Comprehensive Chemistry AS Study Notes: Edexcel IGCSE Unit 2

Energetics and Thermodynamic Principles in Chemical Reactions

In the study of energetics, it is fundamental that energy cannot be created or destroyed within a chemical reaction, though it can be transferred between forms. Energetics specifically examines the transfer of energy between chemical reactions (the system) and their surroundings. Enthalpy change, denoted by the symbol $\Delta H$, is the heat energy transferred between the system and surroundings at a constant pressure. These measurements are typically conducted under standard conditions, which are defined as a pressure of $100\,kPa$ and a specified temperature, usually $298\,K$.

Chemical reactions involve a two-stage process: bonds are broken and new bonds are formed. Breaking bonds is an endothermic process as it requires an input of energy from the surroundings. Conversely, bond formation is an exothermic process because energy is released. The overall energy change of a reaction, $\Delta H$, is dictated by the net difference between these processes. If more heat is released to the surroundings than is taken in, the reaction is exothermic and $\Delta H$ is negative. If more heat is absorbed from the surroundings than is released, the reaction is endothermic and $\Delta H$ is positive. This can be quantified using the formula $\Delta H = \text{bonds broken} - \text{bonds made}$.

Exothermic processes include the freezing of water, condensing water vapor, dissolving sodium hydroxide in water, the reaction between dilute hydrochloric acid and aqueous sodium hydroxide, and the combustion of petrol. Endothermic processes include melting ice, evaporating water, dissolving ammonium nitrate in water, the reaction between dilute ethanoic acid and solid sodium hydrogencicarbonate, and photosynthesis. On an energy level diagram for endothermic reactions, the products sit higher than the reactants because more energy is required to break bonds than is released during formation. For exothermic diagrams, the products sit lower than the reactants because less energy is needed to break bonds than is released. These diagrams must always have a labeled vertical axis and include the formulas and state symbols for all species, along with the correct sign for $\Delta H$. Activation energy is essential to show in an enthalpy profile diagram, though not strictly required for a basic enthalpy level diagram.

Standard Enthalpy Changes and Definitions

There are several specific standard enthalpy changes categorized by the type of reaction occurring under standard conditions. The Standard Enthalpy Change of Reaction ($\Delta_r H^\ominus$) refers to the enthalpy change when substances react completely according to the molar quantities specified in a balanced equation. For example, in the production of ammonia, $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ has a $\Delta_r H^\ominus = -92\,kJ\,mol^{-1}$, but for $\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g)$, it is $-46\,kJ\,mol^{-1}$. The Standard Enthalpy Change of Formation ($\Delta_f H^\ominus$) is the energy change when one mole of a substance is produced from its constituent elements in their standard states. The Standard Enthalpy Change of Combustion ($\Delta_c H^\ominus$) occurs when one mole of a substance is burned completely in oxygen.

The Standard Enthalpy Change of Neutralisation ($\Delta_{neut} H^\ominus$) is the energy change when one mole of water is produced through the neutralisation of an acid with an alkali. For hydrochloric acid, this is represented as $HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$. For sulfuric acid, it involves $\frac{1}{2}H_2SO_4(aq) + NaOH(aq) \rightarrow \frac{1}{2}Na_2SO_4(aq) + H_2O(l)$. Finally, the Standard Enthalpy Change of Atomisation ($\Delta_{at} H^\ominus$) is the enthalpy change when one mole of gaseous atoms is formed from an element in its standard state. Examples include $C(s) \rightarrow C(g)$ ($\Delta_{at} H^\ominus = +717\,kJ\,mol^{-1}$) and $\frac{1}{2}Cl_2(g) \rightarrow Cl(g)$ ($\Delta_{at} H^\ominus = +122\,kJ\,mol^{-1}$).

Calorimetry and Experimental Measurement of Enthalpy

Calorimetry is the experimental method used to find enthalpy changes by measuring temperature changes over time. By plotting data and extrapolating the line of best fit, an accurate value for the temperature change ($\Delta T$) at the start of a reaction can be determined. The energy change ($Q$) is proportional to the temperature change following the equation $Q = m \times c \times \Delta T$, where $m$ is the mass of the substance in grams, $c$ is the specific heat capacity ($4.18\,J\,g^{-1}\,K^{-1}$ for water), and $\Delta T$ is the change in Kelvin or Celsius. The specific heat capacity is defined as the energy required to raise the temperature of $1\,g$ of a substance by $1\,K$ without changing its state.

To find the enthalpy change of combustion of a liquid, such as ethanol, a known mass of the liquid is burned to heat a known volume of water. In a typical experiment where $0.420\,g$ of ethanol ($M_r = 46$) heats $100.0\,cm^3$ of water by $+24.5\,^{\circ}C$, the energy released is $Q = 100\,g \times 4.18\,J\,g^{-1}\,K^{-1} \times 24.5\,K = 10241\,J$ or $10.24\,kJ$. The amount of ethanol burned is $n = \frac{0.420}{46} = 9.13 \times 10^{-3}\,mol$. The $\Delta_c H^\ominus = -\frac{Q}{n} = -\frac{10.24}{9.13 \times 10^{-3}} = -1120\,kJ\,mol^{-1}$. Sources of error in this method include heat loss to the air or the copper can, incomplete combustion of ethanol, and evaporation of the fuel. For neutralisation, temperatures of the acid and alkali are measured, the two are mixed, and the maximum temperature reached is recorded. If $25.0\,cm^3$ of $1.00\,mol\,dm^{-3}\,HCl$ is mixed with $25.0\,cm^3$ of $1.2\,mol\,dm^{-3}\,NaOH$ and the temperature rises from a mean of $18.7\,^{\circ}C$ to $25.4\,^{\circ}C$ ($\Delta T = +6.7\,K$), the mass heated is $50.0\,g$ (assuming density is $1.00\,g\,cm^{-3}$). The heat produced is $1.4003\,kJ$. Since $0.0250\,mol$ of water is formed, the enthalpy change is calculated by dividing energy by moles.

Hess’s Law and Enthalpy Cycles

Hess’s Law states that the enthalpy change of a reaction is independent of the path taken to convert reactants into products, provided the initial and final conditions are identical. It is used to calculate enthalpy changes for reactions that cannot be measured directly. This involves building a triangular cycle with an intermediate product. For example, the enthalpy change of a reaction can be calculated from formation data using $\Delta H_{\text{reaction}} = \sum \Delta_f H_{\text{products}} - \sum \Delta_f H_{\text{reactants}}$. Alternatively, formation changes can be found from combustion data using $\sum \Delta_c H_{\text{reactants}} = \Delta_f H + \sum \Delta_c H_{\text{products}}$.

Examples of application: calculating the $\Delta_f H$ of methanol ($CH_3OH$) requires $\Delta_c H$ data for carbon ($-394\,kJ\,mol^{-1}$), hydrogen ($-286\,kJ\,mol^{-1}$), and methanol ($-726\,kJ\,mol^{-1}$. Using Hess’s Law: $\Delta_f H = (-394) + (-286 \times 2) - (-726) = -240\,kJ\,mol^{-1}$. Similarly, the thermal decomposition of calcium carbonate ($CaCO_3$) can be calculated using its reaction with $HCl$. If $CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$ has $\Delta H = -17\,kJ\,mol^{-1}$ and $CaO(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l)$ has $\Delta H = -195\,kJ\,mol^{-1}$, then the decomposition $\Delta_r H = -17 - (-195) = +178\,kJ\,mol^{-1}$.

Bond Enthalpy and Bond Strengths

Bond enthalpy is the energy needed to break one mole of a specific bond in the gaseous state. Bond breaking is endothermic ($\Delta H$ is positive), while bond making is exothermic ($\Delta H$ is negative). For polyatomic molecules like methane ($CH_4$), bond enthalpy varies based on the chemical environment. Therefore, mean (average) bond enthalpy is used. For $CH_4$, the four successive C-H bonds require $+423$, $+480$, $+425$, and $+335\,kJ\,mol^{-1}$ respectively, giving a mean value of $415.75\,kJ\,mol^{-1}$.

To calculate reaction enthalpy using mean bond enthalpies: $\Delta_r H = \sum (\text{bond broken}) - \sum (\text{bonds made})$. For the reaction $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$, with $E(H-H) = 436$, $E(Cl-Cl) = 244$, and $E(H-Cl) = 432$, the calculation is $(436 + 244) - (2 \times 432) = -184\,kJ\,mol^{-1}$. There is often a difference between measured values and those calculated from bond enthalpies because bond enthalpies are standardized for the gaseous state, whereas reactants or products might be liquids in the actual reaction. Bonds with high bond enthalpies require more energy to break and are less likely to break first in a reaction; such reactions often require heat or catalysts.

Intermolecular Forces: London Forces, Dipoles, and Hydrogen Bonding

Intermolecular forces are interactions between molecules that are weaker than covalent or ionic bonds. There are three main types: London Forces, Permanent Dipoles, and Hydrogen Bonding. London forces occur in all molecules, including non-polar ones like nitrogen. They arise from fluctuations in electron density that create instantaneous dipoles, which in turn induce dipoles in neighboring molecules. The strength of London forces increases with the number of electrons in a molecule and the number of contact points between molecules. This explains why noble gas boiling points increase down the group and why branched alkanes have lower boiling points than unbranched isomers (e.g., pentane at $309\,K$ vs. 2,2-dimethylpropane at $283\,K$).

Permanent dipoles act between molecules with polar bonds due to electronegativity differences. Surprisingly, London forces are often more significant than permanent dipole-dipole interactions because induced dipoles are always favorably aligned, while permanent dipoles in random liquid motion may not be. Hydrogen bonding is the strongest intermolecular force, occurring when hydrogen is bonded to highly electronegative nitrogen, oxygen, or fluorine. The positive charge on the hydrogen atom is high enough to attract a lone pair of electrons from a neighboring $N, O, \text{or } F$ atom. Water, for instance, forms an average of two hydrogen bonds per molecule because it has two hydrogens and two lone pairs, allowing for an extensive three-dimensional network.

Physical Properties and Solubility

Intermolecular forces dictate physical properties like boiling temperature and volatility. Methanol has a higher boiling point than ethane despite having the same number of electrons ($338\,K$ vs. $184\,K$) because of hydrogen bonding. Alcohols also exhibit lower volatility as the chain length increases. Water displays anomalous properties: it has unusually high melting and boiling points for its size, and ice is less dense than liquid water. In ice, molecules are held in rings of six by hydrogen bonds, creating open space; melting destroys this structure, allowing molecules to get closer, increasing density.

Solubility follows the "like dissolves like" rule. Polar solvents dissolve polar/ionic solutes, while non-polar solvents dissolve non-polar solutes. For an ionic solid to dissolve in water, the ion-dipole interactions between the water molecules and the ions must overcome the lattice energy. This arrangement of water molecules around an ion is called hydration, releasing hydration energy. Alcohols are soluble in water due to the -OH group forming hydrogen bonds, but solubility decreases as the carbon chain length increases because London forces begin to predominate. Non-polar molecules like alkenes or ethoxyethane are immiscible with water because they cannot replace the strong hydrogen bonds between water molecules.

Redox Chemistry and Oxidation States

Redox reactions involve simultaneous oxidation and reduction through electron transfer. Historically defined as the addition/removal of oxygen or hydrogen, modern definitions focus on electrons: Oxidation is the loss of electrons (OIL), and Reduction is the gain of electrons (RIG). An oxidising agent gains electrons and is reduced, while a reducing agent loses electrons and is oxidised. Disproportionation is a specific redox reaction where the same element is both oxidised and reduced, such as $Cu_2O(s) + H_2SO_4(aq) \rightarrow CuSO_4(aq) + Cu(s) + H_2O(l)$.

Oxidation numbers denote the oxidation state of an element. Key rules include: uncombined elements are zero; the sum of numbers in a neutral compound is zero; the sum in an ion equals the charge; fluorine is always $-1$; hydrogen is $+1$ (except in metal hydrides where it is $-1$); oxygen is $-2$ (except in peroxides, $-1$, or with fluorine, $+2$). Systematic names use Roman numerals to indicate oxidation states, such as iron(II) chloride ($FeCl_2$) vs. iron(III) chloride ($FeCl_3$). Full ionic equations are constructed by balancing the electrons in two half-equations. For example, combining $Fe^{2+} \rightarrow Fe^{3+} + e^-$ and $Cl_2 + 2e^- \rightarrow 2Cl^-$ requires multiplying the iron half-equation by two to balance electrons.

Trends in Group 1, 2, and 7 Elements

In Groups 1 and 2, atomic radius increases down the group while first ionisation energy decreases due to greater shielding and atomic radius, making outer electrons easier to lose and increasing reactivity. Group 1 metals react with oxygen to form oxides ($4M + O_2 \rightarrow 2M_2O$) and with water to form hydroxides and hydrogen ($2M + 2H_2O \rightarrow 2MOH + H_2$). Group 2 metals react similarly, though magnesium's reaction with cold water is very slow. Magnesium reacts vigorously with steam to form magnesium oxide. Group 2 hydroxide solubility increases down the group ($Mg(OH)_2$ is used as an antacid), while sulfate solubility decreases down the group (barium sulfate is insoluble and used to test for sulfate ions).

Thermal stability of Group 1 and 2 carbonates and nitrates depends on the cation's charge and size. Group 2 cations (higher charge, smaller size) polarize the complex anions more, leading to easier decomposition on heating. All Group 2 carbonates decompose to the oxide and $CO_2$. Group 1 carbonates do not decompose, except for lithium carbonate. Group 2 nitrates decompose to the oxide, $NO_2$ (brown fumes), and $O_2$, while most Group 1 nitrates only decompose to the nitrite and $O_2$ (no brown fumes), again with the exception of lithium. Flame tests identify these metals: Lithium (Red), Sodium (Orange/Yellow), Potassium (Lilac), Calcium (Brick red), Strontium (Crimson red), and Barium (Pale green). Magnesium shows no color because its electron transition energy is outside the visible spectrum.

Group 7 (halogens) elements are non-metals that gain an electron to form $-1$ ions. Atomic radius and boiling points increase down the group (fluorine is a gas, iodine a solid) due to increasing London forces. Electronegativity and reactivity decrease down the group. More reactive halogens displace less reactive ones from halides; chlorine can displace bromine and iodine. Group 7 reactions with concentrated sulfuric acid reveal reducing power trends: halides become stronger reducing agents down the group. Chloride and fluoride only produce misty fumes of $H-X$ (non-redox). Bromide reduces $H_2SO_4$ to $SO_2$ and $Br_2$ (brown fumes). Iodide is strong enough to reduce $H_2SO_4$ to $SO_2$, sulfur (yellow solid), and $H_2S$ (rotten egg smell).

Practical Chemistry: Titrations and Uncertainties

A standard solution has an accurately known concentration, prepared using a primary standard that is pure, stable, and has a high molar mass (e.g., sulfamic acid, $NH_2SO_3H$). Sodium hydroxide is unsuitable as it absorbs water and reacts with $CO_2$ from the air. Preparation involves "weighing by difference," dissolving the solid in a beaker, transferring to a volumetric flask, and filling to the meniscus graduation mark. Titration find the concentration of an unknown by reacting it with the standard until the equivalence point. Indicators like phenolphthalein or methyl orange are used to signal the end point. Accuracy measures closeness to the true value, while precision measures the consistency of repeated results.

Measurement uncertainties are potential errors determined by the precision of apparatus. Class A glassware is more precise and expensive than Class B. Total percentage uncertainty is the sum of the percentage uncertainties of each piece of equipment used. For example, if a titration involves uncertainties from a balance ($\pm 0.09\%$), volumetric flask ($\pm 0.12\%$), pipette ($\pm 0.24\%$), and burette ($\pm 0.47\%$), the overall uncertainty is $\pm 0.92\%$. If the calculated concentration is $0.118\,mol\,dm^{-3}$, the final quoted value is $0.118 \pm 0.001\,mol\,dm^{-3}$.

Kinetics: Rates of Reaction and Collision Theory

The rate of reaction is the change in reactant or product concentration per unit time: $\text{Rate} = \frac{\text{Change in concentration}}{\text{Time}}$. It can be measured by the rate of reactant depletion, product formation (e.g., gas volume), or monitoring color changes. Collision theory states that for a reaction to occur, particles must collide with sufficient energy (Activation Energy, $E_a$) and in the correct orientation. $E_a$ is the minimum energy required for particles to react; otherwise, they simply bounce apart.

Factors affecting rate include concentration and pressure (increasing collision frequency), surface area (increasing available sites for collision), and temperature. Temperature uniquely increases rate because particles move faster and a much greater proportion have energy $\ge E_a$. This is visualized via the Maxwell-Boltzmann distribution, which shows the energy spread of molecules. Raising temperature shifts the curve right and down, increasing the area beyond the $E_a$ line. Catalysts increase the rate without being consumed by providing an alternative pathway with a lower $E_a$. In industry, catalysts reduce energy costs and allow for higher atom economy by enabling reactions at lower temperatures and pressures.

Chemical Equilibrium and Industrial Applications

Reversible reactions can reach dynamic equilibrium in a closed system when the forward and backward reaction rates are equal and concentrations remain constant. Le Chatelier's Principle predicts how equilibrium shifts in response to changes: increasing reactant concentration shifts the equilibrium to the right; increasing pressure shifts it toward the side with fewer gas moles; and increasing temperature shifts it toward the endothermic direction. A catalyst increases both rates equally and does not change the position of equilibrium.

In industry, the Haber Process produces ammonia: $N_2 + 3H_2 \rightleftharpoons 2NH_3$ ($\Delta H = -92\,kJ\,mol^{-1}$). Compromise conditions of $450\,^{\circ}C$ and $250\,atm$ are used to balance high yield (favored by low temp and high pressure) with an acceptable rate and cost. The Contact Process produces sulfur trioxide for sulfuric acid: $2SO_2 + O_2 \rightleftharpoons 2SO_3$. It uses a $V_2O_5$ catalyst at $450\,^{\circ}C$ and $1\text{-}2\,atm$, as the yield is already high at low pressures.

Organic Chemistry: Mechanisms and Functional Groups

Organic reactions are categorized into types: Addition (two species form one), Elimination (groups removed to form a double bond), Substitution (one group replaces another), Oxidation (loss of $H$ or gain of $O$), Reduction (gain of $H$ or loss of $O$), Hydrolysis (water/hydroxide replaces a group), and Polymerisation. Halogenoalkanes include a polar $C-X$ bond, making the carbon electron-deficient ($\delta+$) and susceptible to nucleophilic attack. Nucleophiles like $OH^-$, $CN^-$, and $NH_3$ have lone pairs. Primary halogenoalkanes typically undergo substitution ($S_N2$), while tertiary ones hydrolyze faster due to different mechanisms. Elimination occurs when halogenoalkanes react with ethanolic $KOH$, acting as a base rather than a nucleophile, forming alkenes.

Alcohols are classified as primary, secondary, or tertiary. They can be halogenated using chemicals like $PCl_5$ (chlorination), $KBr/H_2SO_4$ (bromination), or $P/I_2$ (iodination). Alcohols dehydrate to form alkenes when heated with acid. Primary alcohols oxidize to aldehydes (using distillation with addition) or carboxylic acids (using heating under reflux) in the presence of acidified potassium dichromate(VI). Secondary alcohols oxidize to ketones, while tertiary alcohols do not oxidize. Purification of organic liquids involves simple or fractional distillation, solvent extraction using a separating funnel, and drying with anhydrous salts like $MgSO_4$. Purity is confirmed by checking the boiling point against known data.

Analytical Techniques: Mass Spectrometry and IR Spectroscopy

Mass spectrometry identifies isotopes and molecular structures. The molecular ion peak ($M^+$) represents the relative molecular mass. Smaller peaks result from fragmentation, where a bond breaks to form a positive ion and a neutral radical. The base peak is the tallest peak, representing the most stable ion. In the mass spectrum of propane ($C_3H_8$), peaks at $m/z = 15$ and $29$ represent methyl ($CH_3^+$) and ethyl ($CH_3CH_2^+$) cations respectively. Infrared (IR) spectroscopy uses radiation to induce bond vibrations (stretching and bending). The spectrum plots transmittance vs. wavenumber ($cm^{-1}$). Specific functional groups absorb characteristic frequencies: $C=O$ at $1670\text{-}1740\,cm^{-1}$, and alcohol $O-H$ at $3200\text{-}3750\,cm^{-1}$. The fingerprint region ($1500\text{-}500\,cm^{-1}$) is unique to every molecule, allowing for definitive identification.