced in B compared to A.
(b) (i) Describe and explain the change, if any, to the percentage yield of NH₃(g) produced in C compared to A.
(ii) Describe and explain the change to the rate of the forward reaction that occurs to establish the equilibrium in C compared to A. You do not need to refer to the Boltzmann distribution in your answer.

(c) (i) Write an expression for the equilibrium constant, \(K_p\), for reaction 1. State the units.
(ii) Equilibrium mixture D is made when 1.0mol of \(N_2(g)\) and 3.0mol of \(H_2(g)\) are added to a sealed container at 750°C and 1000 atm and left to reach equilibrium. This mixture contains 1.16mol of NH₃(g). Calculate the mole fraction of NH₃(g) in D.
(iii) The mole fraction of \(N_2(g)\) is 0.625 in a new equilibrium mixture, E. Calculate the partial pressure of \(N_2(g)\) in E when the total pressure is 1000 atm.

(d) When oxides of nitrogen escape into the atmosphere they may be involved in:
• formation of acid rain from sulfur dioxide
• formation of photochemical smog.
(i) Identify the role of NO and \(NO_2\) in the formation of \(H_2SO_4\) from \(SO_2\) in the atmosphere to produce acid rain. Use relevant equations to support your answer.
(ii) Outline how NO and \(NO_2\) may contribute to the formation of photochemical smog.

▶️ Answer/Explanation

Solution

(a) The percentage yield of NH₃ remains unchanged (58%) in B compared to A.

Explanation: A catalyst (iron in this case) does not affect the position of equilibrium or the yield; it only speeds up the attainment of equilibrium.

(b)(i) The percentage yield of NH₃ decreases in C compared to A.

Explanation: The forward reaction is exothermic (\(\Delta H = -92 \text{ kJ/mol}\)). Increasing the temperature (from 450°C to 550°C) shifts the equilibrium to the left (Le Chatelier’s principle), reducing NH₃ yield.

(b)(ii) The rate of the forward reaction increases in C compared to A.

Explanation: Higher temperature provides more molecules with energy greater than the activation energy (\(E > E_a\)), increasing the frequency of effective collisions and thus the reaction rate.

(c)(i) \(K_p = \frac{(p_{\text{NH}_3})^2}{(p_{\text{N}_2})(p_{\text{H}_2})^3}\), Units: \(\text{atm}^{-2}\)

Explanation: The equilibrium constant \(K_p\) is expressed in terms of partial pressures. The units are derived from the stoichiometric coefficients.

(c)(ii) Mole fraction of NH₃ = 0.41

Explanation: At equilibrium, moles of \(N_2 = 0.42\), \(H_2 = 1.26\), and \(NH_3 = 1.16\). Total moles = \(0.42 + 1.26 + 1.16 = 2.84\). Mole fraction of NH₃ = \(\frac{1.16}{2.84} = 0.41\).

(c)(iii) Partial pressure of \(N_2\) = 625 atm

Explanation: Partial pressure = Mole fraction × Total pressure = \(0.625 \times 1000 \text{ atm} = 625 \text{ atm}\).

(d)(i) NO and NO₂ act as catalysts in the oxidation of SO₂ to H₂SO₄.

Explanation: The catalytic cycle involves:
\(\text{SO}_2 + \text{NO}_2 \rightarrow \text{NO} + \text{SO}_3\)
\(\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4\)
\(\text{2NO} + \text{O}_2 \rightarrow \text{2NO}_2\) (regenerates NO₂).

(d)(ii) NO and NO₂ contribute to photochemical smog by reacting with VOCs.

Explanation: In sunlight, NO₂ photodissociates to form NO and O atoms, which react with hydrocarbons to form ozone and PAN (peroxyacyl nitrates), key components of smog.

Question 3

Topic: 3.1

(a) Write an equation to show the reaction for the standard enthalpy change of formation of \(H_2O\). Include state symbols.

(b) Water is one of the products in the reaction of \(B_2O_3\) and \(NH_3\), as shown in reaction 2.
reaction 2 \(B_2O_3 + 2NH_3 → 2 BN + 3H_2O\)
Table 3.1 shows information about the standard enthalpy change of formation, \(ΔH _f^{\theta}\), of some substances.

Table 3.1: Standard enthalpy changes of formation

Calculate the enthalpy change, ∆H, for reaction 2 using the data from Table 3.1.

(c) Boron carbide is a hard crystalline solid that has a melting point greater than 2000°C.
(i) Suggest the structure and bonding in boron carbide.
(ii) 100g of pure boron carbide contains 78.26g of boron. Calculate the empirical formula of boron carbide. Show your working.

▶️ Answer/Explanation

Solution

(a) \(H_2 (g) + \frac{1}{2}O_2 (g) → H_2O(l)\)

Explanation: The standard enthalpy change of formation (\(\Delta H_f^\theta\)) of \(H_2O\) is defined as the enthalpy change when 1 mole of \(H_2O(l)\) is formed from its elements in their standard states (\(H_2(g)\) and \(O_2(g)\)).

(b) \(\Delta H = +230 \, \text{kJ mol}^{-1}\)

Explanation: Using Hess’s Law, \(\Delta H = \sum \Delta H_f^\theta (\text{products}) – \sum \Delta H_f^\theta (\text{reactants})\):
\(\Delta H = [2 \times (-134) + 3 \times (-286)] – [(-1264) + 2 \times (-46)] = -1074 + 1304 = +230 \, \text{kJ mol}^{-1}\).

(c)(i) Giant covalent structure.

Explanation: Boron carbide’s high melting point (>2000°C) and hardness suggest a giant covalent lattice with strong covalent bonds between boron and carbon atoms.

(c)(ii) Empirical formula: \(B_4C\).

Explanation: – Moles of boron: \(\frac{78.26 \, \text{g}}{10.8 \, \text{g/mol}} = 7.246 \, \text{mol}\).
– Moles of carbon: \(\frac{21.74 \, \text{g}}{12 \, \text{g/mol}} = 1.812 \, \text{mol}\).
– Simplest ratio: \(\frac{7.246}{1.812} \approx 4\) (Boron) : \(1\) (Carbon).
Thus, the empirical formula is \(B_4C\).

Question 4

Topic: 15.1

(a) M reacts to form R by the addition of one reagent, as shown in Fig. 5.1.

Identify the reagent and conditions for this reaction.

(b) R is also made from M by two steps, as shown in Fig. 5.2.

(i) Identify the reagents and conditions for steps 1 and 2 in Fig. 5.2.
(ii) Name the mechanism for step 1 in Fig. 5.2.

Use the absorptions in the region above \(1500cm^{–1}\) in Table 5.1 when answering this question.
• Add F to Fig. 5.3 to identify the peak that is present in an infrared spectrum of both Q and R. Identify the bond that corresponds to the absorption for F.
• Add G to Fig. 5.3 to identify the peak that is not present in an infrared spectrum of Q. Identify the bond that corresponds to the absorption for G.

(d) Y is made from Q in a three-step reaction.

(i) Draw the structure of W in the box in Fig. 5.4. [1]
(ii) In step 2, W is heated with HCl (aq) to produce X and an inorganic product. Identify the formula of the inorganic product.
(iii) In step 3, X reacts with reducing agent Z to produce Y. Complete the equation for the reaction of X with Z. Use a molecular formula to represent the organic product. Use [H] to represent one atom of hydrogen from Z.
…… \(C_8H_{12}O_4 + ……[H]\)
(iv) Identify Z

▶️ Answer/Explanation

Solution

(a) The reagent is cold dilute \(KMnO_4\) (potassium manganate(VII)).

Explanation: Cold dilute \(KMnO_4\) oxidizes alkenes to diols (e.g., M to R) without breaking the C=C bond further.

(b)(i)
• Step 1: \(Br_2\) (in the dark).
• Step 2: NaOH(aq) + heat.
Explanation: Step 1 involves electrophilic addition of bromine to form a dibromoalkane. Step 2 uses NaOH(aq) under heat to substitute Br with OH via nucleophilic substitution.

(b)(ii) The mechanism for step 1 is electrophilic addition.

Explanation: Bromine (\(Br_2\)) acts as an electrophile, adding across the C=C double bond.

(c)

Explanation:
• F: The peak at ~2900 cm\(^{-1}\) corresponds to C-H bonds (present in both Q and R).
• G: The peak at ~3400 cm\(^{-1}\) corresponds to O-H bonds (absent in Q but present in R due to the diol group).

(d)(i) W = Structure of W (Note: Replace with actual structure if available).

(d)(ii) The inorganic product is NH₄Cl.

Explanation: Hydrolysis of W (amide) with HCl produces a carboxylic acid (X) and ammonium chloride.

(d)(iii) The balanced equation is:
\[ C_8H_{12}O_4 + 8[H] \rightarrow C_8H_{16}O_2 + 2H_2O \]

Explanation: Reducing agent Z adds hydrogen atoms to X, converting the carboxylic acid groups to alcohols (Y).

(d)(iv) Z is LiAlH₄ (lithium aluminium hydride).

Explanation: LiAlH₄ is a strong reducing agent capable of reducing carboxylic acids to primary alcohols.

Question 5

Topic: 14.2

Hydrocarbon molecules contain covalent bonds.
(a) Define covalent bond.
(b) A C=C bond in an alkene is made from a σ bond and a π bond.
(i) Identify the hybridisation of the carbon atoms in a C=C bond in an alkene.
(ii) Draw labelled diagrams to show, in terms of orbital overlap, how the σ and π bonds are made in a C=C bond.
σ bond
π bond

(c) In electrophilic reactions involving alkenes the π bond of C=C is broken.
(i) Suggest one difference between σ and π bonds that explains why the π bond is broken in electrophilic addition reactions involving alkenes.
(ii) Complete Fig. 5.1 to show the mechanism for the electrophilic addition of hydrogen bromide to 2-methylpropene to produce the major organic product. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.

▶️ Answer/Explanation

Solution

(a) (electrostatic) attraction between nuclei of two atoms and shared pair of electrons

Explanation: A covalent bond is formed when two atoms share a pair of electrons, resulting in an electrostatic attraction between the positively charged nuclei and the negatively charged shared electrons.

(b)(i) \(sp^2\)

Explanation: In a C=C bond, each carbon atom forms three σ bonds (two with hydrogen and one with the other carbon) using \(sp^2\) hybrid orbitals, leaving one unhybridized p orbital for the π bond.

(b)(ii)

Explanation: The σ bond is formed by head-on overlap of \(sp^2\) hybrid orbitals, while the π bond is formed by sideways overlap of the remaining p orbitals perpendicular to the plane of the σ bonds.

(c)(i) EITHER (pair of) electrons in 𝜋 bond are further away from the nuclei so weaker attraction OR (pair of) electrons in 𝜎 bond are closer to the two nuclei so stronger attraction

Explanation: The π bond is weaker because its electron density is distributed above and below the internuclear axis, making it more accessible to electrophiles compared to the σ bond which has direct overlap along the axis.

(c)(ii)

Explanation: The mechanism shows the π electrons attacking HBr, forming a carbocation intermediate on the more substituted carbon (Markovnikov’s rule), followed by bromide ion attack to form 2-bromo-2-methylpropane.

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