Chem - R2

2.1 How much? The amount of chemical change

• Central idea: stoichiometry links the amounts of reactants and products in a chemical reaction.
• Why it matters: to quantify how much reactant is needed, how much product will form, and to maximize yield in industrial processes.
• Core principle: mass is conserved in chemical change; atoms are rearranged but not created/destroyed.
• Key term: stoichiometry – study of reacting ratios in chemical reactions.
• Etymology: stoicheion (element) + metron (measure).
• Stoichiometric coefficients denote the number of units of each species in the balanced equation.
• Chemical equation structure:
  • Left-hand side: reactants
  • Right-hand side: products
  • State symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous
• Balancing rule: do not change subscripts; only adjust coefficients.
• Example: balancing H–O reaction (water formation): ext{H}2 + ext{O}2 ightarrow 2 ext{H}_2 ext{O}
  • Check: atoms balance on both sides (H: 4, O: 2 on left; H: 4, O: 2 on right).
• Mass balance check example: decomposition of sodium hydrogencarbonate (NaHCO₃) to Na₂CO₃, H₂O, CO₂.
  • Unbalanced form: NaHCO₃ → Na₂CO₃ + H₂O + CO₂
  • Balanced with coefficient 2 on the left: 2 ext{NaHCO}3 ightarrow ext{Na}2 ext{CO}3 + ext{H}2 ext{O} + ext{CO}_2
  • Atom counts: Na 2; H 2; C 2; O 6 on both sides.
• Role of state symbols: provide extra detail about phases.
• Distinctions:
  • Decomposition, combustion, neutralization, synthesis, etc., are all described by chemical equations.
  • The equation expresses a fixed reacting ratio; actual amounts depend on limiting factors.
• Worked worked-through concept: mass balance check is essential after writing a guessed equation to ensure equality of each element’s count on both sides.
• Gas relationships and Avogadro’s law (linking moles to volumes):
  • Avogadro’s hypothesis: Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
  • Molar volume, Vm, ties moles to volume: $V = n \, V_m$
  • Under STP (standard temperature and pressure): $V_m = 22.7 ext{ dm}^3 ext{ mol}^{-1}$ at $T = 273 ext{ K}$, $P = 100 ext{ kPa}$.
  • STP volume example: 1 mole of any gas occupies 22.7 dm³ at STP.
• Practical use: relate gas volumes in reactions to mole ratios via the balanced equation.
• Worked example: complete combustion of butane, C₄H₁₀, with O₂.
  • Balanced equation for complete combustion:
    2 ext{C}4 ext{H}{10}(g) + 13 ext{O}2(g) ightarrow 8 ext{CO}2(g) + 10 ext{H}_2 ext{O}(g)
  • Stoichiometric O₂ required per 1 mole C₄H₁₀: 6.5 mol O₂.
  • At STP, the volume of O₂ needed per mole of C₄H₁₀ is $6.5 imes 22.7 = 147.55 ext{ dm}^3$.
• Important note: when balancing, the coefficient of 1 is not always written explicitly; the balance is still valid.

2.1.1 Chemical equations

• Chemical equations summarize the reacting ratios in a reaction.
• Balancing ensures the same total number of atoms of each element on both sides (law of conservation of mass).
• Example illustrated: copper + chlorine forms copper(II) chloride (CuCl₂):
  • Unbalanced: Cu + Cl₂ → CuCl₂
  • Balanced: ext{Cu} + ext{Cl}2 ightarrow ext{CuCl}2
• Role of coefficients: denote the number of units in the equation (stoichiometric coefficients).
• Coefficient 1 can be omitted for brevity, but it remains understood.
• Practice task: deduce a chemical equation when given reactants/products; include state symbols.
• Purpose: universal language of chemistry; enables calculation of masses, volumes, concentrations, and process efficiency.

2.1.2 Using mole ratios in equations

• Mole ratio: derived from the coefficients in the balanced equation; expresses the proportional relationship between reactants and products in moles.
• Example using methane combustion:
  • Balanced equation: ext{CH}4(g) + 2 ext{O}2(g)
    ightarrow ext{CO}2(g) + 2 ext{H}2 ext{O}(g)
  • Mole ratio for reactants and products is read directly from coefficients: 1 CH₄ : 2 O₂ : 1 CO₂ : 2 H₂O.
• Converting between mole and mass:
  • Mass to moles: $n = \frac{m}{M}$
  • Molar mass (M) is used for each species.
  • Mass ratio approach shows how to get masses of reactants/products from mole ratios.
• Example 1: from CH₄ + 2 O₂ → CO₂ + 2 H₂O, the mass of CO₂ produced per mole of CH₄ is:
  • 1 mol CH₄ (16.04 g) yields 1 mol CO₂ (44.01 g) and 2 mol H₂O (36.02 g).
  • This demonstrates mass conservation and the direct link between moles and masses via M.
• Example 2: mass-to-volume considerations for gases rely on Vm and Avogadro’s law.
• Quick check approach: do not convert every species if the question asks for specific terms; focus on the terms identified in the prompt.
• Important conversion note: when Avogadro’s law is used, volumes are proportional to moles at fixed T and P: $\frac{V<em>1}{V</em>2} = \frac{n<em>1}{n</em>2}$ for gases under identical conditions.
• Relevance: this approach allows determining masses/volumes/concentrations from a single equation’s mole ratio.

2.1.3 The limiting reactant and theoretical yield

• Concept: the limiting reactant determines the maximum amount of product that can form; other reactants are in excess.
• Theoretical yield: maximum amount of product that could be formed from given reactants if the reaction goes to completion (100% conversion of limiting reactant).
• Practical significance: design experiments and industrial syntheses to ensure desired product quantity is achieved.
• Steps to identify limiting reactant and theoretical yield:
  • Write the balanced equation.
  • Convert all given reactants to moles (n = m/M).
  • Compare the mole ratios from the data to the coefficients in the balanced equation.
  • Identify the limiting reactant as the one that would run out first based on the comparison.
  • Use the stoichiometry to calculate the moles (and thus mass) of product from the limiting reactant.
• Worked example: NH₃ + CuO → N₂ + Cu + H₂O (example for limiting reactant). The limiting reactant determines the amount of N₂ formed; compute moles of N₂ via the stoichiometric ratio from CuO amount and then convert to mass.
• Worked example: NO + CO → NO₂ + CO₂ (example of a two-step mechanism). Use the limiting reactant to determine the amount of NO₂ formed.
• Practical tip: to determine the excess reactant and its leftover amount after reaction, calculate the remaining moles using the limiting reactant’s consumption and the initial amounts. This helps in planning purification and product recovery.

2.1.4 Percentage yield

• Real-world reactions are not 100% efficient due to side reactions, incomplete conversion, losses during purification, and reversibility.
• Definitions:
  • Theoretical yield: maximum amount of product predicted by stoichiometry from limiting reagent.
  • Actual (experimental) yield: amount of product actually formed.
  • Percentage yield: measures efficiency of the process.
    $ext{Percentage yield} = \frac{ ext{experimental yield}}{ ext{theoretical yield}} imes 100$
• Factors affecting yield (potentially reducing yield):
  • Side reactions, decomposition of reactants/products, incomplete drying, impurities, and reversible reactions.
• Factors that can cause higher measured yield than theoretical? Impurities mistaken as product, incomplete drying, measurement errors, etc.
• Worked example: Synthesis of N₂ from NH₃ and CuO; given theoretical yield and experimental yield, compute percentage yield as:
  ext{percentage yield} = rac{8.35}{10.65} imes 100 = 78.4 ext{%}
• Conceptual linkage: yield is a measure of process efficiency; atom economy and other metrics give broader sustainability context.

2.1.5 Atom economy

• Green Chemistry perspective: atom economy measures how effectively reactants’ atoms are incorporated into the desired product, minimizing waste.
• Formula (as given in data booklet):
  ext{% atom economy} = rac{ ext{molar mass of desired product}}{ ext{molar mass of all reactants}} imes 100
• High atom economy indicates fewer wasted atoms; a simple addition polymerization (monomer → polymer) can have 100% atom economy since all atoms from reactants end up in the product.
• Trade-offs: some high-yield processes may have low atom economy due to excess reagents or by-products; atom economy complements percentage yield for evaluating process efficiency.
• Worked example: nickel production from NiO + C → Ni + CO; atom economy =
  • M(Ni) = 58.69, M(O) = 16.00, M(C) = 12.01, total reactants = 58.69 + 16.00 + 12.01 = 86.70, so
  • ext{atom economy} = rac{58.69}{86.70} imes 100 \\approx 67.7 ext{%}
• Practical note: atom economy and percentage yield are two complementary metrics for process efficiency; other factors include energy usage, catalysts, and waste management.

2.2 How fast? The rate of chemical change

• Definition: rate of reaction is the speed at which reactants are converted into products.
  • For a reaction, rate can be expressed using either the increase in product concentration or the decrease in reactant concentration:
  • Rate of appearance of product P: $ext{rate} = \frac{d[P]}{dt}$
  • Rate of disappearance of reactant R: $ext{rate} = -\frac{d[R]}{dt}$
• Units: typically mol dm⁻³ s⁻¹ (or equivalent per unit time depending on data).
• Determining rate at a given time: use a tangent to a concentration–time curve to obtain the instantaneous rate: gradient of the tangent at time t.
• Initial rate: rate determined at t = 0, useful for comparing effects of changing concentrations on rate.
• Graphs: concentration vs time (R vs t) and rate vs time graphs illustrate how rate changes as reactant concentration changes during the reaction.
• The rate is not constant; it is often highest at the start when concentrations are highest and decreases as reactants are consumed.
• Collision theory (brief): reaction occurs when particles collide with sufficient energy and proper orientation; rate depends on the frequency of such successful collisions.

2.2.1 Rate of reaction and measurement

• Methods to measure rate (non-exhaustive):
  • Change in volume of gas (gas evolution): use gas syringe or displacement methods.
  • Change in mass (gas evolution or solid formation): balance can track mass loss/gain.
  • Change in colour (spectrophotometry/colorimetry): track absorbance changes for coloured species; convert absorbance to concentration via a calibration curve.
  • Conductivity (ions in solution): monitor conductivity changes with time.
  • Titration (volumetric analysis): withdraw samples, quench the reaction, and titrate to determine concentration changes; requires careful handling to freeze the reaction state (quenching).
• Quenching: a method to stop the reaction in a sample so its concentration can be measured; essential for non-continuous measurement of concentration changes.
• Clock reactions: measure time to reach a fixed end point; provides a simple comparative rate indicator, though not continuous.
• Signal-based methods: use signals (color, conductivity, volume, etc.) that correlate with concentration changes to determine rate.
• In reporting data, rate is typically expressed as a positive quantity, even though reactant concentrations decrease (hence the negative sign is often used in formulas to indicate decrease).

2.2.2 Collision theory, activation energy, and temperature effects

• Collision theory essentials:
  • Reactants must collide to react.
  • Collisions must have sufficient energy to overcome an activation barrier: the activation energy, $E_a$.
  • Collisions must occur with proper orientation (geometry).
• Maxwell–Boltzmann energy distribution explains why temperature affects reaction rate:
  • At higher temperature, more molecules have energy greater than $E_a$, increasing the fraction of successful collisions.
  • Increasing temperature also increases total collision frequency.
  • An Arrhenius-type discussion links temperature to rate constants via the Arrhenius equation (see next subsection).
• Activation energy (Ea): minimum energy required for a collision to lead to a reaction.
• Temperature effects: higher temperature generally increases rate by increasing both collision frequency and the fraction of collisions with energy > Ea.
• Catalysts and Ea: catalysts provide an alternate reaction pathway with lower Ea; this increases the rate for both forward and reverse directions equally and does not change the equilibrium position (K) but accelerates attainment of equilibrium.

2.2.3 The rate equation and reaction order

• Rate equation (rate law): expresses rate in terms of concentrations with exponents (orders) that are determined experimentally, not by coefficients in the balanced equation.
  • General form: $ext{rate} = k [A]^m [B]^n \, ext{…}$
  • The exponents m, n are the orders of reaction with respect to each reactant.
  • Overall order: $m + n + ext{(orders for all reactants)}$.
• Determination of order from data requires experiments with varying concentrations and the corresponding initial rates (often via the initial rates method).
• Example: If rate doubles when [A] doubles (with B constant), rate is first order in A; if rate quadruples when [B] doubles, rate is second order in B, etc.
• Important distinction: the order with respect to a reactant is not necessarily equal to its coefficient in the overall equation; it depends on the mechanism and the rate-determining step.
• Rate-determining step: the slowest elementary step controls the rate of the overall reaction; the rate equation for the overall reaction reflects the rate equation of this step.
• For a one-step unimolecular process A → products, rate = k[A]. For a bimolecular A + B → products, rate = k[A][B]. For a unimolecular step like 2A → products, rate = k[A]^2, etc.

2.2.4 The Arrhenius equation and activation energy (Ea)

• Arrhenius equation: $k = A e^{-\frac{E_a}{RT}}$
  • A is the Arrhenius (frequency) factor, reflecting collision frequency and orientation.
  • Ea is the activation energy; R is the gas constant; T is absolute temperature (K).
• Linear form (Arrhenius plot): taking natural logs gives $ext{ln}(k) = ext{ln}(A) - \frac{E_a}{R}\frac{1}{T}$
  • Plot ln(k) vs 1/T; the slope equals $-E_a/R$ and the intercept equals $ext{ln}(A)$.
• Activation energy from data: Ea = - (slope) × R.
• Using two temperatures to estimate Ea: for temperatures T1, T2 and rate constants k1, k2, $\frac{ ext{ln}(k<em>2/k</em>1)}{\frac{1}{T<em>2} - \frac{1}{T</em>1}} = -\frac{E_a}{R}$
  • Solve for Ea accordingly.
• Practical note: Ea can be large or small; temperature sensitivity of k depends on Ea (higher Ea → greater sensitivity).
• Arrhenius factor A: obtained from the intercept of the Arrhenius plot via $A = e^{ ext{intercept}}$.
• Applications: use Arrhenius analysis to understand how temperature controls rates in industrial and biological systems; discuss the “rule of thumb” that increasing temperature by ~10 K roughly doubles many rates when Ea is moderate (~50 kJ/mol).

2.2.5 Graphical representations and data analysis (zero, first, second order)

• Concentration–time graphs for zero-, first-, and second-order reactions:
  • Zero order: rate is independent of [A], rate = k; concentration vs time is a straight line with constant slope; rate is constant; $ext{rate} = k$.
  • First order: rate ∝ [A]; concentration decays exponentially; rate-concentration graph is a straight line through the origin with slope k; $ext{rate} = k [A]$.
  • Second order: rate ∝ [A]²; concentration-time curve is steeper initially and flattens; rate-concentration graph is a parabola; $ext{rate} = k [A]^2$.
• Rate–concentration graphs help identify order more clearly than concentration-time graphs, especially for distinguishing first vs second order.
• The order of a reaction must be determined experimentally; coefficients in the balanced equation do not determine the rate law.
• For more complex mechanisms, rate equations may be determined by the rate-determining step, which may involve intermediates; the observed rate law reflects the elementary step(s) in the slow stage.
• Radioactive decay is an example of a first-order process with a constant half-life; t½ is independent of initial concentration.

2.3 How far? The extent of chemical change

• Dynamic equilibrium: a state in a closed system where forward and backward reaction rates are equal; concentrations of reactants and products remain constant over time.
• Key features of equilibria (physical and chemical):
  • Equilibrium is dynamic: both forward and backward reactions occur.
  • Occurs in a closed system; no exchange of matter with surroundings (energy can be exchanged).
  • Concentrations of reactants and products remain constant at equilibrium.
  • At equilibrium, there is no observable macroscopic change, though microscopic processes continue.
  • Equilibrium can be reached from either direction (start with reactants or products).
• Equilibrium constant (K): describes the position of equilibrium for a given reaction at a fixed temperature. Expressed as: $K = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$
  • Products in the numerator; reactants in the denominator; each concentration raised to the power of its coefficient.
  • For homogeneous equilibria (same phase), K is dimensionless if activities are used; in practice, K is often quoted without units in IB contexts.
• Interpreting K:
  • If K >> 1, equilibrium lies to the right (products favored).
  • If K << 1, equilibrium lies to the left (reactants favored).
  • If K ≈ 1, appreciable amounts of both reactants and products exist at equilibrium.
• Relationship with Gibbs energy: the standard Gibbs energy change ΔG° and K are related by $riangle G^ ext{o} = -RT \, ext{ln} K$
  • If ΔG° < 0, K > 1 and the forward reaction is favored at that temperature; if ΔG° > 0, K < 1 and the reverse is favored.
• Reaction quotient Q: similar to K, but uses instantaneous (non-equilibrium) concentrations.
  • If Q = K, the system is at equilibrium.
  • If Q < K, the reaction proceeds to the right (toward products).
  • If Q > K, the reaction proceeds to the left (toward reactants).
• Le Châtelier’s principle: qualitative predictions about how equilibria respond to changes in concentration, pressure, temperature, and catalysts.
  • Concentration: increase of a reactant shifts equilibrium to the right; increase of a product shifts left; K remains unchanged.
  • Pressure/volume (gases): increasing pressure shifts to the side with fewer gas molecules; decreasing pressure shifts to the side with more gas molecules; K remains unchanged (at fixed temperature).
  • Temperature: changing temperature changes K; the direction depends on whether the forward reaction is exothermic or endothermic (Ea values explain the temperature sensitivity).
  • Catalyst: does not change K or equilibrium composition; it speeds up attainment of equilibrium by lowering Ea for both forward and reverse reactions equally.
• Practical case studies relevant to Green Chemistry and industrial processes (e.g., Haber process for ammonia synthesis, Contact process for sulfuric acid production).
  • Haber process: N₂ + 3H₂ ⇌ 2NH₃; high pressure, moderate temperature, and a catalyst increase rate but not the equilibrium yield; per-pass conversion is limited; recycle unreacted gas is used to achieve overall high yield.
• The extent of reaction can be quantified by various methods:
  • Determine K from initial and equilibrium concentrations.
  • Determine equilibrium concentrations from K and either initial concentrations or one known equilibrium concentration.
  • For small K (<< 1), approximations may be used to simplify calculations (e.g., assume [reactant] ≈ [reactant]₀ when K is very small).
• The equilibrium law and Gibbs energy framework provide a complementary quantitative and qualitative understanding of equilibrium positions and thermodynamics.

Quick-reference formulas (LaTeX)

• Stoichiometric balance and atom counting:
  ext{Balance: } ext{aA} + ext{bB}
  ightarrow ext{cC} + ext{dD} \ ext{Ensure: } aA = cC, ext{ etc.}
• Molar mass to moles:
  $n = \frac{m}{M}$
• Mole ratio (from coefficients in balanced equation):
  • For reaction aA + bB → cC + dD, from coefficients: ratio A:B = a:b, etc.
• Gas volume–moles relationship (Avogadro):
  $V = n V<em>m, ext{ with } V</em>m = 22.7 ext{ dm}^3 ext{ mol}^{-1} ext{ at STP}$
• Molar volume at STP: $V_m = 22.7 ext{ dm}^3 ext{ mol}^{-1}$
• Mass-to-mole conversion in solutions: $n = cV ext{ (units: mol, L)}$
• Balance check for NaHCO₃ decomposition example: 2 ext{NaHCO}3 ightarrow ext{Na}2 ext{CO}3 + ext{H}2 ext{O} + ext{CO}_2
• Equilibrium constant expression (generic):
  $K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$
• Gibbs energy–equilibrium relation:
  $riangle G^ ext{o} = -RT \ln K$
• Arrhenius equation:
  $k = A e^{-\frac{E_a}{RT}}$
• Linear form of Arrhenius plot:
  $ext{ln } k = ext{ln } A - \frac{E_a}{R} \frac{1}{T}$
• Rate equations (example forms):
  • Zero order: $ext{rate} = k$
  • First order: $ext{rate} = k [A]$
  • Second order: $ext{rate} = k [A]^2$
  • General: $ext{rate} = k [A]^m [B]^n \text{(overall order } = m+n ext{)}$
• Relationship for limiting yield (theoretical yield) and percentage yield:
  $ext{percentage yield} = \frac{ ext{experimental yield}}{ ext{theoretical yield}} imes 100$
• Atom economy:
  ext{% atom economy} = rac{ ext{molar mass of desired product}}{ ext{molar mass of all reactants}} imes 100
• Dynamic equilibrium conditions (qualitative Le Châtelier’s predictions):
  • If concentration of reactants increases, shift to products; if products increase, shift to reactants; K constant at fixed T.
  • If pressure increases for gas-phase with fewer moles on one side, shift toward fewer moles; K constant at fixed T.
  • If temperature changes, K changes; direction depends on ΔH of forward vs backward reaction.
  • Catalyst changes rates but not equilibrium position (K).

Connections and real-world relevance

• Stoichiometry is foundational to all chemistry; it connects microscopic particle counts to macroscopic measurements (masses, volumes, concentrations).
• Green Chemistry: atom economy and optimized yields reduce waste and energy consumption; catalysts enable more sustainable industrial processes.
• The Haber process illustrates the balance between kinetics (rate) and thermodynamics (equilibrium yield); catalysts can improve rate without altering the equilibrium position.
• Arrhenius analysis helps design processes by predicting how temperature changes affect production rates and energy efficiency.
• Understanding equilibrium and Le Châtelier’s principle informs adjustments to industrial conditions to maximize yields, and highlights the trade-offs between yield and rate.

Note on how to study from this material

• Practice: write balanced equations for a variety of reactions, paying attention to oxidation states and half-equations when appropriate.
• Practice: perform mole-mass and mole-volume conversions using the data booklet values for M and Vm.
• Practice: determine limiting reagents and theoretical yields from given masses or volumes.
• Practice: compute and interpret K, Q, and ΔG° to predict the direction and extent of reactions.
• Practice: apply Le Châtelier’s principle to qualitatively predict shifts in equilibrium under changes in concentration, pressure, and temperature.
• Practice: derive rate laws from initial-rate data and understand the role of the rate-determining step in multi-step mechanisms.
• Practice: construct and interpret Arrhenius plots to extract Ea and A from experimental data.
• Practice: compare atom economy and percentage yield as metrics of process efficiency and sustainability.