Senior High School Chemistry: Comprehensive Year 2 Study Notes

THERMODYNAMICS AND ENERGY CHANGES

  • Fundamental Definitions     - Atomisation Energy: The energy required to break all the bonds in one mole of a substance to form individual atoms in the gas phase under standard conditions.     - Bond Dissociation Energy: The energy required to break a specific chemical bond within a compound.     - Bond Enthalpy: The average energy needed to split one mole of a specific type of covalent bond in a gaseous molecule.     - Enthalpy (H): A thermodynamic quantity describing the energy of a system, calculated as: H=U+PVH = U + PV, where UU is internal energy, PP is pressure, and VV is volume.     - Enthalpy Change (\Delta H): Defined as: ΔH=HproductsHreactants\Delta H = H_{products} - H_{reactants}.     - Hess’s Law of Constant Heat Summation: States that the total enthalpy change of a chemical reaction is equal to the sum of all individual enthalpy changes, regardless of the pathway taken.     - Lattice Energy: The energy required to separate 1 mole of an ionic solid into gaseous ions.

  • Types of Chemical Systems     - Open System: Can exchange both heat and matter with the surroundings (e.g., heating water in an open beaker).     - Closed System: Can exchange heat but not matter with the surroundings (e.g., heating water in a sealed beaker).     - Isolated System: Cannot exchange heat or matter with surroundings (e.g., well-insulated thermos or calorimeter). Note: No system is truly ideal; some heat/matter leakage always occurs.

  • Exothermic and Endothermic Reactions     - Exothermic Reactions: Release heat to the surroundings. The enthalpy of the system decreases (ΔH\Delta H is negative). Example: Methane combustion: CH4(g)+2O2(g)CO2(g)+2H2O(l)CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l), ΔH=890.4kJ/mol\Delta H = -890.4\,kJ/mol.     - Endothermic Reactions: Absorb heat from surroundings. The enthalpy of the system increases (ΔH\Delta H is positive). Example: Decomposition of calcium carbonate: CaCO3(s)CaO(s)+CO2(g)CaCO_3(s) \rightarrow CaO(s) + CO_2(g), ΔH=+177.8kJ/mol\Delta H = +177.8\,kJ/mol.

  • Standard Enthalpy Changes (\Delta H^\circ)     - Standard Conditions: Pressure of 1atm1\,atm (101.3kPa101.3\,kPa) and Temperature of 298K298\,K (25C25\,^\circ C).     - Standard State: The most stable form of an element/compound at standard conditions (e.g., O2(g)O_2(g), H2O(l)H_2O(l), C(graphite)C(graphite)).     - Standard Enthalpy of Formation (\Delta H_f^\circ): Enthalpy change when one mole of a compound is formed from its elements in their standard states. Elements in their standard states have ΔHf=0\Delta H_f^\circ = 0.     - Standard Enthalpy of Reaction (\Delta H_{rxn}^\circ): Calculated as: ΔHrxn=nΔHf(products)mΔHf(reactants)\Delta H_{rxn}^\circ = \sum n\Delta H_f^\circ(products) - \sum m\Delta H_f^\circ(reactants).     - Standard Enthalpy of Combustion (\Delta H_c^\circ): Enthalpy change when one mole of a substance is burned completely in oxygen. Example: H2(g)+12O2(g)H2O(l)H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l), ΔHc=285.8kJ/mol\Delta H_c^\circ = -285.8\,kJ/mol.     - Standard Enthalpy of Neutralisation (\Delta H_n^\circ): Enthalpy change when 1 mole of water is produced from acid-alkali reaction. For strong acids/bases, this value is approximately ΔH=57.4kJ/mol\Delta H = -57.4\,kJ/mol.     - Standard Enthalpy of Solution (\Delta H_{soln}^\circ): Heat change when one mole of ionic substance dissolves in excess water.     - Standard Enthalpy of Hydration (\Delta H_{hyd}^\circ): Energy evolved when one mole of gaseous ions is surrounded and stabilized by water molecules (infinite dilution).

  • Experimental Determination and Calorimetry     - Energy Transfer Formula: Q=mwater×c×ΔTQ = m_{water} \times c \times \Delta T, where c=4.18J/gCc = 4.18\,J/g^\circ C.     - Heat of Combustion Calculation: Per mole = Qmassburned×MolarMass\frac{Q}{mass\,burned} \times Molar\,Mass.

  • Hess's Law Manipulations and Born-Haber Cycles     - Rules for Equations: Reversing an equation negates the sign of ΔH\Delta H. Multiplying coefficients by a factor requires multiplying ΔH\Delta H by the same factor.     - Born-Haber Cycle for LiF: Shows steps: Sublimation of Li(s)Li(s) (+155.2kJ/mol+155.2\,kJ/mol), Dissociation of 12F2(g)\frac{1}{2}F_2(g) (+75.3kJ/mol+75.3\,kJ/mol), Ionisation of Li(g)Li(g) (+520kJ/mol+520\,kJ/mol), Electron affinity of F(g)F(g) (328kJ/mol-328\,kJ/mol), and Formation of LiF(s)LiF(s) (594.1kJ/mol-594.1\,kJ/mol). Calculated Lattice Energy x=1016.6kJ/molx = -1016.6\,kJ/mol.

CHEMICAL KINETICS AND REACTION RATES

  • Rate Principles     - Rate of Reaction: Change in concentration of reactant or product per unit time. Unit: moldm3s1mol\,dm^{-3}\,s^{-1}.     - Mathematical Rate for aA+bBcC+dDaA + bB \rightarrow cC + dD: Rate=1aΔ[A]Δt=1bΔ[B]Δt=1cΔ[C]Δt=1dΔ[D]ΔtRate = -\frac{1}{a} \frac{\Delta[A]}{\Delta t} = -\frac{1}{b} \frac{\Delta[B]}{\Delta t} = \frac{1}{c} \frac{\Delta[C]}{\Delta t} = \frac{1}{d} \frac{\Delta[D]}{\Delta t}.     - Initial Rate: Rate at t=0t = 0.     - Instantaneous Rate: Rate at a specific instant, determined by the tangent slope of the concentration-time curve.     - Average Rate: Δ[Concentration]Δt\frac{\Delta[Concentration]}{\Delta t}.

  • Factors Affecting Reaction Rate     - Temperature: Higher temperature increases average kinetic energy; a greater proportion of molecules exceed Activation Energy (EaE_a).     - Concentration/Pressure: More particles in a given volume increases collision frequency.     - Surface Area: More contact points in solids lead to more frequent effective collisions.     - Catalyst: Lowers EaE_a by providing an alternative reaction pathway; it is not consumed.

  • Collision Theory and Maxwell-Boltzmann     - Collision Theory Requirements: Particles must collide with (1) energy Ea\ge E_a and (2) correct orientation.     - Maxwell-Boltzmann Distribution: At higher temperatures (T_2 > T_1), the peak shift right and lowers; the area under the curve beyond EaE_a increases significantly.

  • Rate Equations and Reaction Order     - Rate Law: Rate=k[A]x[B]yRate = k[A]^x[B]^y, where kk is the rate constant, and x,yx, y are orders determined experimentally.     - Zero Order: Rate is independent of concentration (kk units: moldm3s1mol\,dm^{-3}\,s^{-1}).     - First Order: Rate doubles when concentration doubles (kk units: s1s^{-1}). Half-life t1/2=0.693kt_{1/2} = \frac{0.693}{k}.     - Second Order: Rate quadruples when concentration doubles (kk units: mol1dm3s1mol^{-1}\,dm^3\,s^{-1}). Half-life t1/2=1k[A]0t_{1/2} = \frac{1}{k[A]_0}.     - Rate-Determining Step (RDS): The slowest elementary step in a reaction mechanism.

DYNAMIC EQUILIBRIUM IN CHEMICAL SYSTEMS

  • The Nature of Equilibrium     - Reversible Reactions: Products can convert back to reactants (e.g., N2+3H22NH3N_2 + 3H_2 \rightleftharpoons 2NH_3).     - Dynamic Equilibrium: A state in a closed system where forward and reverse rates are equal, and concentrations remain constant.

  • Equilibrium Constants     - Law of Mass Action: For aA+bBcC+dDaA + bB \rightleftharpoons cC + dD, Kc=[C]c[D]d[A]a[B]bK_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}.     - Partial Pressure Constant (KpK_p): For gases, Kp=PCc×PDdPAa×PBbK_p = \frac{P_C^c \times P_D^d}{P_A^a \times P_B^b}.     - Relation Formula: Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}, where Δn=(c+d)(a+b)\Delta n = (c+d) - (a+b).     - Solubility Product (KspK_{sp}): For sparingly soluble salt AxBy(s)xAy+(aq)+yBx(aq)A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq), Ksp=[Ay+]x[Bx]yK_{sp} = [A^{y+}]^x[B^{x-}]^y. Pure solids/liquids are excluded from expressions.

  • Le Chatelier’s Principle     - Concentration: Adding reactants shifts equilibrium to use them up (forward).     - Pressure/Volume: Increasing pressure (decreasing volume) shifts equilibrium to the side with fewer gas moles.     - Temperature: Increasing temperature shifts in the endothermic direction (\Delta H > 0) to absorb heat.     - Catalyst: Reaches equilibrium faster but does not shift the position or change KK.

  • Industrial Applications     - Haber Process: N2(g)+3H2(g)2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), ΔH=92kJ/mol\Delta H = -92\,kJ/mol. Optimized by high pressure, iron catalyst, and removing ammonia. Lower temperature favors yield but higher (compromised) temperature favors rate.     - Contact Process: 2SO2(g)+O2(g)2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g), ΔH=196kJ/mol\Delta H = -196\,kJ/mol. Uses Vanadium(V) oxide (V2O5V_2O_5) catalyst.     - Catalytic Converters: Reduces NONO to N2N_2 and oxidizes COCO to CO2CO_2 using platinum catalysts.

THE CHEMISTRY OF ACIDS, BASES, AND SALTS

  • Acid-Base Theories     - Arrhenius: Acids produce H+/H3O+H^+/H_3O^+ in water; Bases produce OHOH^- in water. Limit: Only applies to aqueous solutions.     - Brønsted-Lowry: Acid is a proton (H+H^+) donor; Base is a proton acceptor. Conjugate pairs differ by a single H+H^+. Water is amphiprotic.     - Lewis: Acid is an electron-pair acceptor (e.g., BF3BF_3, Al3+Al^{3+}); Base is an electron-pair donor (e.g., NH3NH_3, H2OH_2O). Forms coordinate covalent bonds.

  • Properties and Applications     - Physical: Acids (sour, pH < 7, turn blue litmus red); Bases (bitter, slippery, pH > 7, turn red litmus blue).     - Chemical: Acids + Carbonates \rightarrow Salt + H2OH_2O + CO2CO_2; Bases + Ammonium salts + heat \rightarrow Salt + H2OH_2O + NH3(g)NH_3(g).     - Neutralisation In Life: Antacids (Mg(OH)2Mg(OH)_2, Al(OH)3Al(OH)_3) for reflux; Agriculture (lime/CaCO3CaCO_3) for soil acidity.

  • Salts and Solubility     - Classification:         - Acidic Salts: Partial replacement of acid hydrogens (e.g., NaHSO4NaHSO_4).         - Double Salts: Two different cations (e.g., Potash alum).         - Complex Salts: Central metal with ligands (e.g., [Cr(NH3)6]Cl3[Cr(NH_3)_6]Cl_3).         - Deliquescent: Absorb enough air moisture to dissolve (e.g., NaOHNaOH, CaCl2CaCl_2).         - Efflorescent: Lose water of crystallization to air (e.g., Na2CO310H2ONa_2CO_3 \cdot 10H_2O).

  • Acid-Base Titration     - Key Terms: Titrant (burette, known conc); Analyte (flask, unknown conc); Equivalence Point (molar equivalence); Endpoint (indicator colour change).     - Direct Titration Calculation: Mole ratio nbasenacid=CbVbCaVa\frac{n_{base}}{n_{acid}} = \frac{C_bV_b}{C_aV_a}.     - Back Titration: Used when direct measurement is difficult (e.g., Ecotrin aspirin analysis). Add excess known reagent, then titrate the excess.     - Double-Indicator Titration: Uses phenolphthalein and methyl orange to analyze mixtures of Na2CO3Na_2CO_3 and NaHCO3NaHCO_3.

PERIODICITY AND PROPERTIES OF ELEMENTS

  • Period 3 Trends     - Metallic Character: Decreases from NaNa to ArAr. Na,Mg,AlNa, Mg, Al are metals; SiSi is a metalloid; P,S,Cl,ArP, S, Cl, Ar are non-metals.     - Melting/Boiling Points: Increase from NaNa to SiSi (SiSi has a giant covalent network), then decrease. Trend for non-metals: S_8 > P_4 > Cl_2 > Ar (based on Van der Waals forces/molecular size).     - Electrical Conductivity: Decreases across. AlAl is the best conductor (3 delocalized electrons). SiSi is a semiconductor.

  • Period 3 Compounds     - Oxides: Na2O,MgONa_2O, MgO (Basic, ionic); Al2O3Al_2O_3 (Amphoteric); SiO2,P4O10,SO3SiO_2, P_4O^{10}, SO_3 (Acidic, covalent).     - Chlorides: NaCl,MgCl2NaCl, MgCl_2 (Ionic, neutral in water); AlCl3,SiCl4,PCl5AlCl_3, SiCl_4, PCl_5 (Covalent, hydrolysis in water to produce acidic HClHCl fumes).     - Thermal Stability: Carbonates and nitrates become more stable down Group 1/2 as cation size increases and polarizing power decreases.

THE HALOGENS (GROUP 17)

  • Physical and Chemical Trends     - Physical State: F2,Cl2F_2, Cl_2 (Gas); Br2Br_2 (Liquid); I2,AtI_2, At (Solid). Colour darkens down the group.     - Oxidising Strength: Decreases down (F_2 > Cl_2 > Br_2 > I_2). Standard reduction potential for F2/2F=+2.87VF_2/2F^- = +2.87\,V.     - Reducing Power of Halides: Increases down (I^- > Br^- > Cl^- > F^-).

  • Reactions with H2SO4(conc)H_2SO_4(conc)     - Chlorides: Produce HCl(g)HCl(g) only.     - Bromides: Produce HBrHBr, which further reduces H2SO4H_2SO_4 to SO2SO_2 and Br2(g)Br_2(g).     - Iodides: Strongest reducer; reduces H2SO4H_2SO_4 to SO2,S,H2SSO_2, S, H_2S, and I2(g)I_2(g).

  • Hydrogen Halides (HX)     - Acid Strength: HF \ll HCl < HBr < HI. HFHF is weak due to high bond strength (565kJ/mol565\,kJ/mol). HIHI is the strongest acid (Ka5.0×1010K_a \approx 5.0 \times 10^{10}).     - Thermal Stability: Decreases down as bond length increases and bond energy decreases.

MOLECULAR STRUCTURE AND BONDING

  • Electronegativity and Polarity     - Pauling Scale: F=4.0F = 4.0, Cs=0.7Cs = 0.7. Difference \Delta EN > 1.7 is ionic; 0.51.70.5 - 1.7 is polar covalent.     - Dipole Moment (DD): Symmetrical molecules (e.g., CCl4,CO2CCl_4, CO_2) have no net dipole even if bonds are polar. Asymmetrical molecules (e.g., H2O,NH3H_2O, NH_3) are polar.

  • VSEPR Theory     - Linear: CO2CO_2, 180180^\circ.     - Trigonal Planar: BF3BF_3, 120120^\circ.     - Tetrahedral: CH4CH_4, 109.5109.5^\circ.     - Trigonal Pyramidal: NH3NH_3, 107107^\circ.     - Bent: H2OH_2O, 104.5104.5^\circ.

  • Sigma(σ)Sigma (\sigma) and Pi(π)Pi (\pi) Bonds     - Sigma (\sigma): Head-on orbital overlap (ss,sp,pps-s, s-p, p-p). Allows free rotation. Found in single bonds.     - Pi (\pi): Lateral/sideways p-orbital overlap. Restricted rotation. Found in double (1 \sigma, 1 \pi) and triple (1 \sigma, 2 \pi) bonds.

  • Hybridization in Carbon     - sp3sp^3: 4 equivalent orbitals, tetrahedral (CH4,C2H6CH_4, C_2H_6).     - sp2sp^2: 3 equivalent orbitals + 1 unhybridized p-orbital, trigonal planar (C2H4C_2H_4, benzene).     - spsp: 2 equivalent orbitals + 2 unhybridized p-orbitals, linear (C2H2C_2H_2).

ORGANIC FUNCTIONAL GROUPS

  • Alkanes (CnH2n+2C_n H_{2n+2})     - Reactivity: Low due to strong, non-polar \sigma bonds. Undergo free-radical halogenation via initiation, propagation, and termination steps.     - Cracking/Reforming: Thermal or catalytic processes to produce shorter fuels or branched isomers (higher octane).

  • Alkenes (CnH2nC_n H_{2n})     - Reactions: Electrophilic addition.         - Hydrogenation: Uses NiNi catalyst, 130C130\,^\circ C.         - Markovnikov’s Rule: In unsymmetrical additions (e.g., HXHX to propene), HH adds to the carbon with more hydrogens.     - Bromine Water Test: Alkenes decolorize reddish-brown bromine water; alkanes do not.

  • Benzene (C6H6C_6H_6)     - Bonding: sp2sp^2 hybridized with delocalized \pi-electrons above and below the ring. All CCC-C bond lengths are equal.     - Reactions: Prefers Electrophilic Substitution (Nitration, Halogenation, Friedel-Crafts) to preserve aromatic stability.

  • Alkanols (CnH2n+1OHC_n H_{2n+1}OH)     - Classification: Primary (1\circ), Secondary (2\circ), Tertiary (3\circ).     - Oxidation: 1\circ \rightarrow Aldehyde \rightarrow Carboxylic Acid; 2\circ \rightarrow Ketone; 3\circ \rightarrow No reaction.     - Lucas Test: Tertiary alcohols turn cloudy instantly; secondary in 5-10 mins; primary only on heating.

  • Alkanoic Acids (RCOOHR-COOH)     - Acidity: Weak acids that form dimers via hydrogen bonding (high boiling points).     - Esterification: Carboxylic Acid + Alkanol \rightleftharpoons Ester + H2OH_2O (H2SO4H_2SO_4 catalyst).     - Sodium Bicarbonate Test: Carboxylic acids release CO2(g)CO_2(g) (effervescence); alkanols do not.

  1. Question 1: Consider the following reversible reaction: N<em>2(g)+3H</em>2(g)2NH3(g)N<em>2(g) + 3H</em>2(g) \rightleftharpoons 2NH_3(g).

    • Describe what happens to the equilibrium position if the concentration of H2(g)H_2(g) is increased.

  2. Question 2: For the exothermic reaction: A(g)+B(g)C(g)+D(g)A(g) + B(g) \rightleftharpoons C(g) + D(g), explain how increasing the temperature will affect the yield of products at equilibrium.

  3. Question 3: Given the equilibrium constant expression Kc=[C][D][A][B]K_c = \frac{[C][D]}{[A][B]}, what does it signify about the relationship between the concentrations of products and reactants at equilibrium?

  4. Question 4: In the Contact Process: 2SO<em>2(g)+O</em>2(g)2SO<em>3(g)2SO<em>2(g) + O</em>2(g) \rightleftharpoons 2SO<em>3(g), it is known that an increase in pressure shifts the equilibrium towards the production of SO</em>3SO</em>3. Describe the reasoning behind this shift in equilibrium position.

  5. Question 5: How does the addition of a catalyst affect the position of equilibrium in a chemical reaction?