A Level H2 Chemistry Practice Paper 3

TuitionGoWhere Practice Paper - Chemistry H2 A-Level

Answer Key (Version 3)

Section A: Physical Chemistry

Question 1 (a)(i) Concordant results: $23.10, 23.20, 23.15$. Mean $= (23.10 + 23.20 + 23.15) / 3 = 23.15\text{ cm}^3$. [2] (a)(ii) $\text{n}(\text{NaOH}) = 0.100 \times (23.15/1000) = 2.315 \times 10^{-3}\text{ mol}$. $\text{n}(\text{HA}) = 2.315 \times 10^{-3}\text{ mol}$ (1:1 ratio). $\text{Conc}(\text{HA}) = (2.315 \times 10^{-3}) / (25.0/1000) = 0.0926\text{ mol dm}^{-3}$. [2] (b) At equivalence point, the solution contains the conjugate base $\text{A}^-$. $\text{A}^-$ undergoes hydrolysis: $\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-$, increasing $[\text{OH}^-]$ and thus $\text{pH} > 7$. [2]

Question 2 (a) A solution that resists significant changes in pH upon the addition of small amounts of acid or base. [1] (b)(i) $\text{pH} = \text{p}K_a + \log([\text{salt}]/[\text{acid}]) = 4.76 + \log(0.20/0.10) = 4.76 + 0.301 = 5.06$. [2] (b)(ii) $\text{n}(\text{salt}) = 0.20 - 0.01 = 0.19\text{ mol}$; $\text{n}(\text{acid}) = 0.10 + 0.01 = 0.11\text{ mol}$. $\text{New pH} = 4.76 + \log(0.19/0.11) = 4.76 + 0.238 = 4.998 \approx 5.00$. $\Delta\text{pH} = 5.06 - 5.00 = 0.06$. [3]

Question 3 (a) $\text{Mg}^{2+}$ has a higher charge ($+2$ vs $+1$) and a smaller ionic radius than $\text{Na}^+$. This results in a much stronger electrostatic attraction between the ions and the lattice, leading to a more exothermic lattice energy. [3] (b) $\text{Al}_2\text{O}_3(\text{s}) + 2\text{OH}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow 2[\text{Al}(\text{OH})_4]^-(\text{aq})$. [2]

Question 4 (a) $K_p = \frac{(P_{\text{SO}_3})^2}{(P_{\text{SO}_2})^2(P_{\text{O}_2})}$. Since $P_i = c_i RT$: $K_p = \frac{(c_{\text{SO}_3} RT)^2}{(c_{\text{SO}_2} RT)^2(c_{\text{O}_2} RT)} = \frac{c_{\text{SO}_3}^2}{c_{\text{SO}_2}^2 c_{\text{O}_2}} \times \frac{(RT)^2}{(RT)^3} = K_c(RT)^{-1}$. [3] (b) Equilibrium shifts to the right (towards $\text{SO}_3$). According to Le Chatelier's principle, increasing pressure shifts the equilibrium to the side with fewer moles of gas (3 moles $\rightarrow$ 2 moles). [2]

Question 5 (a) $Q = I \times t = 2.50 \times (45 \times 60) = 6750\text{ C}$. $\text{n}(e^-) = 6750 / 96500 = 0.0699\text{ mol}$. $\text{n}(\text{Cu}) = 0.0699 / 2 = 0.03495\text{ mol}$. $\text{Mass} = 0.03495 \times 63.5 = 2.22\text{ g}$. [4] (b) Effervescence of a pale green gas (chlorine). [1]

Section B: Inorganic Chemistry

Question 6 (a) $\text{Al}^{3+}$: White ppt., insoluble in excess $\text{NH}_3$. $\text{Cu}^{2+}$: Blue ppt., insoluble in excess $\text{NaOH}$. $\text{Fe}^{3+}$: Brown ppt., insoluble in excess $\text{NH}_3$. $\text{Zn}^{2+}$: White ppt., soluble in excess $\text{NaOH}$. [4] (b) Transition metals have partially filled d-orbitals. Ligands cause d-orbital splitting. Electrons absorb visible light to transition between these split levels; the complementary colour is observed. Main group elements lack partially filled d-orbitals. [3]

Question 7 (a) Solubility increases down the group. While both lattice energy and hydration energy decrease as ionic radius increases, the lattice energy decreases more significantly, making the enthalpy of solution more exothermic. [3] (b) $\text{MgCO}_3(\text{s}) \xrightarrow{\Delta} \text{MgO}(\text{s}) + \text{CO}_2(\text{g})$. [2]

Question 8 (a) $\text{Cl}_2 > \text{Br}_2 > \text{I}_2$. Oxidizing power depends on the ability to attract electrons. $\text{Cl}_2$ has the smallest atomic radius and highest electronegativity among the three, making it the strongest oxidizing agent. [3] (b) The orange/brown solution of $\text{Br}_2$ becomes darker/brown and a dark purple/black precipitate of iodine may form (or solution turns dark brown). [2]

Question 9 (a) Boron has an empty p-orbital and an incomplete octet, allowing it to accept a pair of electrons from a donor. [2] (b) [Structure showing $\text{B}$ bonded to three $\text{F}$ and one $\text{N}$ from $\text{NH}_3$, with a coordinate bond $\text{N} \rightarrow \text{B}$ and formal charges $\text{N}^+$, $\text{B}^-$]. [2]

Section C: Organic Chemistry

Question 10 (a) [Mechanism: $\text{CN}^-$ attacks carbonyl carbon $\rightarrow$ $\text{C}=\text{O}$ pi bond breaks to $\text{O}^-$ $\rightarrow$ $\text{O}^-$ attacks $\text{H}$ of $\text{HCN}$]. [4] (b) Ketones are more sterically hindered (two alkyl groups vs one in aldehydes), making the approach of the nucleophile more difficult. Additionally, the two alkyl groups provide more inductive stabilization to the carbonyl carbon, making it less electrophilic. [3]

Question 11 (a) $\text{S}_{\text{N}}1$. The substrate is a tertiary haloalkane, which forms a stable tertiary carbocation intermediate and is too sterically hindered for a direct $\text{S}_{\text{N}}2$ attack. [3] (b) 2-methylpropan-2-ol. [2]

Question 12 (a) $\text{C}_6\text{H}_5\text{NH}_2 < \text{NH}_3 < \text{CH}_3\text{NH}_2$. Aniline is least basic due to delocalisation of the lone pair into the benzene ring. Methylamine is most basic due to the $+I$ effect of the methyl group increasing electron density on the $\text{N}$ atom. [4] (b) $\text{CH}_3\text{CH}_2\text{NH}_2 + \text{HCl} \rightarrow \text{CH}_3\text{CH}_2\text{NH}_3^+\text{Cl}^-$. [2]

Question 13 (a) [Mechanism: $\text{Br}_2$ approaches $\text{C}=\text{C}$ $\rightarrow$ formation of cyclic bromonium ion $\rightarrow$ $\text{Br}^-$ attacks from opposite side]. [4] (b) According to Markovnikov's rule, the $\text{H}^+$ adds to the carbon with more hydrogens to form the more stable secondary carbocation rather than a primary carbocation. [3]

Question 14 (a) Compounds with the same molecular formula but different structural arrangements of atoms. [1] (b) 2-methylpropanoic acid. [2]